WA DERI G WORLD SMARA DACHE UMBERS

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2 WA DERI G I THE WORLD OF SMARA DACHE UMBERS A.A.K. MAJUMDAR Ritsumeikan Asia-Pacific University Jumonjibaru, Beppu-shi , Japan aakmajumdar@gmail.com / majumdar@apu.ac.jp & Guest Professor (September 00 March 0) Department of Mathematics, Jahangirnagar University Savar, Dhaka 34, Bangladesh Telephone : Ext

3 Wandering in the World of Smarandache umbers This book can be ordered in a paper bound reprint from : Books on Demand ProQuest Information & Learning (University of Microfilm International) 300 N. Zeeb Road P.O. Box 346, Ann Arbor MI , USA Tel.: (Customer Service) Copyright 00 by InProQuest and the Author Many books can be downloaded from the following Digital Library of Sciences : Peer Reviewers :. Professor Dr. Anwarul Haque Sharief, Department of Mathematics, Jahangirnagar University, Savar, Dhaka 34, Bangladesh. Professor Dr. M.A. Matin, Department of Mathematics, Dhaka University, Dhaka 000, Bangladesh 3. Professor Dr. M. Kaykobad, Department of CSE (Computer Science & Engineering), BUET (Bangladesh University of Engineering & Technology), Dhaka 000, Bangladesh Front cover design K.M. Anwarus Salam and A.A.K. Majumdar Printed in the United States of America ISB :

4 3 Chapter 4 : The Pseudo Smarandache Function PREFACE God created man And man created numbers And F. Smarandache created, literally an infinite array of problems, known after him. It was in mid-nineties of the last century when I received a letter from Professor Ion Patrascu of the Fratii Buzesti College, Craiova, Romania, with lots of enclosures, introducing me with this new branch of Mathematics. Though my basic undergraduate degree is in Mathematics, my research field at that time was Operations Research and Mathematical Programming. But soon the materials sent by Professor Ion Patrascu caught my interest. And as I collected more materials from the on-line journal on Smarandache Notions, I got more and more fascinated by the simplicity of the problems. Some of the problems are simple and straight-forward, some are really challenging, and the rest are thought-provoking. Now, we have Smarandache sequences, Smarandache functions,smarandache Fuzzy, Smarandache Anti- Geometry, Smarandache Groupoid, Quantum Smarandache Paradoxes, Smarandache Philosophy, and what not. And more importantly, mathematicians throughout the world are contributing continuously to this new field of research, enriching and diversifying it in many different directions. At the request of Professor Ion Patrascu, I spent some time on the Smarandache Prime Product Sequence, and my first paper was published in 998 in Smarandache Notions Journal, but the second one took a long gap till 004. But since then, I became a regular contributor to the new journal Scientia Magna. Then some time in 008, Professor Ion Patrascu requested me to write a book on the Smarandache Notions. The present book is an effort to respond to the request of him. In writing the book, the first problem I faced is to choose the contents of the book. Today the Smarandache Notions is so vast and diversified that it is indeed difficult to choose the materials. Then, I fixed on five topics, namely, Some Smarandache Sequences, Smarandache Determinant Sequences, the Smarandache Function and its generalizations, the Pseudo Smarandache Function and its generalizations, and Smarandache Number Related Triangles. Most of the results appeared before, but some results, particularly in Chapter 5, are new. In writing the book, I took the freedom of including the more recent results, found by other researchers till 009, to keep the expositions up-to-date. At the end of Chapter, Chapter 3, Chapter 4 and Chapter 5, some open problems are given in the form of conjectures and questions. I would like to take this opportunity to thank Dr. Perez for his keen interest in publishing the book. A.A.K. Majumdar Ritsumeikan Asia-Pacific University, Japan

5 4 Wandering in the World of Smarandache umbers To Yumiko for her constant encouragement and enthusiasm during the preparation of the manuscript

6 5 Chapter 4 : The Pseudo Smarandache Function TABLE OF CO TE TS Chapter 0 Introduction 7 8 Chapter Some Smarandache Sequences Some Preliminary Results 0. Smarandache Odd Sequence 3.3 Smarandache Even Sequence Smarandache Prime Product Sequence Smarandache Square Product Sequences 0.6 Smarandache Higher Power Product Sequences.7 Smarandache Permutation Sequence Smarandache Circular Sequence Smarandache Reverse Sequence Smarandache Symmetric Sequence Smarandache Pierced Chain Sequence Series Involving Smarandache Sequences 36.3 Some Remarks 37 References 38 Chapter Smarandache Determinant Sequences Smarandache Cyclic Arithmetic Determinant Sequence Smarandache Bisymmetric Arithmetic Determinant Sequence Series with Smarandache Determinant Sequences Smarandache Bisymmetric Determinant Natural Sequence Smarandache Bisymmetric Arithmetic Determinant Sequence 50 5 References 5 Chapter 3 The Smarandache Function Some Explicit Expressions for S(n) Some Generalizations The Smarandache Dual Function S * (n) The Additive Analogues of S(n) and S * (n) Miscellaneous Topics Some Observations and Remarks References 88

7 6 Wandering in the World of Smarandache umbers Chapter 4 The Pseudo Smarandache Function Some Preliminary Results Some Explicit Expressions for Z(n) (Expressions for Z(p. k ), Z(np), Z(3. k p), Z(pq)) 4.3 Some Generalizations The Pseudo Smarandache Dual Function Z * (n) The Additive Analogues of Z(n) and Z * (n) Miscellaneous Topics Some Observations and Remarks References 68 Chapter 5 Smarandache umber Related Triangles Some Preliminary Results Families of 60 Degrees Smarandache Function Related and Pseudo Smarandache Function Related Triangles 5.3 Families of 0 Degrees Smarandache Function Related and Pseudo Smarandache Function Related Triangles 5.4 Families of Smarandache Function Related and Pseudo Smarandache Function Related Pythagorean Triangles 5.5 Smarandache Function Related and Pseudo Smarandache Function Related Triangles in Terms of Their Angles 5.5. Smarandache Function Related Triangles Pseudo Smarandache Function Related Triangles Some Observations and Remarks 89 9 References 9 APPE DICES 93 3 Appendix A. : Values of Z(n), n = () Appendix A. : Values of Z(np); 07 n = 6, 7, 8, 0,,, 3, 6, 4, 3, 48, 96; p 5 Appendix A.3 : Values of Z(p. k ), p =, 3, 7, 9, 3, 9, 3 3 Acknowledgements 4 SUBJECT I DEX 5

8 7 Chapter 4 : The Pseudo Smarandache Function Chapter 0 Introduction When we talk of Smarandache sequences, they are literally innumerable in number. For our book, we have chosen ten Smarandache sequences. They are : () Smarandache odd sequence, () Smarandache even sequence, (3) Smarandache prime product sequence, (4) Smarandache square product sequences, (5) Smarandache higher power product sequences, (6) Smarandache permutation sequence, (7) Smarandache circular sequence, (8) Smarandache reverse sequence, (9) Smarandache symmetric sequence, and (0) Smarandache pierced chain sequence. All of these sequences share the common characteristic that they are all recurrence type, that is, in each case, the n th term can be expressed in terms of one or more of the preceding terms. Some of the properties of these ten recurrence type Smarandache sequences are studied in Chapter. In case of the Smarandache odd, even, circular and symmetric sequences, we show that none of these sequences satisfies the recurrence relationship for Fibonacci or Lucas numbers. We prove further that none of the Smarandache prime product and reverse sequences contains Fibonacci or Lucas numbers (in a consecutive row of three or more). For the Smarandache even, prime product, permutation and square product sequences, the question is : Are there any perfect powers in each of these sequences? We have a partial answer for the first of these sequences, while for each of the remaining sequences, we prove that no number can be expressed as a perfect power. We also prove that no number of the Smarandache higher power product sequences is square of a natural number. In Chapter, we consider four Smarandache determinant sequences, namely, the Smarandache (a) cyclic determinant natural sequence, (b) cyclic arithmetic determinant sequence, (c) bisymmetric determinant natural sequence, and (d) bisymmetric arithmetic determinant sequence. Actually, the sequence (a) is a particular case of the sequence (b), and (c) is a special case of (d). In each case, we derive the explicit forms of the n th term. We also find the sequences of the n th partial sums for the sequences (c) and (d). Two of the most widely studied Smarandache arithmetical functions, namely, the Smarandache function S(n) and the pseudo Smarandache function Z(n), are the subject matters of Chapter 3 and Chapter 4 respectively, where we give some elementary properties of these two functions and their generalizations. Chapter 5 deals with the Smarandache number related triangles. Besides the Smarandache number related dissimilar Pythagorean triangles, detailed analysis on the Smarandache number related dissimilar 60 0 and 0 0 triangles is provided. We may recall that an arithmetic function f(n) has, as its domain, the set of all positive integers n. In the classical Theory of Numbers, the most commonly encountered arithmetical functions are d(n), φ(n) and σ(n). Recently, the researchers are getting interested in studying equations involving S(n) or Z(n) and one or more of the classical arithmetical functions. With this in mind, we review below the relevant definitions and theorems. The classic book, An Introduction to the Theory of Numbers, by G.H. Hardy and E.M. Wright (Clarendon Press, Oxford, UK, 5 th Edition (00)), is recommended for the definitions and the corresponding formulas.

9 8 Wandering in the World of Smarandache umbers Definition 0. : Given an integer n, the divisor function d(n) is the number of all divisors of n, including and n itself. Theorem 0. : Let α α αi αi+ αs n = p p... p p... p (0.) i i+ s be the representation of n in terms of its prime factors p, p,, p k. Then, d(n) = (α + )(α + ) (α k + ). Definition 0. : Given an integer n, the Euler phi function, denoted by φ(n), is the number of positive integers not exceeding n, which are prime to n. Theorem 0. : Let n have the prime factor representation of the form (0.). Then, φ(n) = n... p p p k. Definition 0.3 : Given an integer n, σ(n) is the sum of all divisors of n (including and n itself). Theorem 0.3 : Let n have the prime factor representation of the form (0.). Then, α + α + α k + p p p k σ (n) =..... p p p k Definition 0.4 : An integer n is called perfect if and only if it is the sum of all its proper divisors (that is, all divisors including but excluding n itself). Thus, n is perfect if and only if n = σ(n). Theorem 0.4 : Any even perfect number n is of the form n = k ( k+ ), where the integer k is such that k+ is prime. Definition 0.5 : An arithmetic function f(.) is called multiplicative if and only if f(mn) = f(m)f(n) for all integers m and n with (m, n) =. Theorem 0.5 : The functions d(n), φ(n) and σ(n) are multiplicative. Definition 0.6 : An integer n > 0 is called f perfect if k n = f (d i ), i= where d, d,, d k are the proper divisors of n, and f(.) is an arithmetical function. Chapter, Chapter, Chapter 4 and Chapter 5 are based on our previous papers, published in different journals. In the meantime, some new results have been found, and those results are also included in this book. Moreover, to keep the book up-to-date, we did not hesitate to include the results of other researchers as well, very frequently with simplified proofs.

10 9 Chapter 4 : The Pseudo Smarandache Function Chapter Some Smarandache Sequences For this chapter, we have chosen the following ten Smarandache sequences : () Smarandache odd sequence () Smarandache even sequence (3) Smarandache prime product sequence (4) Smarandache square product sequences (5) Smarandache higher power product sequences (6) Smarandache permutation sequence (7) Smarandache circular sequence (8) Smarandache reverse sequence (9) Smarandache symmetric sequence (0) Smarandache pierced chain sequence. The above ten sequences have one property in common : They are all recurrence type. It is conjectured that there are no Fibonacci or Lucas numbers in any of the Smarandache odd, even, circular and symmetric sequences. We confirm the conjectures by showing that none of these sequences satisfies the recurrence relationship for Fibonacci or Lucas numbers. We prove further that none of the Smarandache prime product and reverse sequences contains Fibonacci or Lucas numbers (in a consecutive row of three or more). We recall that the sequence of Fibonacci numbers, { F(n) } n=, and the sequence of Lucas numbers { L(n) } n=, are defined through the recurrence relations satisfied by their terms : F() =, F() = ; F(n + ) = F(n + ) + F(n), n, (.) L() =, L() = 3; L(n + ) = L(n + ) + L(n), n. (.) For the Smarandache even, prime product, permutation and square product sequences, the question is : Are there any perfect powers in each of these sequences? We have a partial answer for the first of these sequences, while for each of the remaining sequences, we prove that no number can be expressed as a perfect power. We also prove that no number of the Smarandache higher power product sequences is square of a natural number. For the Smarandache odd, prime product, circular, reverse and symmetric sequences, the question is : How many primes are there in each of these sequences? For the Smarandache even sequence, the question is : How many elements of the sequence are twice a prime? These questions still remain open. After giving some preliminary results in., we study some elementary properties of the ten sequences separately in the next ten sections from. through.. In each case, we give the recurrence relation satisfied by the sequence.. gives some series involving the Smarandache sequences. The chapter ends with some remarks in.3.

11 0 Wandering in the World of Smarandache umbers. Some Preliminary Results Lemma.. : 3 divides (0 m + 0 n + ) for all integers m, n 0. Proof : We consider the following three possible cases separately : Case () : When m = n = 0. In this case, the result is clearly true. Case () : When m = 0, n. Here also, the result is seen to be true, since 0 m + 0 n + = 0 n + = (0 n ) + 3 = 9( n ) + 3. Case (3) : When m, n. In this case, the result follows, since 0 m + 0 n + = (0 m ) + (0 n ) + 3. Lemma.. : The only non negative integer solution of the Diophantine equation x² y² = is x =, y = 0. Proof : The given Diophantine equation is equivalent to (x y)(x + y) =, where both x y and x + y are integers. Therefore, the only two possibilities are () x y = = x + y, () x y = = x + y, the first of which gives the desired non negative solution. Corollary..: Let N ( > ) be a fixed number. Then, () the Diophantine equation x² N = has no (positive) integer solution x; () the Diophantine equation N y² = has no (positive) integer solution y. Lemma..3 : Let n ( ) be a fixed integer. Then, the only non negative integer solution of the Diophantine equation x² + = y n is x = 0, y =. Proof : By Lemma.., it is sufficient to consider the case when n > odd : If n is even, say, n = m for some integer m >, then rewriting the given Diophantine equation as (y m )² x² =, we see that, by Lemma.., the only non negative integer solution is y m =, x = 0, that is x = 0, y =, as required. So, let n be odd, say, n = k + for some integer k. Then, the given Diophantine equation can be written as x = y k+ = (y )(y k + y k ). () From (), we see that x = 0 if and only if y =, since y k + y k > 0. Now, if x 0, from (), the only possibilities are the following two : Case () : y = x, y k + y k = x. But then y = x +, and we are led to the equation (x + ) k + (x + ) k (x + ) + = 0, which is impossible. Case () : y =, y k + y k = x. Then, y =, x = k+. Rewriting this equation in the following equivalent form (x )(x + ) = ( k )( k + ), we see that the l.h.s. is divisible by 4, while the r.h.s. is not divisible by 4 (since both k and k + are odd). And we reach to a contradiction. The contradiction in both the cases establishes the lemma.

12 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Corollary.. : Let n ( ) and N ( > 0) be two fixed integers. Then, the Diophantine equation N + = y n has no integer solution y. Corollary..3 : Let n ( ) and N ( > ) be two fixed integers. Then, the Diophantine equation x + = N n has no (positive) integer solution x. Lemma..4 : Let n ( ) be a fixed integer. Then, the only non negative integer solutions of the Diophantine equation x y n = are () x =, y = 0; () x = 3, y =, n = 3. Proof : For n =, the lemma reduces to Lemma... So we consider the case when n 3. From the given Diophantine equation, we see that, y = 0 if and only if x = ±, giving the only non negative integer solution x =, y = 0. To see if the given Diophantine equation has any non zero integer solution, we assume that x. If n is even, say, n = m for some integer a, then x y n x (y m ) =, which has no integer solution y for any x > (by Corollary..()). Next, let n be odd, say, n = k + for some integer k. Then, x y k+ =, that is, (x )(x + ) = y k+. We now consider the following three cases that may arise : Case () : When x =, x + = y k+. Here, x = together with the equation y k+ = 3, and the latter has no integer solution y. Case () : When x = y, x + = y k. Rewriting the second equation in the equivalent form (y k )(y k + ) = x, we see that (y k + ) x. But this contradicts the first equation x = y + if k >, since for k >, y k + > y + = x. If k =, then (y )(y + ) = x y =, y + = x, so that y =, x = 3, n = 3, which is a solution of the given Diophantine equation. Case (3) : When x = y t for some integer t with t k, k (so that x + = y k t+ ). In this case, we have x = y t [ + y (k t)+ ]. Since x does not divide y, it follows that + y (k t)+ = Cx, for some integer C. Thus, x = y t (Cx) Cy t =. If C =, then y =, and the resulting equation x = has no integer solution. On the other hand, if C, the equation Cy t = has no integer solution. Thus, case (3) cannot occur. All these complete the proof of the lemma. Corollary..4 : The only non negative integer solution of the Diophantine equation x y 3 = is x = 3, y =. Corollary..5 : Let n ( > 3) be a fixed integer. Then, the Diophantine equation x y n = has x =, y = 0 as its only non negative integer solution. Corollary..6 : Let n ( > 3) and N ( > 0) be two fixed integers. Then, the Diophantine equation x N n = has no integer solution x. Corollary..7 : Let n ( 3) and N (> ) be two fixed integers with N 3. Then, the Diophantine equation N y n = has no integer solution.

13 Wandering in the World of Smarandache umbers. Smarandache Odd Sequence { OS(n) } n = The Smarandache odd sequence, denoted by { OS(n) } n =, is the sequence of numbers formed by repeatedly concatenating the odd positive integers (Ashbacher []). The first few terms of the sequence are, 3, 35, 357, 3579, 3579, 35793, ,. In general, the n th term of the sequence is given by OS(n) = (n ), n. (..) For any n, OS(n + ) can be expressed in terms of OS(n) as follows : For n, OS(n + ) =35... (n )(n + ) = 0 s OS(n) + (n + ) for some integer s. (..) More precisely, s = number of digits in (n + ). Thus, for example, OS(5) = 0 OS(4) + 7, while, OS(6) = 0² OS(5) +. By repeated application of (..), we get OS(n + 3) = 0 s OS(n+) + (n + 5) for some integer s = 0 s [0 t OS(n + ) + (n + 3)] + (n + 5) for some integer t (..3a) =0 s+t [0 u OS(n) + (n +)] + (n + 3)0 s + (n + 5) (..3b) for some integer u, so that s t u. Therefore, OS(n + 3) = 0 s+t+u OS(n) + (n + )0 s+t + (n + 3)0 s + (n + 5). (..4) From (..3), we see that, for any integer n, OS(n + 3) = 0 s+t OS(n + ) + (n + 3)(n + 5) = 0 s+t+u OS(n) + (n + )(n + 3)(n + 5). Lemma.. : 3 divides OS(n) if and only if 3 divides OS(n + 3). Proof : For any s, t with s t, by Lemma.., 3 [(n + )0 s+t + (n + 3)0 s + (n + 5)] = (n + )(0 s+t + 0 s + ) + (0 s + ). The result now follows by virtue of (..4). Lemma.. : 9 divides OS(3n) for any integer n. Proof : First note that, the sum of digits of OS(n) is (n + ) = n. Since the sum of digits of OS(3n) is (3n) = 9n, the result follows.

14 3 Chapter Chapter 4 : The : Pseudo Some Smarandache Sequences Function In proving Lemma.., we have used the fact that an integer is divisible by 9 if and only if its sum of digits is divisible by 9 (see, for example, Bernard and Child []). Lemma..3 : 5 divides OS(5n + 3) for any integer n 0. Proof : From (..), for any arbitrary but fixed n 0, OS(5n + 3) = 0 s OS(5n + ) + (0n + 5) for some integer s. The r.h.s. is clearly divisible by 5, and hence 5 divides OS(5n + 3). Based on computer findings with all numbers from 35 through OS(999), Ashbacher [] conjectures that (except for the trivial cases of n =,, for which OS() = = F() = F() = L(), OS() = 3 = F(7)), there are no numbers in the Smarandache odd sequence that are also Fibonacci (or, Lucas) numbers. In Theorem.. and Theorem.., we prove the conjectures of Ashbacher in the affirmative. The proof of the theorems relies on the following results. Lemma..4 : For any integer n, OS(n + ) > 0 OS(n). Proof : From (..), for any n, OS(n + ) = 0 s OS(n) + (n + ) > 0 s OS(n) > 0 OS(n), where s is an integer. We thus get the desired inequality. Corollary.. : For any integer n, OS(n + ) OS(n) > 9[OS(n + ) + OS(n)]. Proof : From Lemma..4, OS(n + ) OS(n) > 9 OS(n) for all n. (..5) Now, using the inequality (..5), we get OS(n + ) OS(n) = [OS(n + ) OS(n + )] + [OS(n + ) OS(n)] > 9[OS(n + ) + OS(n)], which establishes the result. Theorem.. : (Except for n =, for which OS() = = F() = F(), OS() = 3 = F(7)) there are no numbers in the Smarandache odd sequence that are also Fibonacci numbers. Proof : Using Corollary.., we see that, for all n, OS(n + ) OS(n) > 9[OS(n + ) + OS(n)] > OS(n + ). Thus, no numbers of the Smarandache odd sequence satisfy the recurrence relation (.) satisfied by the Fibonacci numbers. By similar reasoning, we have the following result. Theorem.. : (Except for n = for which OS() = = L()) there are no numbers in the Smarandache odd sequence that are Lucas numbers Thus, no numbers of the Smarandache odd sequence satisfy the recurrence relation (.) satisfied by the Fibonacci numbers. Ashbacher [] reports that, among the first elements of the sequence, only OS(), OS(0) and OS(6) are primes. Marimutha [3] conjectures that there are infinitely many primes in the Smarandache odd sequence, but the conjecture still remains open. In order to search for primes in the Smarandache odd sequence, by virtue of Lemma.. and Lemma..3, it is sufficient to check the terms of the forms OS(3n ± ) (n ), where neither 3n + nor 3n is of the form 5k + 3 for some integer k.

15 4 Wandering in the World of Smarandache umbers.3 Smarandache Even Sequence { ES(n) } n = The Smarandache even sequence, denoted by { ES(n) } n=, is the sequence of numbers formed by repeatedly concatenating the even positive integers (Ashbacher []). The n th term of the sequence is given by ES(n) = 4... (n), n. (.3.) The first few terms of the sequence are, 4, 46, 468, 4680, 4680, 46804,, of which only the first is a prime number. We note that, for any integer n, ES(n + ) = 4... (n)(n + ) = 0 s ES(n) + (n + ) for some integer s, (.3.) where, more precisely, s = number of digits in (n + ). Thus, for example, ES(4) = 468 = 0 ES(3) + 8, while, ES(5) = 4680 = 0² ES(4) + 0. From (.3.), the following result follows readily. Lemma.3. : For any integer n, ES(n + ) > 0 ES(n). Using Lemma.3., we can prove that ES(n + ) ES(n) > 9[ES(n + ) + ES(n)] for all n. (.3.3) The poof is similar to that of Corollary.., and is omitted here. By repeated application of (.3.), we see that, for any integer n, ES(n + ) = 0 t ES(n + ) + (n + 4) for some integer t = 0 t [0 u ES(n) + (n + )] + (n + 4) for some integer u = 0 u+t ES(n) + (n + )0 t + (n + 4), and ES(n + 3) = 0 s ES(n + ) + (n + 6) for some integer s = 0 s [0 u+t ES(n) + (n + )0 t + (n + 4)] + (n + 6) = 0 s+t+u ES(n) + (n + )0 s+t + (n + 4)0 s + (n + 6), (.3.4) for some integers s, t and u with s t u. From (.3.4), we see that

16 5 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function ES(n + 3) = 0 s+t ES(n + ) + (n + 4)(n + 6) = 0 s+t+u ES(n) + (n + )(n + 4)(n + 6). Using (.3.4), together with Lemma.., we can prove the following result. Lemma.3. : 3 divides ES(n) if and only if 3 divides ES(n + 3). Lemma.3.3 : For any integer n, 3 divides ES(3n). Proof : The proof is by induction on n. Since ES(3) = 46 is divisible by 3, the lemma is true for n =. We now assume that the result is true for some n, that is, 3 ES(3n) for some n. Now, by Lemma.3., together with the induction hypothesis, we see that 3 divides ES(3n + 3) = ES(3(n + )). Thus the result is true for n +. Corollary.3. : For any integer n, 3 divides ES(3n ). Proof : Let n ( ) be any arbitrary but fixed integer. From (.3.), ES(3n) = 0 s ES(3n ) + 6n for some integer s. Now, by Lemma.3.3, 3 divides ES(3n). Therefore, 3 must also divide ES(3n ). Since n is arbitrary, the lemma is proved. Corollary.3. : For any integer n, 3 does not divide ES(3n + ). Proof : Let n ( ) be any arbitrary but fixed integer. From (.3.), ES(3 n + ) = 0 s ES(3n) + (6n + ) for some integer s. Since 3 divides ES(3n), but 3 does not divide (6n + ), the result follows. Lemma.3.4 : 4 divides ES(n) for any integer n. Proof : Since 4 divides ES() = 4 and 4 divides ES(4) = 468, we see that the result is true for n =,. Now, from (.3.), for n, ES(n) = 0 s ES(n ) + 4n, where s is the number of digits in 4n. Clearly, s for all n 3. Thus, 4 divides 0 s if n 3, and we get the desired result. Corollary.3.3 : For any integer n 0, 4 does not divide ES(n + ). Proof : Clearly the result is true for n = 0, since ES() = is not divisible by 4. For n, from (.3.), ES(n + ) = 0 s ES(n) + (4n + ) for some integer s. By Lemma.3.4, 4 divides ES(n). Since 4 does not divide (4n + ), the result follows. Lemma.3.5 : For any integer n, 0 divides ES(5n). Proof : For any arbitrary but fixed n, from (.3.), ES(5n) = 0 s ES(5n ) + 0n for some integer s. The result is now evident from the above expression of ES(5n). Corollary.3.4 : 0 divides ES(0n) for any integer n. Proof : follows by virtue of Lemma.3.4 and Lemma.3.5.

17 6 Wandering in the World of Smarandache umbers Based on the computer findings with numbers up through ES(499) = , Ashbacher [] conjectures that (except for the case of ES() = = F(3)) there are no numbers in the Smarandache even sequence that are also Fibonacci (or, Lucas) numbers. The following two theorems establish the validity of Ashbacher s conjectures. Theorem.3. : (Except for ES() = = F(3)) there are no numbers in the Smarandache even sequence that are Fibonacci numbers. Proof : By virtue of (.3.3), for all n, ES(n + ) ES(n) > 9[ES(n + ) + ES(n)] > ES(n + ). Thus, no numbers of the Smarandache odd sequence satisfy the recurrence relation (.) satisfied by the Fibonacci numbers. By similar reasoning, we have the following result. Theorem.3. : There are no numbers in the Smarandache even sequence that are Lucas numbers. Ashbacher [] raises the question : Are there any perfect powers in ES(n)? The following theorem gives a partial answer to the question. Theorem.3.3 : None of the terms of the subsequence { ES(n ) } n = is a perfect square or higher power of an integer ( > ). Proof : Let, for some n, ES(n) = 4... (n) = x² for some integer x >. Now, since ES(n) is even for all n, x must be even. Let x = y for some integer y. Then, ES(n) = (y)² = 4y², which shows that 4 divides ES(n). Now, if n is odd of the form k, k, by Corollary.3.3, ES(k ) is not divisible by 4, and hence, the numbers of the form ES(k ), k, can not be perfect squares. By same reasoning, none of the terms ES(n ), n, can be expressed as a cube or higher powers of an integer. Remark.3. : It can be seen that, if n is of the form k 0 s + 4 or k 0 s + 6, where k ( k 9) and s ( ) are integers, then ES(n) cannot be a perfect square (and hence, cannot be the power of an even integer). The proof is as follows : If ES(n) = x² for some integer x >, () then x must be an even integer. The following table gives the possible trailing digits of x and the corresponding trailing digits of x. Now, if n is of the form n = k 0 s + 4, then the trailing digit of ES(k 0 s + 4) is 8 for all admissible values of k and s, and hence, it follows that the representation of ES(k 0 s + 4) in the form () is not possible. Again, since the trailing digit of ES(k 0 s + 6) is for all admissible values of k and s, the numbers of the form ES(k 0 s + 6) can also be not expressible in the form ().

18 7 Chapter Chapter 4 : The : Pseudo Some Smarandache Sequences Function Trailing digit of x Trailing digit of x It thus remains to consider the cases when n is one of the three forms : (i) n = k 0 s, (ii) n = k 0 s +, (iii) n = k 0 s + 8, (where k ( k 9) and s ( ) are integers). Smith [4] conjectures that none of the terms of the sequence { ES(n) } n= is a perfect power..4 Smarandache Prime Product Sequence { PPS(n) } n = Let { p } n n= be the (infinite) sequence of primes in their natural order, so that p =, p = 3, p 3 = 5, p 4 = 7, p 5 =, p 6 = 3,. The Smarandache prime product sequence, denoted by { } n (Smarandache [5]) : PPS(n), is defined as follows = PPS(n) = p p p n +, n. (.4.) The following lemma gives a recurrence relation in connection with the sequence. Lemma.4. : PPS(n + ) = p n + PPS(n) (p n + ) for all n. Proof : By definition, PPS(n + ) = p p p n p n + + = (p p p n + )p n + p n + +, which now gives the desired relationship. From Lemma.4., we get Corollary.4. : PPS(n + ) PPS(n) = [PPS(n) ](p n + ) for all n. Lemma.4. : The values of PPS(n) satisfy the following inequalities : () PPS(n) < (p n ) n for all n 4, () PPS(n) < (p n ) n for all n 7, (3) PPS(n) < (p n ) n 3 for all n 0, (4) PPS(n) < (p n + ) n for all n 3, (5) PPS(n) < (p n + ) n for all n 6, (6) PPS(n) < (p n + ) n 3 for all n 9. Proof : We prove parts (3) and (6) only, the proof of the other parts is similar. To prove part (3) of the lemma, we note that the result is true for n = 0, since PPS(0) = < (p 0 ) 7 = 9 7 = Now, assuming the validity of the result for some integer k ( 0), and using Lemma.4., we see that,

19 8 Wandering in the World of Smarandache umbers PPS(k + ) = p k + PPS(k) (p k + ) < p k + PPS(k) < p k + (p k ) n 3 (by the induction hypothesis) < p k + (p k + ) n 3 = (p k + ) n, where the last inequality follows from the fact that the sequence of primes, { p } increasing in n ( ). Thus, the result is true for k + as well. To prove part (6) of the lemma, we note that the result is true for n = 9, since PPS(9) = < (p 0 ) 6 = 9 6 = n n=, is strictly Now to appeal to the principle of induction, we assume that the result is true for some integer k ( 9). Then using Lemma.4., together with the induction hypothesis, we get PPS(k + ) = p k + PPS(k) (p k + ) < p k + PPS(k) < p k + (p k + ) k 3 = (p k + ) k. Thus the result is true for k +. All these complete the proof by induction. The lemma below improves the results of Prakash [6] and Majumdar [7]. Lemma.4.3 : Each of PPS(), PPS(), PPS(3), PPS(4) and PPS(5) is prime, and for n 6, PPS(n) has at most n 4 prime factors, counting multiplicities. Proof : Clearly, the first five terms of the sequence, namely, PPS() = 3, PPS() = 7, PPS(3) = 3, PPS(4) = and PPS(5) = 3, are all primes. Also, since PPS(6) = 3003 = , PPS(7) = 505 = , PPS(8) = = , we see that the lemma is true for 6 n 8. Now, if p is a prime factor of PPS(n), then p p n +. Therefore, if for some n 9, PPS(n) has n 3 (or more) prime factors (counted with multiplicity), then PPS(n) (p n + ) n 3, and this is in contradiction with part (6) of Lemma.4.. Hence the lemma is established. It has been proved that, for each n, PPS(n) and PS(n + ) are relatively prime (see, Majumdar [8]). The following lemma gives the more general result. Lemma.4.4 : For any n and k, PPS(n) and PPS(n + k) can have at most k number of prime factors (counting multiplicities) in common. Proof : For any n and k, PPS(n + k) PPS(n) = p p p n (p n + p n + p n + k ). (.4.) If p is a common prime factor of PPS(n) and PPS(n + k), since p p n + k, it follows from (.4.) that p divides (p n + p n + p n + k ). Now if PPS(n) and PPS(n + k) have k (or more) prime factors in common, then the product of these common prime factors is greater than (p n + k) k, which can not divide p n + p n + p n + k < (p n + k) k. This contradiction proves the lemma.

20 9 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Corollary.4. : For any integers n ( ) and k ( ), if all the prime factors of p n + p n + p n + k are less than p n + k, then PPS(n) and PPS(n + k) are relatively prime. Proof : If p is any common prime factor of PPS(n) and PPS(n + k), then p divides p n + p n + p n + k. Also, for such p, p > p n + k, contradicting the hypothesis of the corollary. Thus, if all the common prime factors of PPS(n) and PPS(n + k) are less than p n + k, then PPS(n) and PPS(n + k) are relatively prime. The following result has been proved by Prokash [6] and Majumdar [8]. Here we give a simpler proof. Theorem.4. : For any n, PPS(n) is never a square or higher power of an integer (>). Proof : Clearly, none of PPS(), PPS(), PPS(3), PPS(4) and PPS(5) can be expressed as powers of integers (by Lemma.4.3). Now, if possible, let for some n 6, PPS(n) = x k for some integers x ( > 3) and k ( ). () Without loss of generality, we may assume that k is a prime (if k is a composite number, letting k = pr where p is prime, we have PPS(n) = (x r ) p = N p, where N = x r ). By Lemma.4.3, k n 4 and so k cannot be greater than p n 5 (k p n 4 k > n 4, since p n > n for all n ). Hence, k must be one of the primes p, p,, p n 5. Also, since PPS(n) is odd, x must be odd. Let x = y + for some integer y > 0. Then, from (), p p... p n = (y + ) k = (y) k + k k (y)k k (y). () If k =, we see from (), 4 divides p p... p n, which is absurd. On the other hand, for k 3, since k divides p p... p n, it follows from () that k divides y, and consequently, k divides p p... p n, which is impossible. Hence, the representation of PPS(n) in the form () is not possible. Using Corollary.4. and the fact that PPS(n + ) PPS(n) > 0, we get PPS(n + ) PPS(n) = [PPS(n + ) PPS(n + )] + [PPS(n + ) PPS(n)] > [PPS(n + ) ](p n + ) > [PPS(n + ) ] for all n. Hence, PPS(n + ) PPS(n) > PPS(n + ) for all n. (.4.) The inequality (.4.) shows that no elements of the Smarandache prime product sequence satisfy the recurrence relation for Fibonacci (or, Lucas) numbers. This leads to the following theorem. Theorem.4. : There are no numbers in the Smarandache prime product sequence that are Fibonacci (or Lucas) numbers (except for the trivial cases of PPS() = 3 = F(4) = L(), PPS() = 7 = L(4)).

21 0 Wandering in the World of Smarandache umbers.5 Smarandache Square Product Sequences { SPS (n) } and { } n= SPS (n) The Smarandache square product sequence of the first kind, denoted by { SPS } (n) n= n= Smarandache square product sequence of the second kind, denoted by { SPS } (n) n=, and the, are defined by (Russo [9]) SPS (n) = (²)(²) (n²) +, n, SPS (n) = (²)(²) (n²), n. Alternative expressions for SPS (n) and SPS (n) are SPS (n) = (n!)² +, n, (.5.a) SPS (n) = (n!)², n. (.5.b) The first few terms of the sequence { } (n) n= SPS are SPS () =, SPS () = 5, SPS (3) = 37, SPS (4) = 577, SPS (5) = 440, SPS (6) = 5840 = , SPS (7) = = 0 550, SPS (8) = = , SPS (9) = , of which the first five terms are prime numbers. The first few terms of the sequence { } (n) n= SPS are SPS () = 0, SPS () = 3, SPS (3) = 35, SPS (4) = 575, SPS (5) = 4399, SPS (6) = 58399, SPS (7) = , SPS (8) = , SPS (9) = , of which, disregarding the first term, the second term is prime, and the remaining terms of the sequence are all composite numbers. Theorem.5. proves that, for any n, neither of SPS (n) and SPS (n) is a square of an integer ( > ). Theorem.5. shows that no terms in these sequences can be cubes.

22 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Theorem.5. : For any n, none of SPS (n) and SPS (n) is a square of an integer (>). Proof : If possible, let SPS (n) (n!)² + = x² for some integers n, x >, SPS (n) (n!)² = y² for some integers n, y >. But these contradict Corollary..() and Corollary..() respectively. Theorem.5. : For any integer n, none of the SPS (n) and SPS (n) is a cube or higher power of an integer ( > ). Proof : is by contradiction. Let SPS (n) (n!) + = y m for some integers n, y >, m 3, SPS (n) (n!) = z s for some integer n, z, s 3. We then have contradictions to Corollary.. and Corollary..7 respectively. Lemma.5. : For any integer n, () SPS (n + ) = (n + ) SPS (n) n(n + ), () SPS (n + ) = (n + ) SPS (n) n(n + ). Proof : The proof is for part () only, and is given below. SPS (n + ) = [(n + )!] + = (n + ) [(n!) + ] (n + ) +. Lemma.5. : For any integer n, () SPS (n + ) SPS (n) > SPS (n + ), () SPS (n + ) SPS (n) > SPS (n + ). Proof : It is straightforward, using (.5.), to prove that SPS (n + ) SPS (n) = SPS (n + ) SPS (n) = (n!) [(n + ) (n + ) ]. Some algebraic manipulations give the desired inequalities. Lemma.5. can be used to prove the following results. Theorem.5.3 : (Except for the trivial cases, SPS () = = F(3), SPS () = 5 = F(5)) there are no numbers of the Smarandache square product sequence of the first kind that are Fibonacci (or Lucas) numbers. Theorem.5.4 : (Except for the trivial case, SPS () = 3 = F(4) = L()) there are no numbers of the Smarandache square product sequence of the second kind that are Fibonacci (or Lucas) numbers. The question raised by Iacobescu [0] is : How many terms of the sequence { SPS } (n) n= are prime? The following theorem, due to Le [], gives a partial answer to this question. Theorem.5.5 : If n ( > ) is an even integer such that n + is prime, then SPS (n) is not a prime. Russo [9], based on values of SPS (n) and SPS (n) for n 0, conjectures that each of the sequences contains only a finite number of primes.

23 Wandering in the World of Smarandache umbers.6 Smarandache Higher Power Product Sequences { HPPS (n) }, { } n= HPPS (n) n = Let m (>3) be a fixed integer. The Smarandache higher power product sequence of the first kind, HPPS (n), and the Smarandache higher power product sequence of the second denoted by { } n= kind, denoted by { } (n) n= HPPS, are defined as follows : HPPS (n) = ( m )( m ) (n m ) +, n, HPPS (n) = ( m )( m ) (n m ), n. Alternative expressions for HPPS (n) and HPPS (n) are HPPS (n) = (n!) m +, n, (.6.a) HPPS (n) = (n!) m, n. (.6.b) Lemma.6. : For any integer n, () HPPS (n + ) = (n + ) m HPPS (n) [(n + ) m + ], () HPPS (n + ) = (n + ) m HPPS (n) + [(n + ) m + ]. Theorem.6. : For any integer n, none of HPPS (n) and HPPS (n) is a square of an integer ( > ). Proof : If possible, let HPPS (n) (n!) m + = x for some integer x >. But, by Corollary..6, this permits no solution if m > 3. Again, letting, for some integer n (HPPS () = 0) HPPS (n) (n!) m = y for some integer y, we have a contradiction to Corollary..3. Thus, we reach to a contradiction in either case, establishing the theorem. The following two theorems are due to Le [, 3]. Theorem.6. : If m is not a number of the form m = k for some integer k, then the HPPS (n) contains only one prime, namely, HPPS () =. sequence { } n= Theorem.6.3 : If both m and m are primes, then the sequence { } (n) n= HPPS contains only one prime, namely, HPPS () = m ; otherwise, the sequence contains no prime. Remark.6. : We have defined the Smarandache higher power product sequences under the restriction that m > 3, and under such restriction, as has been proved in Theorem.6., none of HPPS (n) and HPPS (n) is a square of an integer ( > ) for any n. However, if m = 3, the situation is a little bit different : For any n, HPPS (n) = (n!) 3 still cannot be a perfect square of an integer ( > ), by Corollary..3, but since HPPS (n) = (n!) 3 +, we see that HPPS () = (!) 3 + = 3, that is, HPPS () is a perfect square. However, this is HPPS (n) that can be expressed as a perfect square. the only term of the sequence { } n=

24 3 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function.7 Smarandache Permutation Sequence { PS(n) } n = The Smarandache permutation sequence, denoted by { PS(n) } n=, is defined by (Dumitrescu and Seleacu [4]) PS(n) =35... (n )(n)(n )... 4, n. (.7.) The first few terms of the sequence are, 34, 3564, , ,, all of which are even numbers. For the Smarandache permutation sequence, the question raised (Dumitrescu and Seleacu [4]) is : Is there any perfect power among these numbers? Smarandache conjectures that there are none. In Theorem.7., we prove the conjecture in the affirmative. To prove the theorem, we need the following results. Lemma.7. : For n, PS(n) is of the form (k + ) for some integer k >. Proof : Since for n, PS(n) = (n )(n)(n )... 4, (.7.) we see that PS(n) is even and after division by, the last digit of the quotient is. An immediate consequence of the above lemma is the following. Corollary.7. : k divides PS(n) (for some integer n ) if and only if k =. Theorem.7.: For n, PS(n) is not a perfect power. Proof : The result is clearly true for n =, since PS() = 3 is not a perfect power. The proof for the case n is by contradiction. Let, for some integer n, PS(n) = x k for some integers x >, k. Since PS(n) is even, so is x. Let x = y for some integer y >. Then, PS(n) = (y) = k y k, which shows that k divides PS(n), contradicting Corollary.7.. To get more insight into the numbers of the Smaradache permutation sequence, we define a new sequence, called the reverse even sequence, and denoted by { RES(n) } n=, as follows : RES(n) = (n)(n )... 4, n. (.7.3) The first few terms of the sequences are, 4, 64, 864, 0864, 0864,. We note that, for all n, RES(n + ) = (n + )(n)(n )... 4 = (n + )0 s + RES(n) for some integer s n, (.7.4)

25 4 Wandering in the World of Smarandache umbers where, more precisely, s = number of digits in RES(n). Thus, for example, RES(4) = RES(3), RES(5) = RES(4), RES(6) = RES(5). Lemma.7. : For any integer n, 4 divides [RES(n + ) RES(n)]. Proof : Since from (.7.4), RES(n + ) RES(n) = (n + )0 s for some integer s ( n ), the result follows. Lemma.7.3 : For n, RES(n) is of the form (k + ) for some integer k 0. Proof : The proof is by induction on n. The result is true for n =. So, we assume that the result is true for some n, that is, RES(n) = (k + ) for some integer k 0. But, by virtue of Lemma.7., RES(n + ) RES(n) = 4k' for some integer k' > 0, which, together with the induction hypothesis, gives, RES(n + ) = 4k' + RES(n) = 4(k + k') +. Thus, the result is true for n + as well, completing the proof. Lemma.7.4 : 3 divides RES(3n) if and only if 3 divides RES(3n ). Proof : The result follows by noting that RES(3n) = (6n) 0 s + RES(3n ) for some integer s n. By repeated application of (.7.4), we get successively RES(n + 3) = (n + 6)0 s + RES(n + ) for some integer s n + = (n + 6)0 s + (n + 4)0 t + RES(n + ) for some integer t n + = (n + 6)0 s + (n + 4)0 t + (n + )0 u + RES(n) for some integer u n, (.7.5) so that, for some integers s, t and u with s > t > u n, RES(n + 3) RES(n) = (n + 6)0 s + (n + 4)0 t + (n + )0 u. (.7.6) Lemma.7.5 : 3 divides [RES(n + 3) RES(n)] for any integer n. Proof : is evident from (.7.6), since 3 (n + 6)0 s + (n + 4)0 t + (n + )0 u = 0 u [(n + 6)(0 s u + 0 t u + ) (0 s u + )]. Corollary.7. : 3 divides RES(3n) for any integer n. Proof : The result is true for n =, since RES(3) = 64 is divisible by 3. Now, assuming the validity of the result for n, so that 3 divides RES(3n), we get, from Lemma.7.5, 3 divides RES(3n + 3) = RES(3(n + )), so that the result is true for n + as well. This completes the proof by induction. Corollary.7.3 : 3 divides RES(3n ) for any integer n. Proof : follows from Lemma.7.4, together with Corollary.7..

26 5 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Corollary.7.4 : For any integer n 0, 3 does not divide RES(3n + ). Proof : Clearly, the result is true for n = 0. For n, from (.7.4), RES(3n + ) = (6n + )0 s + RES(3n) for some integer s 3n. Now, 3 RES(3n) (by Corollary.7.) but 3 does not divide (6n + ). Hence the result. Lemma.7.6 : RES(n + ) RES(n) > RES(n + ) for any integer n. Proof : Using (.7.5), we see that, for all n, RES(n + ) RES(n) = (n + 4)0 t + (n + )0 u, where t is the number of digits in RES(n + ). Hence the lemma. PS(n) can be expressed in terms of OS(n) and RES(n) as follows : For any n, PS(n) = 0 s OS(n) + RES(n) for some integer s n, (.7.7) where, more precisely, s = number of digits in RES(n). From (.7.7), together with Lemma.7.3, we observe that, for n, (since 4 divides 0 s for s n ), PS(n) is of the form 4k + for some integer k > (see Lemma.7.). Lemma.7.7 : 3 divides PS(n) if and only if 3 divides PS(n + 3). Proof : follows by virtue of Lemma.. and Lemma.7.5. Lemma.7.8 : 3 divides PS(3n) for any integer n. Proof : follows by virtue of Lemma.. and Corollary.7., since PS(3n) = 0 s OS(3n) + RES(3n) for some integer s 3n. Lemma.7.9 : 3 divides PS(3n ) for any integer n. Proof : Since 3 divides PS() =, the result is true for n =. To prove by induction, we assume that the result is true for some n, that is, 3 divides PS(3n ). But, then, by Lemma.7.7, 3 divides PS(3n + ), showing that the result is true for n + as well. Lemma.7.0 : For any integer n, PS(n + ) PS(n) > PS(n + ). Proof : By repeated application of (.7.7), we get PS(n + ) = 0 t OS(n + ) + RES(n + ) for some integer t n +, PS(n + ) = 0 u OS(n + ) + RES(n + ) for some integer u n +, where u > t > s n. Therefore, PS(n + ) PS(n) = [0 u OS(n + ) 0 s OS(n)] + [RES(n + ) RES(n)] > 0 u [OS(n + ) OS(n)] + [RES(n + ) RES(n)] >0 t OS(n + ) + RES(n + ) = PS(n + ), where the last inequality follows from (..6), Lemma.7.6, and the fact that 0 u > 0 t. Lemma.7.0 can be used to prove the following result (see also Le [5]). Theorem.7. : There are no numbers in the Smarandache permutation sequence that are Fibonacci (or, Lucas) numbers.

27 6 Wandering in the World of Smarandache umbers.8 Smarandache Circular Sequence { CS(n) } n = The Smarandache circular (or, consecutive) sequence, denoted by { CS(n) } n=, is obtained by repeatedly concatenating the positive integers (Dumitrescu and Seleacu [4]). The first few terms of the sequence are,, 3, 34, 345, 3456,. The n th tem of the circular sequence is given CS(n) = 3... (n )(n), n. (.8.) Since CS(n + ) = 3... (n )(n)(n + ), we see that, for any integer n, CS(n + ) = 0 s CS(n) + (n + ) for some integer s, CS() = ; (.8.) where, more precisely, s = number of digits in (n + ). Thus, for example, CS(9) = 0 CS(8) + 9, and CS(0) = 0 CS(9) + 0. From (.8.), we get the following result : Lemma.8. : For any integer n, CS(n + ) CS(n) > 9 CS(n). Using Lemma.8., we get, following the proof of Corollary.., CS(n + ) CS(n) > 9[CS(n + ) + CS(n)] for all n. (.8.3) Thus, CS(n + ) CS(n) > CS(n + ), n. (.8.4) Based on computer search for Fibonacci (and Lucas) numbers from up through CS(999) = , Asbacher [] conjectures that (except for the trivial case, CS() = = F() = F() = L()), there are no Fibonacci (and Lucas) numbers in the Smarandache circular sequence. The following theorem confirms the conjectures of Ashbacher. Theorem.8. : There are no Fibonacci (and Lucas) numbers in the Smarandache circular sequence (except for the trivial cases of CS() = = F() = F() = L(), CS(3) = 3 = L(0)). Proof : is evident from (.8.4). Remark.8. : As has been pointed out by Ashbacher [], CS(3) is a Lucas number. However, note that, CS(3) CS() + CS(). Lemma.8. : Let 3 divide n. Then, 3 divides CS(n) if and only if 3 divides CS(n ). Proof : follows readily from (.8.).

28 7 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function By repeated application of (.8.), we get, where s t u. CS(n + 3) = 0 s CS(n + ) + (n + 3) for some integer s = 0 s [0 t CS(n + ) + (n + )] + (n + 3) for some integer t = 0 s+t [0 u CS(n) + (n + )] + (n + )0 s + (n + 3) for some integer u = 0 s+t+u CS(n) + (n + )0 s+t + (n + )0 s + (n + 3), (.8.5) Lemma.8.3 : 3 divides CS(n) if and only if 3 divides CS(n + 3). Proof : follows from Lemma.., together with (.8.5), since 3 [(n + )0 s+t + (n + )0 s + (n + 3)] = [(n + )(0 s+t + 0 s + ) + (0 s + )]. Lemma.8.4 : 3 divides CS(3n) for any integer n. Proof : The proof is by induction on n. The result is clearly true for n =, since 3 divides CS(3) = 3. So, we assume that the result is true for some n, that is, we assume that 3 divides CS(3n) for some n. But then, by Lemma.8.3, 3 divides CS(3n + 3) = CS(3(n + )), showing that the result is true for n + as well, completing induction. Corollary.8. : 3 divides CS(3n ) for any integer n. Proof : From (.8.), for n, CS(3n) = 0 s CS(3n ) + 3n for some integer s. Since, by Lemma.8.4, 3 divides CS(3n), the result follows. Corollary.8. : 3 does not divide CS(3n + ) for any integer n 0. Proof : For n = 0, CS() = is not divisible by 3. By Lemma.8.4, 3 divides CS(3n) for any integer n. Since (from (.8.)), CS(3n + ) = 0 s CS(3n) + (3n + ), and since 3 does not divide (3n + ), we get desired the result. Lemma.8.5 : For any integer n, 5 divides CS(5n). Proof : For n, from (.8.), CS(5n) = 0 s CS(5n ) + 5n for some integer s. Clearly, the r.h.s. is divisible by 5. Hence, 5 divides CS(5n). For the Smarandache circular sequence, the question is : How many terms of the sequence are prime? Fleuren [6] gives a table of prime factors of CS(n) for n = ()00, which shows that none of these numbers is a prime. In the Editorial Note following the paper of Stephan [7], it is mentioned that, using a supercomputer, no prime has been found in the first 307 terms of the Smarandache circular sequence. This gives rise to the conjecture that there is no prime in the Smarandache circular sequence. This conjecture still remains to be resolved. We note that, in order to check for prime numbers in the Smarandache circular sequence, it is sufficient to check the terms of the form CS(3n + ) (n ), where 3n + is odd and is not divisible by 5.

29 8 Wandering in the World of Smarandache umbers.9 Smarandache Reverse Sequence { RS(n) } n = The Smarandache reverse sequence, denoted by { RS(n) } n=, is the sequence of numbers formed by concatenating the consecutive integers on the left side, starting with RS() =. The first few terms of the sequence are (Ashbacher []),, 3, 43, 543, 6543,, and in general, the n th term is given by RS(n) = n(n )..., n. (.9.) From (.9.), we see that, for all n, RS(n + ) = (n + )n(n )..., n, = (n + )0 s + RS(n) for some integer s n (with RS() = ), (.9.) where, more precisely, Thus, for example, s = number of digits in RS(n). RS(9) = RS(8), RS(0) = RS(9), RS() = 0 + RS(0). By repeated application of (.9.), we get, for all n, RS(n + 3) = (n + 3)0 s + RS(n + ) for some integer s n + = (n + 3)0 s + (n + )0 t + RS(n + ) for some integer t n + = (n + 3)0 s + (n + )0 t + (n + )0 u + RS(n) for some integer u n, (.9.3) where s > t > u n. Thus, RS(n + 3) = 0 u [(n + 3)0 s u + (n + )0 t u + (n + )] + RS(n). (.9.4) Lemma.9. : For any integer n, () 4 divides [RS(n + ) RS(n)], () 0 divides [RS(n + ) RS(n)]. Proof : For all n, from (.9.), RS(n + ) RS(n) = (n + )0 s (with s n), and the r.h.s. is divisible by both 4 and 0. Corollary.9. : For any n, RS(n) is of the form 4k + for some integer k 0. Proof : Since RS() = = 4 5 +, the result is clearly true for n =. So, we assume the validity of the result for n, that is, we assume that RS(n) = 4k + for integer k. Now, by Lemma.9., for some integer k', RS(n + ) RS(n) = 4k' RS(n + ) = RS(n) + 4k' = 4(k + k') +, where the expression on the right follows by virtue of the induction hypothesis. Thus, the result is true for n + as well, completing induction.

30 9 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Lemma.9. : Let 3 n for some n ( ). Then, 3 RS(n) if and only if 3 RS(n ). Proof : follows immediately from (.9.). Lemma.9.3 : 3 divides [RS(n + 3) RS(n)] for any integer n. Proof : is immediate from (.9.4), together with Lemma... A consequence of Lemma.9.3 is the following. Corollary.9. : 3 divides RS(n) if and only if 3 divides RS(n + 3). Using Corollary.9., the following result can be established by induction on n. Corollary.9.3 : 3 divides RS(3n) for all n. Corollary.9.4 : 3 divides RS(3n ) for any integer n. Proof : follows from Corollary.9.3, together with Lemma.9.. Lemma.9.4 : 3 does not divide RS(3n + ) for any integer n 0. Proof : The result is true for n = 0. For n, by (.9.), RS(3n + ) = (3n + )0 s + RS(3n). This gives the desired result, since 3 divides RS(3n) but 3 does not divide (3n + ). Lemma.9.5 : For any integer n, RS(n + ) > RS(n). Proof : Using (.9.), we see that RS(n + ) = (n + )0 s + RS(n) > RS(n) if and only if RS(n) < (n + )0 s, which is true since RS(n) is an s digit number while 0 s is an (s + ) digit number. Corollary.9.5 : For any integer n, RS(n + ) RS(n) > RS(n + ). Proof : Using (.9.3), we have (for some integers t and u with t > u) RS(n + ) RS(n) = [RS(n + ) RS(n + )] + [RS(n + ) RS(n)] = [(n + )0 t (n + )0 u ] + [RS(n + ) RS(n)] > [RS(n + ) RS(n)] > RS(n + ), by Lemma.9.5. This gives the desired inequality. Theorem.9. : There are no numbers in the Smarandache reverse sequence that are Fibonacci or Lucas numbers (except for RS() = = F() = F() = L(), RS() = = F(8)). Proof : follows from Corollary.9.5. The following expression for RS(n) is due to Alexander [8]. RS(n) = + i *0 n i= i j= ( + [log j]) ([x] denotes the integer part of x). For the Smarandache reverse sequence, the question is : How many terms of the sequence are prime? It is conjectured, based on the first 739 terms, that RS(8) is the only prime in the sequence (see Stephan [7]). The conjecture still remains to be resolved.

31 30 Wandering in the World of Smarandache umbers.0 Smarandache Symmetric Sequence { SS(n) } n = The Smarandache symmetric sequence, denoted by { SS(n) } n=, is defined by (Ashbacher []),,, 3, 343, 34543, ,. More precisely, the n th term of the Smarandache symmetric sequence is SS(n) =... (n )(n )(n )..., n 3; SS() =, SS() =. (.0.) The numbers in the Smarandache symmetric sequence can be expressed in terms of the numbers of the Smarandache circular sequence (in.8) and the Smarandache reverse sequence (in.9) as follows : For all n 3, SS(n) = 0 s CS(n ) + RS(n ) for some integer s, (.0.) with SS() =, SS() =, where more precisely, s = number of digits in RS(n ). Thus, for example, SS(3) = 0 CS() + RS(), SS(4) = 0 CS(3) + RS(). Lemma.0. : 3 divides SS(3n + ) for any integer n. Proof : Let n ( ) be any arbitrary but fixed number. Then, from (.0.), SS(3n + ) = 0 s CS(3n) + RS(3n ). Now, by Lemma.8.4, 3 divides CS(3n), and by Corollary.9.4, 3 divides RS(3n ). Therefore, 3 divides SS(3n + ). Lemma.0. : For any integer n, () 3 does not divide SS(3n), () 3 does not divide SS(3n + ). Proof : Using (.0.), we see that SS(3n) = 0 s CS(3n ) + RS(3n ), n. By Corollary.8., 3 divides CS(3n ), and by Lemma.9.4, 3 does not divide RS(3n ). Hence, CS(3n) cannot be divisible by 3. Again, since SS(3n + ) = 0 s CS(3n + ) + RS(3n), n, and since 3 does not divide CS(3n + ) (by Corollary.8.) and 3 divides RS(3n) (by Corollary.9.3), it follows that SS(3n + ) is not divisible by 3. Using (.8.4) and Corollary.9.5, we can prove the following lemma. The proof is similar to that used in proving Lemma.7.0, and is omitted here. Lemma.0.3 : For any integer n, SS(n + ) SS(n) > SS(n + ). By virtue of the inequality in Lemma.0.3, we have the following. Theorem.0. : (Except for the trivial cases, SS() = = F() = L(), SS() = = L(5)), there are no members of the Smarandache symmetric sequence that are Fibonacci (or, Lucas) numbers.

32 3 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function The following lemma gives the expression of the difference SS(n + ) SS(n) in terms of the difference CS(n) CS(n ). Lemma.0.4 : SS(n + ) SS(n) = 0 s+t [CS(n) CS(n )] for any integer n 3, where s = number of digits in RS(n ), s + t = number of digits in RS(n ). Proof : By (.0.), for n 3, SS(n) = 0 s CS(n ) + RS(n ), SS(n + ) = 0 s+t CS(n) + RS(n ), so that SS(n + ) SS(n) = 0 s [0 t CS(n) CS(n )] + [RS(n ) RS(n )] = 0 s [0 t CS(n) CS(n ) + (n )], () where the last equality follows from (.9.). But,, if n 9 t = = number of digits in (n ). Therefore, by (.8.) CS(n ) = 0 t CS(n ) + (n ), so that CS(n ) (n ) = 0 t CS(n ). Now, plugging in this expression in (), we get the desired result. m m+ m +, if 0 n 0 for some integer m We observe that SS() = is prime; the next eight terms of the Smarandache symmetric sequence are composite numbers and squares : SS(3) = =, SS(4) = 3 = (3 37) =, SS(5) = 343 = ( 0) =, SS(6) = = (4 7) =, SS(7) = = ( ) =, SS(8) = =, SS(9) = =, SS(0) = = ( ) =. For the Smarandache symmetric sequence, the question is : How many terms of the sequence are prime? The question still remains to be answered.

33 3 Wandering in the World of Smarandache umbers. Smarandache Pierced Chain Sequence { PCS(n) } n = The Smarandache pierced chain sequence, denoted by { PCS(n) } n =, is defined by (Kashihara [9], Ashbacher []) 0, 000, 00000, ,, which is obtained by successively concatenating the string 00 to the right of the preceding terms of the sequence, starting with SPC() = 0. We first observe that the elements of the Smarandache pierced chain sequence, { PCS(n) } n =, satisfy the following recurrence relation : SPC(n + ) = 0 4 SPC(n) + 0, n ; SPC() = 0. (..) We can now prove the following result, giving an explicit expression of SPC(n). Lemma.. : The elements of the sequence { PCS(n) } n = 0, 0(0 4 + ), 0( ), 0( ),, and in general, SPC(n) = 0[0 4(n ) + 0 4(n ) ], n. (..) Proof : The proof of (..) is by induction on n. The result is clearly true for n =. So, we assume that the result is true for some n. Now, from (..), together with the induction hypothesis, we see that SPC(n + ) = 0 4 SPC(n) + 0 are = 0 4 [0(0 4(n ) + 0 4(n ) )] + 0 = 0(0 4n + 0 4(n ) ), which shows that the result is true for n +. From Lemma.., we see that SPC(n) is divisible by 0 for all n. Thus, from the sequence { PCS(n) }, we can form another sequence, namely, the sequence PCS(n) n =, 0 n = whose explicit form is given in the following corollary. Corollary.. : The elements of the sequence, x +, x + x +, x 3 + x + x +,, where x 0 4 ; and in general, PCS(n) 0 PCS(n) = x n + x n + +, n, (..3) 0 n= are

34 33 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function The first few terms of the sequence PCS(n) 0, 000, , , ; and the terms of the sequence satisfy the recurrence relation, given in the following Corollary.. : For any integer n, PCS(n + ) 4n PCS(n) PCS() = 0 +, = Proof : Evident from (..3). n= From Corollary.., we see that, for any n, are PCS(n + ) 4(n+ ) PCS(n + ) 4n 4 PCS(n) = 0 + = 0 (0 + ) +, (..4) PCS(n + 3) 4(n+ ) PCS(n + ) 4n 8 4 PCS(n) = 0 + = 0 ( ) +. (..5) Lemma.. : PCS (n) = (0 8(n ) + 0 8(n ) ) PCS ( ), n. 0 0 Proof : The proof is by induction on n. The result is clearly true for n =. Assuming the validity of the result for some n, we see from (..4) that PCS(n + ) 8n 4 PCS(n) = 0 (0 + ) n 4 8(n ) 8(n ) 8 PCS( ) = 0 (0 + ) + { } 0 8n 8(n ) 8(n ) 8 PCS( ) = { }, 0 which shows that the result is true for n + as well. Lemma..3 : PCS (3n) = (0 (n ) + 0 (n ) ) PCS ( 3 ), n. 0 0 Proof : The result is clearly true for n =. Assuming the validity of the result for some n, we see from (..5) that PCS(3n + 3) n 8 4 PCS(3n) = 0 ( ) n 8 4 (n ) (n ) PCS( 3) = 0 ( ) + { } 0 n (n ) (n ) PCS( 3) = { }, 0 which shows the validity of the result for n +, thereby completing induction.

35 34 Wandering in the World of Smarandache umbers PCS(n) For the sequence, the question is : How many terms are prime? 0 n= Note that the second and third terms of the sequence are not primes, since PCS( ) = 000 = 73 37, PCS ( 3 ) = We now prove the more general result. Theorem.. : For any integer n, PCS (n) is a composite number. 0 Proof : The result is true for n =. In fact, the result is true if n is even as shown below : If n ( 4) is even, let n = m for some integer m ( ). Then, from (..3) (where x 0 4 ), PCS(m) = x m + x m + + x + 0 = x m (x + ) + + (x + ) = (x + )(x m + x m ), (..6) which shows that PCS (m) is a composite number for all m ( ). 0 Thus, it is sufficient to consider the case when n is odd. First we consider the case when n is prime, say n = p, where p ( 3) is a prime. In this case, from (..3) (with x 0 4 ), PCS(p) = x p + x p + + = x p 0 x. () Let y = 0 (so that x = y ). Then, x = y = (y + )(y ), x p = (y ) p = (y p + )(y p ). Using the Binomial expansions of y p + and y p, we get from () so that PCS(p) p p p p = x 0 x = y y = (y )(y + ) (y )(y + ) p p p p {(y )(y + y )}{(y + )(y y )} = (y )(y + ) = (y p y p + y p 3 + )(y p + y p + y p ), PCS(p) = (y p y p + y p 3 + )(y p + y p + y p ). (..7) 0 The above expression shows that PCS (p) 0 is a composite number for each prime p ( 3).

36 35 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function Finally, we consider the case when n is odd but composite. Let n = pr, where p is the largest prime factor of n, and r ( ) is an integer. Then, PCS(n) PCS(pr) = 0 0 = x pr + x pr + + = x p(r ) (x p + x p + + ) + x p(r ) (x p + x p + + ) + + (x p + x p + + ) = (x p + x p + + )[x p(r ) + x p(r ) + + ], (..8) and hence, PCS (n) PCS(pr) = is also a composite number. 0 0 All these complete the proof of the theorem. Following different approaches, Kashihara [9] and Le [0] have proved by contradiction the result of Theorem... On the other hand, we prove the same result by actually finding PCS(n) PCS(n) out the factors of. From (..6), (..7) and (..8), it is clear that can 0 0 not be a square for any n. PCS(n) PCS( ) Lemma.. shows that can be expressed in terms of, and 0 0 PCS(3n) PCS( 3) Lemma..3 shows that can be expressed in terms of. Some 0 0 consequences of these results are given in the following corollaries. Corollary..3 : 3 divides PCS (3n) for all n. 0 Proof : follows from Lemma..3. Corollary..4 : 9 divides PCS (9n) for all n. 0 Proof : By virtue of Lemma..3 and Corollary..3, it is sufficient to show that 3 (0 (3n ) + 0 (3n ) ). The proof is immediate, since, the number of in 0 (3n ) + 0 (3n ) is 3n, and hence, it is divisible by 3. PCS(n) Kashihara [9] raises the question : Is the sequence square-free? Recall that 0 n= an integer is called square-free if it is not divisible by the square of a prime. Corollary..4 shows that an infinite number of terms of the sequence are not square-free (see also Su Gou and Jianghua Li []).

37 36 Wandering in the World of Smarandache umbers. Series Involving Smarandache Sequences From the Smarandache circular sequence (given in.8), we can form the series = (..) CS(n) 3 34 n= The following lemma shows that the series (..) is convergent, and finds its bounds. Lemma.. : The series (..) is convergent with Proof : First note that CS(n) 0 n for all n. Then, = 0. n CS(n) 0 9 n= n= < 0. CS(n) 9 n= Combining together the Smarandache circular sequence (.8) and the Smarandache reverse sequence (.9), we can form the series CS(n) 3 34 = n= RS(n) 3 43 In connection with this series, we have the following result. Lemma.. : The series CS(n) is divergent. RS(n) n= Proof : Let us consider the subsequence few terms of the subsequence are n CS(0 ) n RS(0 ) n=... 9(0)... 99(00) (000),,,.... (0)9... (00)99... (000) This shows that CS(0 n ) > RS(0 n ) for all n, of the given sequence. The first so that n n CS(0 ) lim 0. n RS(0 ) Hence, the series n= n CS(0 ) is divergent, and so also is n RS(0 ) n= CS(n). RS(n)

38 37 Chapter Chapter 4 : The Pseudo : Some Smarandache Sequences Function.3 Some Remarks This chapter gives some elementary properties of ten Smarandache recurrence type sequences. In each case, the recurrence relation, satisfied by the sequence, is given. In case of the Smarandache odd sequence, even sequence, prime product sequence, square product sequences of two types, permutation sequence, circular sequence and the reverse sequence, we have shown that none of these sequences satisfy the recurrence relation of the Fibonacci or Lucas numbers. This shows that, none of these sequences can contain three or more consecutive Fibonacci numbers or Lucas numbers in a row. It should be mentioned here that our theorems don t rule out the possibility of appearance of Fibonacci or Lucas numbers, scattered here and there, in any of these sequences. In case of the Smarandache odd sequence, we find that the terms of the forms OS(3n) and OS(5n + 3) cannot be primes; for the circular sequence, we show that the terms of the form CS(3n), CS(3n ) and CS(5n) cannot be primes; while, only terms of the form SS(3n + ) of the symmetric sequence cannot be primes. On the other hand, it has already been settled that no terms of the pierced chain sequence is a prime, and the higher power sequences of the two kinds can each contain only one prime. Thus, the following old conjectures still remain open. Conjecture.3. (Marimutha [3]) : There are infinitely many primes in the Smarandache odd sequence. Conjecture.3. (Russo [9]) : The Smarandache square product sequences of two kinds each contains only a finite number of primes. Conjecture.3.3 : The Smarandache circular sequence contains no primes. Conjecture.3.4 : Only RS(8) of the Smarandache reverse sequence is prime. In case of the Smarandache even sequence, we find that the terms of the forms ES(k.0 n + 6) and ES(k.0 n + 6) ( k 9) cannot be perfect squares, for the Smarandache higher product sequences of two kinds, our finding is that none contains perfect squares, for the Smarandache permutation sequence, no terms can be perfect powers. So, the following question still remains to be resolved. Question.3. (Ashbacher []) : How many terms of the Smarandache even sequence are perfect powers?

39 38 Wandering in the World of Smarandache umbers References [] Ashbacher, Charles. Pluckings from the Tree of Smarandache Sequences and Functions. American Research Press, Lupton, AZ, USA [] Bernard, S. and Child, J.M. Higher Algebra. Macmillan Book Co. Ltd., U.K [3] Marimutha, H. Smarandache Concatenated Type Sequences. Bulletin of Pure and Applied Sciences, E6, 5 6, 997. [4] Smith, S. A Set of Conjectures on Smarandache Sequences. Smarandache otions Journal,, 86 9, 000. [5] Smarandache, F. Collected Papers Volume П. Tempus publishing House, Bucharest, Romania [6] Prakash, K. A Sequence Free from Powers. Mathematical Spectrum, (3), 9 93, 990. [7] Majumdar, A.A.K. A Note on a Sequence Free from Powers. Mathematical Spectrum, 9 (), 4, 996 / 7. Also, Mathematical Spectrum, 30 (),, 997/98 (Letters to the Editor). [8] Majumdar, A.A.K. A Note on the Smarandache Prime Product Sequence. Smarandache otions Journal, 9, 37 4, 998. [9] Russo, F. Some Results about Four Smarandache U-Product Sequences. Smarandache otions journal,, 4 49, 000. [0] Iacobescu, F. Smarandache Partition Type Sequences. Bulletin of Pure and Applied Sciences, E6, 37 40, 997. [] Le, M. Primes in the Smarandache Square Product Sequence. Smarandache otions Journal, 9, 33, 998. [ 3] Le, M. The Primes in the Smarandache Power Product Sequences. Smarandache otions Journal,, 8 3, 00. [4] Dumitrescu, D. and V. Seleacu. Some Notions and Questions in Number Theory. Erhus University Press, Glendale [5] Le, M. Perfect Powers in the Smarandache Permutation Sequence. Smarandache otions Journal, 0, 48 49, 999. [6] Fleuren, M. Smarandache Factors and Reverse Factors. Smarandache otions Journal, 0, 5 38, 999. [7] Stephan, R.W. Factors and Primes in Two Smarandache Sequences. Smarandache otions Journal, 9, 4 0, 998. [8] Alexander, S. A Note on Smarandache Reverse Sequence. Smarandache otions Journal,, 50, 00. [9] Kashihara, K. Comments and Topics on Smarandache Notions and Problems. Erhus University Press, AZ, U.S.A [0] Le, M. On Primes in the Smarandache Pierced Chain Sequence. Smarandache otions Journal, 0, 54 55, 999. [] Su Gou and Jianghua Li. On the Smarandache Pierced Chain. Scientia Magna, 4(), 44 45, 008. [] Majumdar, A.A.K. and Gunarto, H. On Some Recurrence Type Smarandache Sequences. Smarandache otions Journal, 4, 9, 004.

40 39 Chapter 4 : The Pseudo Smarandache Function Chapter Smarandache Determinant Sequences Murthy [] introduced four determinant sequences, called the Smarandache cyclic determinant natural sequence (SCDNS), the Smarandache cyclic arithmetic determinant sequence (SCADS), the Smarandache bisymmetric determinant natural sequence (SBDNS), and the Smarandache bisymmetric arithmetic determinant sequence (SBADS). In this chapter, we consider the problem of finding the n th terms of these sequences. The expressions of the n th terms for the Smarandache cyclic arithmetic determinant sequence and the Smarandache bisymmetric arithmetic determinant sequence are given in. and. respectively. It may be mentioned that, the SCDNS is a particular case of the SCADS, and the SBDNS is a particular case of the SBADS. In.3, we consider the problem of finding the sum of the first n terms of the Smarandache bisymmetric determinant natural sequence and the Smarandache bisymmetric arithmetic determinant sequence. In. and., we would make use of the following results. Lemma. : Let D d ij ( i, j n) be the determinant of order n ( ) with a, if i = j dij =, otherwise where a is any real number. Then, D L n = (a ). a L a L M L a L a Proof : Performing the indicated column operations (where Ci Ci C indicates the column operation of subtracting the st column from the i th column, i n), we get D L a L a L M L a L a = 0 0 L 0 0 C C C a 0 L 0 0 C3 C3 C 0 a L 0 0 M M Cn Cn C 0 0 L a L 0 a = a 0 L 0 0, 0 a L 0 0 M 0 0 L a L 0 a where the last determinant is diagonal of order n whose diagonal elements are all (a ). Hence, D = (a ) n.

41 40 Wandering in the World of Smarandache umbers Lemma. : Let D (a) Then, d (a) ij a, if i = j =, otherwise (a) d ij ( i, j n) be the determinant of order n ( ) with a L a L a L M L a L a (a) n D = (a ) (a + n ). Proof : We perform the indicated column operations (where C C + C Cn indicates the operation of adding all the columns from through n and then replacing the st column by that sum, and C denotes the operation of taking out the C + C Cn common sum) to get D (a) a L a L a L M L a L a = C C + C C C C + C C n n = (a + n )(a ) n, where the last equality follows by virtue of Lemma.. Lemma.3 : Let A n Then, a ij a ij, if i j =, otherwise (a + n ) L a L a L M L a L a ( i, j n) be the determinant of order n ( ) with n An L =. L L M L L Proof : The proof is by induction on n. Since A = =, the result is true for n =. So, we assume the validity of the result for some integer n ( ).

42 4 Chapter : 4 Smarandache : The Pseudo Smarandache Determinant Sequences Function To prove the result for n +, we consider the determinant of order n +, A n +, and perform the indicated column operations (where C C + Cn indicates the operation of adding the n th column to the st column to get the new st column), to get A n+ L L L M L L Hence, by virtue of the induction hypothesis, A n + = A n = n, so that the result is true for n +. This completes induction. Lemma.4 : For n ( ), = L C C + Cn 0 L 0 L M 0 L 0 L = L = A n. L L L n Bn L n L L M L L = ( ). Proof : We consider the following two cases depending on whether n is even or odd. Case : When n is even, say, n = m for some integer m. In this case, starting with the determinant B n = B m, we perform the indicated column operations. Bn = Bm = L L L M L L m = ( ) L. C Cm L C Cm L M M Cm Cm+ L L

43 4 Wandering in the World of Smarandache umbers Here, C Cm means that the st and the (m) th columns are interchanged, which involves m number of interchanges of columns, changing the sign each time. Similarly, C Cm means that the nd and (m ) st columns are interchanged, involving m number of interchanges of columns, and so on, and finally, the m th and (m + ) st columns are interchanged. Thus, the total number of interchanges of columns is (m ) + (m ) + + = m(m ), which gives the number of times the sign changes. Then, by Lemma.3, B n = B m = ( ) m A m = ( ) m m = ( ) m n. Case : When n is odd, say, n = m + for some integer m. In this case, starting with B n = B m +, we perform the indicated column operations to get Bn = B m+ = L L L M L L m = ( ) L. C Cm+ L C Cm L M M Cm Cm+ L L Note that, in this case, the total number of interchanges of columns is (m) + (m ) + + = m(m + ). Therefore, by virtue of Lemma.3, B n = B m + = ( ) m A m + = ( ) m m = ( ) m n. Since, in either case, m = n ([x] being the floor function), the result is established.. Smarandache Cyclic Arithmetic Determinant Sequence The Smarandache cyclic arithmetic determinant sequence is defined as follows. Definition.. : The Smarandache cyclic arithmetic determinant sequence, denoted by, {SCADS(n)}, is (where a and d are any real numbers) a a + d a + d a + 3d a a + d a + d a a + d a + d a + d a + 3d a a,, a + d a + d a,,... a d a a d a 3d a a d a + d a a + d a + 3d a a + d a + d

44 43 Chapter 4 : Smarandache : The Pseudo Smarandache Determinant Sequences Function Theorem.. : The n th term of the Smarandache cyclic arithmetic determinant sequence, {SCADS(n)}, is SCADS(n) = a a + d a + d L a + (n )d a + (n )d a + d a + d a + 3d L a + (n )d a a + d a + 3d a + 4d L a a + d a + 3d a + 4d a + 5d L a + d a + d M a + (n )d a + (n )d a L a + (n 4)d a + (n 3)d a + (n )d a a + d L a + (n 3)d a + (n )d n = n ( + ) n ( ) a d (nd). Proof : We consider separately the cases when n is even and odd respectively. Case : If n = m for some integer m. In this case, performing the indicated column and row operations, we get successively SCADS(n) = a a + d a + d L a + (m 3)d a + (m )d a + (m )d a + d a + d a + 3d L a + (m )d a + (m )d a a + d a + 3d a + 4d L a + (m )d a a + d a + 3d a + 4d a + 5d L a a + d a + d M a + (m )d a + (m )d a L a + (m 5)d a + (m 4)d a + (m 3)d a + (m )d a a + d L a + (m 4)d a + (m 3)d a + (m )d m = a (m )d a (m )d a (m 3)d a d a d a ( ) L + + C Cm a a + (m )d a + (m )d L a + 3d a + d a + d C Cm a + d a a + (m )d L a + 4d a + 3d a + d M a + d a + d a L a + 5d a + 4d a + 3d Cm Cm+ M a + (m 4)d a + (m 5)d a + (m 6)d L a + (m ) d a + (m )d a + (m 3)d a + (m 3)d a + (m 4)d a + (m 5)d L a a + (m )d a + (m )d a + (m )d a + (m 3)d a + (m 4)d L a + d a a + (m )d C C m m m = C +C +...+C C +C +...+C m m ( ) S m a + (m )d a + (m )d a + (m 3)dL a + d a + d a a + (m )d a + (m )dl a + 3d a + d a + d a a + (m )d L a + 4d a + 3d a + d a + d a L a + 5d a + 4d M a + (m 4)d a + (m 5)d a + (m 6)dL a + (m )da + (m )d a + (m 3)d a + (m 4)d a + (m 5)dL a a + (m )d a + (m )d a + (m 3)d a + (m 4)dL a + d a m = ( ) S a + (m )d a + (m )d a + (m 3)d L a + d a + d m R R R ( m)d d d L d d 0 R3 R3 R d ( m)d d L d d 0 M d d ( m)d L d d 0 Rm Rm Rm M d d d L d d 0 d d d L ( m)d d 0 d d d L d ( m)d 0

45 44 Wandering in the World of Smarandache umbers m m+ m ( ) ( ) d S m m L, = m L m L M L L m L m where the determinant above is of order (m ), and S m = a + (a + d) + (a + d) + + [a + (m )d] = ma + m(m )d. Therefore, by Lemma., m+ m m (m ) SCADS(n) = ( ) { ma + m(m )d } d {( ) (m) } m { } m m m ( ) a d d (m). = + Case : If n = m + for some integer m. In this case, performing the indicated column and row operations, we get successively SCADS(n) = a a + d a + d L a + (m )d a + (m )d a + md a + d a + d a + 3d L a + (m )d a + md a a + d a + 3d a + 4d L a + md a a + d a + 3d a + 4d a + 5d L a a + d a + d M a + (m )d a + md a L a + (m 4)d a + (m 3)d a + (m )d a + md a a + d L a + (m 3)d a + (m )d a + (m )d C C m = a + md a + (m )d a + (m )d a + d a + d a ( ) L C Cm+ a a + md a + (m )d L a + 3d a + d a + d C Cm a + d a a + md L a + 4d a + 3d a + d M a + d a + d a L a + 5d a + 4d a + 3d Cm Cm+ M a + (m 3)d a + (m 4)d a + (m 5)d L a + md a + (m )d a + (m )d a + (m )d a + (m 3)d a + (m 4)d L a a + md a + (m )d a + (m )d a + (m )d a + (m 3)d L a + d a a + md m+ m+ m+ = C +C +...+C C +C +...+C m+ L a a + md a + (m )d L a + 3d a + d a + d a a + md L a + 4d a + 3d a + d a + d a L a + 5d a + 4d M a + (m 3)d a + (m 4)d a + (m 5)d L a + md a + (m )d a + (m )d a + (m 3)d a + (m 4)d L a a + md a + (m )d a + (m )d a + (m 3)d L a + d a m ( ) S m+ a + md a + (m )d a + (m )d a + d a + d (where S m+ = a + (a + d) + (a + d) + + (a + md) = (m + )a + m(m + )d = (m + )(a + md))

46 45 Chapter 4 : Smarandache : The Pseudo Smarandache Determinant Sequences Function m = ( ) S a + md a + (m )d a + (m )d L a + d a + d m+ R R R md d d L d d 0 R3 R3 R d md d L d d 0 M d d md L d d 0 Rm+ Rm+ Rm M d d d L d d 0 d d d L md d 0 d d d L d md 0 { } = ( ) (m + )(a + md) ( ) d m m+ m m L. m L m L M L L m L m Now, noting that the last determinant above is of order m, we get, by Lemma., m m m m SCADS(n) = ( ) (m + )(a + md)d {( ) (m + ) } (m ) + { } = + + m m m ( ) a d d (m ). Now, in either case, m = n. Hence, the theorem is proved. Definition.. : The Smarandache cyclic determinant natural sequence, denoted by {SCDNS(n)}, is ,, 3,, The first few terms of the sequence {SCDNS(n)} are, 3, 8, 60, 875,. Note that the Smarandache cyclic determinant natural sequence is a particular case of the Smarandache cyclic arithmetic determinant sequence, with a =, d =. Hence, as a consequence of Theorem.., we have the following result. Corollary.. : The n th term of the Smarandache cyclic determinant natural sequence is SCDNS(n) = n 3 L n n n n n = ( ) + ( ) n. 3 4 L n n L n L 3 M n n L n 4 n 3 n n L n 3 n n

47 46 Wandering in the World of Smarandache umbers. Smarandache Bisymmetric Arithmetic Determinant Sequence The Smarandache bisymmetric arithmetic determinant sequence is defined as follows. Definition.. : The Smarandache bisymmetric arithmetic determinant sequence, denoted by {SBADS(n)}, is (where a and d are any real numbers) a a + d a + d a a + d a,, a + d a + d a + d,... a d a. + a + d a + d a Theorem.. : The n th term of the Smarandache bisymmetric arithmetic determinant sequence, {SBADS(n)}, n 5, is SBADS(n) = a a + d a + d L a + (n 3)d a + (n )d a + (n )d a + d a + d a + 3d L a + (n )d a + (n )d a + (n )d a + d a + 3d a + 4d L a + (n )d a + (n )d a + (n 3)d M a + (n 3)d a + (n )d a + (n )d L a + 4d a + 3d a + d a + (n )d a + (n )d a + (n )d L a + 3d a + d a + d a + (n )d a + ( n )d a + (n 3)d L a + d a + d a ( ) n n n ( ) a d (d). = + Proof : Starting with SBADS(n), we perform the indicated row and column operations. SBADS(n) = a a + d a + d L a + (n 3)d a + (n )d a + (n )d a + d a + d a + 3d L a + (n )d a + (n )d a + (n )d a + d a + 3d a + 4d L a + (n )d a + (n )d a + (n 3)d M a + (n 3)d a + (n )d a + (n )d L a + 4d a + 3d a + d a + (n )d a + (n )d a + (n )d L a + 3d a + d a + d a + (n )d a + ( n )d a + (n 3)d L a + d a + d a = a a + d a + d L a + (n 3)d a + (n )d a + (n )d R R R d d d L d d d R3 R3 R d d d L d d d M M Rn Rn Rn d d d L d d d d d d L d d d d d d L d d d n = d a a + d a + d L a + (n 3)d a + (n )d a + (n )d Cn Cn + C L 0 L 0 M L 0 L 0 L 0

48 47 Chapter 4 : Smarandache : The Pseudo Smarandache Determinant Sequences Function { } = + n+ n ( ) d a (n )d L, L M L L L where the last determinant above is of order n. Now, appealing to Lemma.4, we get n n+ n n SBADS(n) = ( ) d { a + (n )d } ( ). Now, if n is even, say, n = m for some integer m, then n+ + n n m+ + (m ) m ( ) = ( ) = ( ) = ( ), and if n is odd, say, n = m + for some integer m, then n+ + n Hence, finally, we get n (m+ ) + m m ( ) = ( ) = ( ) = ( ). n n n ( ) d a (n )d, SBADS(n) = { + } which we intended to prove. Definition.. : The Smarandache bisymmetric determinant natural sequence, denoted by{sbdns(n)}, is ,, 3,, The first few terms of the sequence {SBDNS(n)} are, 3, 8, 0, 48,, 56, 576,. Note that the Smarandache bisymmetric determinant natural sequence is a particular case of the Smarandache bisymmetric arithmetic determinant sequence, with a =, d =. Hence, we have the following result as a consequence of Theorem... Corollary.. : The n th term of the Smarandache bisymmetric determinant natural sequence, {SBDNS(n)}, n 5, is SBDNS(n) = 3 L n n n 3 4 L n n n L n n n M n n n L n n n L 4 3 n n n L 3 n n ( ) (n ). = +

49 48 Wandering in the World of Smarandache umbers.3 Series With Smarandache Determinant Sequences.3. Smarandache Bisymmetric Determinant atural Sequence We consider the Smarandache bisymmetric determinant natural sequence, { SBDNS(n) }, for n = which the n th term is, by Corollary.., n n SBDNS(n) = ( ) (n + ). Let { S n } be the sequence of n th partial sums of the sequence { SBDNS(n) } n= n = S n n = SBDNS(k), n. k= In this section, we give an explicit expression for the sequence { } S n n= Theorem.3... To prove the theorem, we would need the following results. Lemma.3.. : For any integer m, 4m (k ) (d) (d) =. 4 k=,3,...,(m ) (d) Proof : Note that, the series is a geometric series with the common ratio (d) 4. Lemma.3.. : For any integer m, m (k ) / m m (y + )(y ) k y = y. k=,3,...,(m ) y (y ), so that. This is given in Proof : Let the series on the left be denoted by s, so that s = + 3y + 5y + + (m )y m. (*) Now, multiplying (*) by y and then subtracting the resulting expression from (*), we get ( y)s = + y + y + + y m (m )y m = [( + y + y + + y m ) ] (m )y m m y = y (m )y m m (y + )(y ) = my m, y which now gives the desired result. From Corollary.., we see that, for any integer k, SBDNS(k) + SBDNS(k + ) = ( ) k (6k + 5) (k ), SBDNS(k + ) + SBDNS(k + 3) = ( ) k+ (6k + ) k, so that SBDNS(k) + SBDNS(k + ) + SBDNS(k + ) + SBDNS(k + 3) = 3( ) k+ (6k + 3) (k ). ()

50 49 Chapter 4 : Smarandache : The Pseudo Smarandache Determinant Sequences Function Theorem.3.. : For any integer m 0, () S 4m + = 3 5 m(m+) m 6 5 = 4m { (60m 3) + 3 }, 5 () S 4m + = 4m+3 m (m+) = 4m { (0m ) }, 5 (3) S 4m + 3 = 3 4(m+) m 6 (m+) = 4m { (60m 6) }, 5 (4) S 4m + 4 = 5 m4m (m+) 6 5 = 4(m ) { (5m 8) }. 5 Proof : The first four terms of the sequence { S n } are : n= S =, S =, S 3 = 0, S 4 = 0. () The sum of the first (4m + ) terms of the sequence, S 4m +, can be written as S 4m + = SBDNS() + SBDNS() + + SBDNS(4m + ) = SBDNS() + k=,3,...,(m ) [SBDNS(k) + SBDNS(k + ) + SBDNS(k + ) + SBDNS(k + 3)], so that, by virtue of (), S 4m + = SBDNS() + 3 ( ) (6k + 3) k=,3,...,(m ) k + (k ) (k ) (k ) = SBDNS() k + 3. k=,3,...,(m ) k=,3,...,(m ) Now, appealing to Lemma.3.. (with d = ) and Lemma.3.. (with y = 4 ), we get 4m 4m S 4m + = [ m 34 ( )] + 3 ( 4m ), which now gives the desired result after some algebraic manipulations. () Since S 4m + = S 4m + + SBDNS(4m + ), from part () above, together with Corollary.., we get S 4m + = [ 3 5 m(m+) m 6 5 ] (4m + 3)4m, which simplifies to the desired expression for S 4m +. (3) Since S 4m + 3 = S 4m + + SBDNS(4m + 3) = [ 4m+3 m (m+) ] (4m + 4)4m+, we get the desired expression for S 4m + 3 after simplifications. (4) Since S 4m + 4 = S 4m SBDNS(4m + 4) = [ 3 4(m+) m 6 (m+) ] (4m + 5)4m+, the result follows after some simple algebraic simplifications. The case when m = 0 can easily be verified. Hence, the proof is complete.

51 50 Wandering in the World of Smarandache umbers.3. Smarandache Bisymmetric Arithmetic Determinant Sequence We consider the Smarandache bisymmetric arithmetic determinant sequence, { SBADS(n) }, n = for which the n th term is given by Theorem... Let { S n } be the sequence of n th partial n= sums of the sequence { SBADS(n) } n =, so that S n n = SBADS(k), n. k= An explicit expression for the sequence { S n } n= so that Theorem.3.. : For any integer m 0, is given below. () m(d + ) a(d + ) d(4d 4d ) S 4m+ = (d) + d (d) 4d + 4d + (4d + ) 4m+ 4m d (4d 4d ) a(d ) +, (4d + ) 4d + 3 () m(d ) a(d ) d(4d + 3d ) S 4m+ = (d) + (d) 4d + 4d + (4d + ) 4m+3 4m+ d (4d 4d ) a(d ) +, (4d + ) 4d + 3 (3) m(d + ) a(d + ) d(6d + 4d + 8d + 3) S 4m+ 3 = (d) d + (d) 4d + 4d + (4d + ) 4(m+) 4m+ d (4d 4d ) a(d ) +, (4d + ) 4d + 3 (4) m(d ) a(d ) d(d 4d + 5d ) S 4m+ 4 = (d) + + (d) 4d + 4d + (4d + ) d (4d 4d ) a(d ) +. (4d + ) 4d + Proof : From Theorem.., for any integer k, 4m+5 4(m+) k d k SBADS(k) + SBADS(k + ) = ( ) a(d ) d(d )k (d), k+ d k+ SBADS(k + ) + SBADS(k + 3) = ( ) a(d ) (4d ) d(d )k (d), SBADS(k) + SBADS(k + ) + SBADS(k + ) + SBADS(k + 3) k+ 3 (k ) = ( ) d a(d + )(4d ) + d(6d + 4d + ) + d(d + )(4d )k (d). (i)

52 5 Chapter 4 : : Smarandache The Pseudo Smarandache Determinant Sequences Function () Since S 4m + can be written as S 4m + = SBADS() + SBADS() + + SBADS(4m + ) = SBADS() + [SBADS(k) + SBADS(k + ) + SBADS(k + ) + SBADS(k + 3)], k=,3,...,(m ) by virtue of (i), Lemma.3.. and Lemma.3.. (with y = (d) 4 ), we get S4m+ = a + d a(d + )(4d ) + d(6d + 4d + ) ( ) (d) 3 k+ (k ) + d (d + )(4d ) ( ) k(d) k+ (k ) k=,3,...,(m ) k=,3,...,(m ) 4m 3 (d) a d a(d )(4d ) d(6d 4d ) (d) 4 = m 4m (d) + 4m + d (d + )(4d ) (d) [(d) ] 4 4 (d) [(d) ] 3 a(d + ) d(6d + 4d + ) m(d + ) = a d + (d) 4 4d (d) + + 4d + 4m+ a(d + ) d(6d + 4d + ) d [(d) + ] + + 4d (d) + (d )(4d + ) Now, collecting together the coefficients of (d) 4m and the constant terms, we get 3 4 4m 4m d (d) [(d) ]. 4 coefficient of (d) 4m is constant term is d 4d + (d) (d )(4d + ) 3 4 a(d + ) d(6d + 4d + ) d(6d + ) + 4 a(d + ) d(4d 4d ) = d, 4d + (4d + ) ad(d + ) 3 d (6d + 4d + ) 4 d (6d + ) 4 a + 4d + (d) (d )(4d + ) d (4d 4d ) a(d ). = (4d + ) 4d + Hence, finally, we get the desired expression for S 4m +. () Since S 4m + = S 4m + + SBADS(4m + ), by virtue Theorem.., 4m S 4m + = S 4m + + SBADS(4m + ) 4m + = S 4m+ (a + + d)(d). Now, using part (), and noting that coefficient of m(d) 4m+ is d + = d d, 4d + 4d +

53 5 Wandering in the World of Smarandache umbers coefficient of (d) 4m is a(d + ) d(4d 4d ) d d(a d + ) 4d (4d ) a(d ) d(4d + 3d ) = 4d +, 4d + (4d + ) we get the desired expression for S 4m +. 4m (3) Since S 4m + 3 = S 4m + + SBADS(4m + 3) 4m + = S 4m+ (a + + d)(d), using part (), we get the desired expression for S 4m + 3, noting that coefficient of m(d) 4m+3 is d = d d +, 4d + 4d + coefficient of (d) 4m+ is 3 a(d ) d(4d + 3d ) + (a + d) 4d + (4d + ) 3 a(d + ) d(6d + 4d + 8d + 3) = d +. 4d + (4d + ) 4m 3 (4) Since S 4m + 4 = S 4m SBADS(4m + 4) 4m 3 + = S 4m+ 3 + (a + + d)(d), by part (3), coefficient of m(d) 4m+4 is d d(d ) + + =, 4d + 4d + coefficient of (d) 4m+ is 3 a(d + ) d(6d + 4d + 8d + 3) d d(a d) 4d (4d ) + + = + 4d + (4d + ) 3 a(d ) d(d 4d + 5d ) 4d, and we get the desired expression for S 4m + 4. Now, corresponding to m = 0, the expressions for S, S, S 3 and S 4 are given as follows : S = a, S = a(d ) d, S 3 = a(4d + d ) d (4d + ), S 4 = a(8d 3 4d d + ) + d (d 4d ). All these complete the proof of the theorem. References [] Amarnath Murthy, Smarandache Determinant Sequences, Smarandache otions Journal,, 75 78, 00. [] Majumdar, A.A.K., On Some Smarandache Determinant Sequences, Scientia Magna, 4(), 80 95, 008. [3] Majumdar, A.A.K., The Smarandache Bisymmetric Arithmetic Determinant Sequence, Scientia Magna, 4(4), 6 67, 008.

54 53 Chapter 4 : The Pseudo Smarandache Function Chapter 3 The Smarandache Function Possibly the most widely studied arithmetical function of Smarandache type is the celebrated Smarandache function, introduced by Smarandache [] himself. In this chapter, we study some of the properties of the Smarandache function, denoted by S(n), as well as its several generalizations. We start with the formal definition of the Smarandache function. Throughout this chapter + and the following ones, we shall denote by Z is the set of all positive integers. Definition 3. : For any integer n, the Smarandache function, S(n), is the smallest positive integer m such that... m m! is divisible by n. That is, S(n) = min{ m : m Z +, n m! }; n. Thus, for examples, (a) S() =, since! =, (b) S(4) = 4, since 4 divides 4!, and 4 does not divide any of!,! and 3!, (c) S(6) = 3, since 6 3! = 6, and 6 divides neither! nor!. Since the function F(m) m!, m Z +, is strictly increasing in m, it follows that the function S(n) is well-defined. By Definition 3., for any integer n, S(n) = m 0 if and only if the following two conditions are satisfied : () n divides m 0!, () n does not divide (m 0 )!. Note that, our definition of S(n) differs slightly from the traditional one in that both the + domain and range of S(n) is Z, while the traditional definition allows the range of S(n) to include the number 0. This makes the difference in the value of S() : According to the traditional definition, S() = 0, whereas by Definition 3., S() =. We prefer Definition 3., like several other researchers, on the following points : () generally, the other arithmetical functions in the Theory of Numbers, have both their + domains and ranges as Z, and so, Definition 3. is in consistence with the traditional approach, () the usual definition of m! is m! =...m for any integer m, and we define 0! = only for the sake of convenience, particularly, in combinatorics. Moreover, by allowing the range to include 0, possibly we gain no extra advantage; rather on the contrary, some of the formulas might become complicated (see Corollary 3..3), (3) according to the classical definition of S(n), though S() is defined, S k () is not defined when k, where S k (.) is the k fold composition of S(.) with itself.

55 54 Wandering in the World of Smarandache umbers Several researchers, including Ashbacher [] and Sandor [3], have studied some of the elementary properties satisfied by S(n). These are summarized in Lemmas The proofs of some of these results depend on Lemmas , some of which are almost trivial. Lemma 3. : Let p be a prime. Then, p divides ab if and only if at least one of a and b is divisible by p. Lemma 3. : If m divides a! then m divides b! for any integer b > a. Proof : Since b! can be written as b! = a!(a + )(a + ) (b), it follows that any integer m dividing a! also divides b!. Lemma 3.3 : For any integer k 3, () k divides k!, () 4k divides (k)!. Proof : The proof of part () is by induction on k. Since 6 divides 3! = 6, we see that the result is true for k = 3. () We assume that the result is true for some integer k, that is, we assume that k k! for some integer k > 3. Then, k(k + ) k! (k + ) = (k + )! (k + ) (k + )!, which shows that the result is true for k + as well. () From part () above, (k).(k) k! (k) (4k ) (k)!. Lemma 3.4 : (n + ) divides (n + )! for any integer n. Proof : If n, then n + > n +, and so the result follows by observing that (n + ) divides (n + )! = (n + )! (n + 3) (n).(n + ). Since 3 divides 3!, the result is true for n = as well. Lemma 3.5 : For any integers a, b, a! b! divides (a + b)!. Proof : Since a + b (a + b)! ( a ) =, we get the desired result. a! b! Lemma 3.6 : For any n Z +, S(n). Moreover, () S(n) = if and only if n =, () S(n) = if and only if n =. Proof : In either case, the proof of the if part is trivial. So, we prove the only if part. () Let S(n) = for some n Z +. Then, by definition, n!, and hence, n =. () Let S(n) = for some n Z +. Then, by definition, n!, and hence, n =. Corollary 3. : S(n) 3 for any integer n 3. The following lemma gives an upper bound of S(n). Lemma 3.7 : S(n) n for all n. Proof : Since n n! for any integer n, the result follows.

56 55 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function S(n) Corollary 3. : 0 < for any integer n. n Proof : Evident from Lemma 3.6 and Lemma 3.7. The lower bound of S(n), given in Corollary 3., can be improved in the case when n is a composite number. This is given in the following Lemma 3.8 : For any composite number n 4, S(n) max{ S(d) : d n }. Proof : Let S(n) = m 0 for some m 0 Z +. By definition, n m 0! (and m 0 is the smallest integer with this property). Then, clearly any divisor d of n also divides m 0!, so that S(d) m 0. Lemma 3.9 : () S(n) n for any integer n 3; () S(4n ) n for any integer n. Proof : follows immediately from Lemma 3.3 as shown below : () (n) n! S(n) n, () (4n ) (n)! S(4n ) n. Lemma 3.0 : For any integers n, n, S(n n ) S(n ) + S(n ). Proof : Let S(n ) = m, S(n ) = m for some integers m, m. Then, n m!, n m! n n m! m!. Now, appealing to Lemma 3.5, so that n n m! m! n n (m + m )!, S(n n ) m + m = S(n ) + S(n ). Lemma 3. : For any integers n, n,, n k, S(n n n k ) S(n ) + S(n ) + + S(n k ). Proof : The proof is by induction on k. By Lemma 3.0, the result is true when k =. So, we assume that the result is true for some integer k, that is, the induction hypothesis is S(n n n k ) S(n ) + S(n ) + + S(n k ). Then, Lemma 3.0, together with the induction hypothesis, gives S((n n n k ) n k+ ) S(n n n k ) + S(n k+ ) [S(n ) + S(n ) + + S(n k )] + S(n k+ ) = S(n ) + S(n ) + + S(n k ) + S(n k+ ), which shows that the result is true for k + as well.

57 56 Wandering in the World of Smarandache umbers Lemma 3. : For any integer m, S(m!) = m. Proof : Since m! m! for any integer m, the result is obtained. Lemma 3. shows that, for any integer m 0, there is at least one integer n, namely, n = m 0!, such that S(n) = m 0 ; moreover, such an n is the maximum number satisfying the equation S(n) = m 0 (since any n > m 0! does not divide m 0!). Lemma 3. leads to various interesting inequalities, as has been derived by Sandor [3]. These are given in the following corollary. Corollary 3.3 : The following inequalities hold : () for any integers n and k, S(n k ) ks(n); () for any integers n and r, r k= k r S(n ) r![s(n)] ; (3) for any integers n and k, S((n!) k ) kn; (4) for any integers n, n,, n r, and k, k,, k r, k k k S( (n!) (n!)... (n!) ) k n + k n k n ; r r r r (5) if n and n are two integers such that n divides n, then n S(n ) S(n ) S( ). n Proof : The proof is given below : () Choosing n = n = = n k = n in Lemma 3., we get the desired result. () From part () with k =,, 3,, r successively, we get S(n) S(n), S(n ) S(n), S(n 3 ) 3S(n),, S(n r ) rs(n). Now, multiplying all these r number of inequalities side-wise, we get the result desired. (3) The inequality follows from part () above, together with Lemma 3. : S((n!) k ) ks(n!) = kn. (4) From part (), we get k k k k k k S( (n!) (n!)... (n!) ) S( (n!) ) S( (n!) )... S( (n!) ). r r r r Now, appealing to part (3), we get the desired result. so that (5) Let n = kn for some integer k. Then, by Lemma 3.0, S(n ) = S(k n ) S(k) + S(n ), S(n ) S(n ) S(k) = S( ). n All these complete the proof of the lemma. n

58 57 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function Lemma 3.3 : p divides S(p k ) for any prime p, and any integer k. Proof : Let S(p k ) = m for some integer m, so that p k divides m!. Then, p must divide m, for otherwise p does not divide m p k divides (m )! S(p k ) m, contradicting the hypothesis. Corollary 3.4 : For any prime p, and any integer k, S(p k ) = αp for some integer α. Proof : follows immediately from Lemma 3.3. Lemma 3.4 : For any integer n and integers k, r with k r, S(n k ) S(n r ). Proof : Let S(n k ) = m 0 for some integer m 0. Then, n r n k = n r.n k r, n k m 0! n r m 0! S(n r ) m 0. Lemma 3.5 : For any integers n and n with n n, S(n k ) S(n k ) for any k. Proof : Let, for any integer k fixed, S(n k ) = m 0 for some integer m 0. Then, n k m 0! n k m 0! S(n k ) m 0. Lemma 3.6 : For any prime p, S(p α+β ) S(p α ) S(p β ) for any integers α, β. Proof : Let, S(p α+β ) = m 0 for some integer m 0. Furthermore, let S(p α ) = m, S(p β ) = m for some integer m, m, where, without loss of generality, we may assume that m m. We want to show that m 0 m m. Let, on the contrary, m 0 > m m for some m 0, m and m. Then, using Lemma 3.0, we get the following chain of inequalities : mm < m0 m + m m m <, which is a contradiction to Corollary 3.4. This contradiction establishes the lemma. In 3., we give some simple explicit expressions for S(n). Two very important results in this respect are given in Theorem 3.. and Theorem 3... Some generalizations of the Smarandache function are given in is devoted to some miscellaneous topics. We conclude this chapter with some remarks about the properties of S(n) in the final 3.4.

59 58 Wandering in the World of Smarandache umbers 3. Some Explicit Expressions for S(n) Let and let In this section, we give the explicit expressions for S(n) and their consequences. First, we state and prove the following theorem, due to Smarandache []. Theorem 3.. : Let α α α α n = p p p... p k k be the representation of n in terms of its distinct prime factors p, p, p,, p k (not necessarily ordered). Then, α α αk { α p ), } S(n) = max S( S(p ), S(p ),..., S(p ). Proof : For definiteness, let k k { } α α α α α S(p ) = max S( p ), S(p ), S(p ),..., S(p ). S(p α ) = m 0 for some integer m0, k α S(p ) = i i mi for some integer mi, i k. Then, by definition, m 0 is the minimum integer such that p α m 0!, () and m i ( i k) is the minimum integer such that αi p i m i! ( i k). () Now, m0 m i for all i k m i! m 0! for all i k. Therefore, from (), α p m! for all i k. i i 0 α α Now, since (p, p ) = for i j, it follows that i i j j α α αk p p... p k m 0!, and by the same reasoning, n p α α α α = p p... p m!. k k 0 Since, by (), m 0 is the minimum integer with this property, it follows that α S(n) = m0 = S(p ), which we intended to prove.

60 59 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function From Theorem 3.., we see that, in order to calculate k p α α α α S(n) = S( p p... p k ), we have to know the values of α α α α k S( p ), S(p ), S(p ),..., S(p k ), which are not available till now. However, we have expressions in some particular cases. Lemma 3.. : For any prime p, S(p) = p. Proof : Let S(n) = m 0 for some m 0 Z +, so that, p m 0!. By Lemma 3., one of the factors of m 0! must be divisible by p. Now, since p p! but p does not divide (p )!, the result follows. Lemma 3.. : Let p, p,, p k be k distinct primes. Then, S(p p p k ) = max{ p, p,, p k }. Proof : By Theorem 3.., S(p p p k ) = max{ S(p ), S(p ),, S(p k )}. Now, we get the desired result by virtue of Lemma 3... Lemma 3..3 : Let n and n be two integers with (n, n ) =. Then, S(n n ) = max{ S(n ), S(n )}. Proof : Let the representations of n and n in terms of their prime factors be α α α αr n = p p p... p, r β β βt n = q q... q t, where the primes p, p, p,, p r as well as the primes q, q,, q t are all distinct. Then, α α α αr β β βt n n = p p p... p r q q... q t. For definiteness, let α αr β t { } S(p ) max S( p ), S(p ),..., S(p ), S(q ),..., S(q ). Then, by Theorem 3.., α = β α r t S(n n ) = S(p ) = S(n ). α An immediate consequence of Lemma 3..3 is the following. Corollary 3.. : The function S(n) is not multiplicative.

61 60 Wandering in the World of Smarandache umbers Theorem 3.. : Let n, n,, n k be k integers with (n, n,, n k ) =. Then, S(n n n k ) = max{ S(n ), S(n ),, S(n k )}. Proof : The proof is by induction on k. The result is true for k =, by virtue of Lemma So, we assume the validity of the result for some k. To prove the validity of the result for k +, let n, n,, n k, n k + be k + integers with (n, n,, n k, n k + ) =. Then, by Lemma 3..3, together with the induction hypothesis, we get S(n n n k n k + ) = max{s(n n n k ), S(n k + )} Hence the theorem. = max{s(n ), S(n ),, S(n k ), S(n k + )}. From Lemma 3.., S(p) = p for any prime p. (3..) We now want to find expressions for S(p k ) for k. To do so, first note that, from (3..), p divides p! but p does not divide (p )!. Therefore, p (p)! = p!(p + )(p + )... [p + (p )](p), (3..) and clearly, p is the minimum integer such that (p)! is divisible by p. Again, 3 p (3p)! = (p)!(p + )(p + )... [p + (p )](3p), (3..3) and 3p is the minimum integer such that (3p)! is divisible by p 3. Continuing the argument p times, we see that (p )p is the minimum integer such that p p [(p )p]! = [(p )p]![(p )p + ][(p )p + ]... [(p )p + p] = [(p )p]![(p )p + ][(p )p + ]... [(p )p]. (3..4) And continuing one more time, p p (p )! = [(p )p]![(p )p + ][(p )p + ]... [(p )p + p] = [(p )p]![(p )p + ][(p )p + ]... (p ). (3..5) From (3..5), we see that p p + (p )!. (3..6) Summarizing the results in (3..) (3..6), we have the following lemma. Lemma 3..4 : For any prime p, () S(p k ) = kp, if k p, () S(p p+ ) = p. By an analysis similar to that leading to Lemma 3..4, we have the following lemma.

62 6 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function Lemma 3..5 : For any prime p, () S(p p+k+ ) = p + kp, if k p, () S(p (p+) ) = p. Lemma 3..6 : S(n n ) S(n ) S(n ) for any integers n, n. Proof : Let the representations of n and n in terms of their prime factors be α α αr β β βs γ γ γr δ δ δk n = p p... p q q... q, n = p p... p t t... t, r s r k where the primes p, p, p,, p r, the primes q, q,, q s as well as the primes t, t,, t k are all distinct, with α, α,, α r 0; β, β,, β s 0 with at least one β ; γ, γ,, γ r 0; δ, δ,, δ k 0 with at least one δ. Then, α + γ α + γ α r + γr β β βs δ δ δk n n = p p... p q q... q t t... t. r s k Therefore, by Theorem 3.., α + γ α + γ β βs δ δ { } r r k S(n n ) = max S(p ),..., S(p ), S(q ),..., S(q ) S(t ),..., S(t ). r s k Now, using Lemma 3.6, we get α γ α γ β βs δ δ { } r r k γ γr δ δ S(n ) max { S(p ),..., S(p ),..., S(t ),..., S(t } k ) S(n n ) max S(p )S(p ),..., S(p )S(p ), S(q ),..., S(q ), S(t ),..., S(t ) r r s k r = S(n )S(n ), αi where we have used the facts that S(n ), and S( p i ) S(n ) for all i r. To complete the proof, note that, for any prime p, S(np) = max{ S(n), p } p S(n). S(nn ) S(n ) S(n ) Corollary 3.. : For any integers n, n, min,. nn n n Proof : follows from the following chain of inequalities (using Lemma 3.7) : S(nn ) S(n ) S(n ) S(n ) S(nn ) S(n ) S(n ) S(n ).,.. n n n n n n n n n n Lemma 3..7 : For any integers a, b, a b a + b. Proof : Letting a = + A, b = + B where A, B 0 are integers, we see that ab = ( + A)( + B) a + b = ( + A)( + B) if and only if A + B + AB 0, which is true for any integers A, B 0, with strict equality sign only when A = 0 = B. Corollary 3..3 : For any integers n, n, () S(n n ) S(n ) + S(n ) S(n ) S(n ), if both n, n, () S(n n ) S(n ) S(n ) S(n ) + S(n ), if n = or n =. Proof : follows from Lemma 3.0 and Lemma 3..6, together with Lemma k

63 6 Wandering in the World of Smarandache umbers 3. Some Generalizations Since its introduction by Smarandache [], the Smarandache function has seen several variations. This section is devoted to two such generalizations, and are given in the following two subsections. 3.. The Smarandache Dual Function S * (n) The Smarandache dual function, denoted by S * (n), introduced by Sandor [3], is defined as follows. Definition 3... : For any integer n, the Smarandache dual function, S * (n), is * + { } S (n) = max m : m Z, m! n. Let S * (n) = m 0 for some integer m 0. Then, by Definition 3..., the following two conditions are satisfied : () m 0! divides n, () n is not divisible by m! for any m < m 0. Lemma 3... : For any integer n, S * (n) S(n) n. Proof : Let S(n) = m 0, S * (n) = m for some integers m 0, m. Then, by definitions of S * (n) and S(n), m! n, n m 0! m! m 0! m0 m. Clearly, divides any integer n S * (n). Finally, the right-most part of the inequality is due to Lemma 3.7. All these complete the proof of the lemma. Lemma 3... : For any prime p, and any integer n p, S * (n! + (p )!) = p. Proof : Writing n! = (p )! p (p + ) n, we see that (p )! divides n! + (p )!; moreover, (p )! is the maximum integer that divides n! + (p )!, since p! divides n! but p! does not divide (p )!. Corollary 3... : The equation S * (n) = p has infinite number of solutions, where p is a prime. Proof : By Lemma 3..., n = k! + (p )! is a solution for any k p.

64 63 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function Lemma : For any integers a, b, S * (ab) max{ S * (a), S * (b)}. Proof : Let S * (a) = m 0 for some integer m 0. Then, by definition, m 0! a m 0! ab S * (ab) m 0 = S * (a). Lemma : S * (n(n ) (n a + )) a for any integers n and a with n a. Proof : follows by virtue of the fact that n! n a! n(n )... (n a + )! = = a!. (n a)! a Lemma : For any integer n, n, if n 3 is a prime S *((n)! (n + )!) = + + n + 3, if n + 3 is not a prime Proof : If n + 3 = p is a prime, then clearly, (n + ) divides (n)!(n + )!, but n + 3 = p does not divide (n)!(n + )! = (p 3)! (p )!. Hence, in this case, S * ((n)!(n + )!) = n +. Next, let n + 3 be not prime, so that n 3. We want to show that, in this case n + 3 divides (n)!. () If n = 3, then () is satisfied, since n + 3 = 3 divides (n)! = 6!. So, let n 4. Case : When n + 3 = ab for some odd integers a and b with b > a 3. In this case, b < n, for otherwise b > n ab 3b > 3n > n + 3, a contradiction. Hence, a and b are distinct integers less than n, and hence, () is satisfied. Case : When n + 3 = a for some odd integer (so that n, a 5). Here, a 3 > a, so that n + 3 = a divides a!(a + )(a + ) (a) (a 3) = (n)!. Now, by virtue of (), (n + 3)! divides (n)!(n + )!, completing the proof. Lemma : For any integer n, S * ((n + )! (n + 3)!) (n + ). Proof : By Lemma 3.4, for any integer n, (n + ) (n + )! [(n + )]! = (n + 3)! [(n + )] (n + )! (n + 3)!. This gives the desired result. Remark 3... : As has been pointed out by Sandor [3], in certain cases, Lemma holds true with equality signs. Some examples are given below : () S * (3! 5!) = 6, () S * (5! 7!) = 8, (3) S * (7! 9!) = 0, (4) S * (9!!) =, (5) S * (3! 5!) = 6, (6) S * (5! 7!) = 8, (7) S * (9!!) =.

65 64 Wandering in the World of Smarandache umbers 3.. The Additive Analogues of S(n) and S * (n) The function, called the additive analogue of S(n), to be denoted by S S (n), has been introduced by Sandor [3]. Definition 3... : For any real number x (, ), S + { } S (x) = min m : m Z, x m!. The dual of S S (x), to be denoted by S S* (x), is defined as follows (Sandor [3]). Definition 3... : For any real number x [, ), S S* (x) = max { m : m +, m! x} Z. From Definition 3... and Definition 3..., it is clear that S S (x) = k if and only if x ((k )!, k!] (for any integer k ), S S* (x) = k if and only if x [(k )!, k!) (for any integer k ). We then have the following results, due to Sandor [3]. Lemma 3... : For any real number x, and any integer k, S S* (x) +, if x (k!, (k + )!) S S (x) = S S* (x), if x = (k + )! Lemma 3... : S ln x S* (x) as x. ln ln x + If the domain is restricted to the set of positive integers, Z, we get the function S S (n), which assume the value k ( ), repeated k! (k )! times. Similarly, we get the function S S* (n), + with domain Z, which can take the value k ( ), repeated k! (k )! times. Lemma : For any integer n, S(n) S S* (n). Proof : Let S(n) = m 0, S S* (n) = m for some integers m 0 and m. Then, by definitions of S(n) and S S* (n), m 0! n and n m! m 0! m!. Thus, m 0 m, which we intended to prove. The lemma below is concerned with the convergence of series involving S S* (n), and is due to Sandor [3]. Lemma : The series is convergent if and only if s >. s n= n[s S* (n)]

66 65 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function 3.3 Miscellaneous Topics In this section, we give some properties of the Smarandache function S(n). The following three lemmas are due to Ashbacher []. Lemma 3.3. : S(n + ) S(n) is unbounded. Proof : Let n + = p, where p 7 is a prime. Then, by Lemma 3.. and Lemma 3.9, S(n + ) = S(p) = p, S(n) = S(p ) p. Therefore, S(n + ) S(n) = S(n + ) S(n) p p p + =, which can be made arbitrarily large (by choosing p large and larger). Lemma 3.3. : For any prime p fixed, and any integer k, (with equality sign only when p = and k = ). Proof : By Corollary 3.4, S(p k ) = rp for some integer r. Now, since p k divides (rp)!, it follows that Therefore, k+ k+ k+ p (rp)!(r + )p p [(r + )p]! S(p ) (r + )p. k+ k S(p ) (r )p S(p ) rp + = r + < = p p p p p k+ k+ k k k k+ k S(p ) S(p ) < k+ k, if and only if r p > r +, which is true for any p and any r, with the exception when p = and r =. In the latter case, the relationship holds with equality sign, as is verified below : S( ) S() = 4 = =. 4 Hence, the proof is complete. Lemma : For any prime p fixed, n n+ S(p ) lim = 0. n+ p n S(p ) n p n= Proof : By Lemma 3.3., the sequence is strictly decreasing for any prime p fixed. Since the sequence is bounded (by Corollary 3.), it follows that it is convergent. Now, for p fixed, given any ε > 0, however small, we can find an integer N >, such that n S(p ) < ε whenever n N. n p This establishes the lemma. p p

67 66 Wandering in the World of Smarandache umbers Corollary 3.3. : Given any real number ε > 0, however small, it is possible to find an integer n > such that S(n). n < ε Proof : Given any real number ε > 0, we choose a prime p. The proof of Lemma shows that there is an integer such that n S(p ) p n < ε whenever n N. Therefore, choosing n = p n, where n is an integer such that n N, we see that the inequality in the corollary is satisfied. Example 3.3. : For p =, the successive values of n S( ) n for n =,,, are,, 4 =, 6 = 3, 8 =, 8 =, 8 =, 0 = 5, = 3, Thus, for example, if ε = 0.05, then choosing n = 8, we get S(n) 8 S( ) 5 8 = = < 0.05 = ε. n 8 Example 3.3. : An alternative proof of Corollary 3.3. is given in Ashbacher []. To find n such that S(n) 0.05, n = ε < we choose a prime q such that = < ε = The choice q = 3 serves the purpose. Now, S(q) q let p be a prime such that p > q. Then, letting n = pq, we get S(n) S(pq) p = = = < ε = n pq pq q The following result is due to Stuparu and Sharpe [4], who mentioned it without any proof. Lemma : Any solution of the equation S(n) = p, where p is a given prime, is of the form n = kp, where k is an integer such that k divides (p )!. Proof : Clearly, n = kp, where k divides (p )!, divides p!. Moreover, p is the minimum integer such that p! is divisible by kp. Hence, S(kp) = p, establishing the lemma.

68 67 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function The proof of Lemma shows that, S(n) = p (p is a prime) has d((p )!) solutions. Example : Some particular cases of Lemma are given below : () The equation S(n) = has only one solution, namely, n =. () The equation S(n) = 3 has two solutions, namely, n = 3, 6. (3) The equation S(n) = 5 has 4! = 8 solutions, namely, n = 5k, where k is a divisor of 4!. The eight solutions are n = 5, 0, 5, 0, 30, 40, 60, 0. Lemma : The only solutions of the equation S(n) = n are () n =, () n = 4, (3) n = p, where p is a prime. Proof : We consider separately the cases when n is odd and n is even. Case () : When n is even, say, n = k for some integer k. In this case, n = k does not divide n ( )! = k! k =,, where the last implication follows by virtue of Lemma 3.9. Thus, n = and n = 4 are the only even integers that satisfy the equation S(n) = n. Case () : When n is odd with n 5. First, we consider the case when n is composite. Let n = ab for some integers a and b with b > a. Then, clearly n = ab b! = a!(a + )(a + )... (b) S(n) b < n. Next, let n = a for some integer a 3. In this case, n = a (a)! = a!(a + )(a + )... (a) S(n) a < a < n. Thus, no composite odd integer can be solution of the equation S(n) = n. And consequently, any odd integral solution of S(n) = n must be an odd prime. From Lemma 3.., any odd prime is a solution the equation S(n) = n. All these complete the proof of the lemma. Corollary 3.3. : The equation S(kn) = n has an infinite number of solutions, for any integer k fixed. Proof : Given the integer k, let p be a prime such that p > k. Then, S(kp) = max{ S(k), S(p) } = max{ S(k), p } = p. Thus, n = p is a solution of the equation S(kn) = n. Since there is an infinite number of primes of the prescribed form, the result follows. Remark 3.3. : Since S(n) n =, it follows that the equation S(n) = k has no solution n n n if k.

69 68 Wandering in the World of Smarandache umbers The following lemma is due to Ashbacher [5]. Lemma : The equation S(n) S(n + ) = n has no solution. Proof : Clearly, neither n nor n + can be a prime. So, let α α α α n p p p... p, β β β r t = r n + = q q q... q t be the representations of n and n + in terms of their prime factors, where the primes p, p, p,, p r and the primes q, q, q,, q t are all distinct. Moreover, let for definiteness S(n) S(p α β = ), S(n + ) = S(q ). Then, the equation S(n)S(n + ) = n takes the form α α α α α β S(p )S(q ) = p p p... p. r r β By Lemma 3.3, S(q ) is divisible by q. Thus, the l.h.s. of () is divisible by q, but q does not divide the r.h.s. of (). This contradiction establishes the lemma. Lemma proves that the equation S(n) S(n + ) = n has no solution. However, the situation is different if we consider the equation S(n) S(n + ) = kn, k. Proceeding as in the proof of Lemma 3.3.6, the above equation becomes α β S(p )S(q ) α α α α = kp p p... p, r r and the argument employed in the proof of Lemma is no longer valid. For example, when k =, besides the trivial solution n =, another solution is n = 5 (with S(5) = 5, S(6) = 6). Again, n =, 5, 8 are three solutions of S(n) S(n + ) = 3n (with S(8) = 4, S(9) = 6). Corresponding to k = 4, besides the prime solutions n = 3, 7,, 3, a composite solution is n = 559 (with S(559) = 853, S(560) = ). For k = 5, in addition to the two prime solutions n = 9, 59, four composite solutions are n = 74, 49, 674, 574 (with S(74) = 37, S(75) = 0, S(49) = 83, S(50) = 5, S(674) = 337, S(675) = 0, S(574) = 787, S(575) = 0). For k = 6, eight prime solutions are n = 7, 47, 7, 79, 89, 79, 39, 359, and n = 4 is a composite solution (with S(4) = 607, S(5) = ). Corresponding to k = 7, n = 3, 4, 83, 39, 67, 5 are six prime solutions and n = 46, 538, 734, 3, 37 are five prime solutions (with S(46) = 73, S(47) = 4, S(538) = 69, S(539) = 4, S(734) = 367, S(735) = 4, S(3) = 66, S(33) = 4, S(37) = 457, S(37) = ). For k = 9, n = 07, 69, 86 are three solutions, only the third one being composite (with S(86) = 093, S(87) = 8). For k = 0, in addition to the prime solutions n = 49, 99, 349, a composite solution is n = 874 (with S(874) = 937, S(875) = 0). For k =, six prime solutions are n = 43, 09, 3, 97, 63, 307, and three composite solutions are n = 36, 84, 66 (with S(36) = 8, S(363) =, S(84) = 907, S(85) =, S(66) = 887, S(66) = 33). For k = 3, five prime solutions are n = 03, 8, 33, 3, 389, and one composite solution is n = 858 (with S(858) = 99, S(859) = 6). Corresponding to k = 7, four solutions are n = 67, 7, 373, 866 (with S(866) = 433, S(867) = 34). For k = 9, n = 37, 3, 5, 7, 379, 08 are solutions (with S(08) = 54, S(083) = 38), and for k = 37, n = 73, 406 are solutions (with S(406) = 053, S(407) = 74). β ()

70 69 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function The problem of convergence of (infinite) series involving S(n) and its different variants has been treated by several researchers. The following two lemmas are due to Ashbacher []. Lemma : The series is divergent. S(n) n= Proof : Since (by Lemma 3.7) S(n) n for all n, it follows that >, S(n) n n= n= since the harmonic series n= n is divergent. Lemma : The series Proof : By Lemma 3.., n= n n= is divergent, where p n is the n th prime. S(p ) n= = >, S(p ) p n n since the series n= p is divergent (Hardy and Wright [6], Theorem 9). n Lemma : The series is divergent. S(n) n= Proof : Letting p n be the n th prime, by Lemma 3.., S(p n ) = max{, p n } = p n, and so > = >. S(n) S(p ) p n= n= n n= n Lemma : Both the series S(n) and n n= Proof : Let n = p n, n (where p n is the n th prime). Then, S(n) S(p n ) pn n pn = = =, = =. n p p S(n) S(p ) n n n n are divergent. S(n) n= Therefore, both the series S(p ) p n n and n= p are divergent. Hence, n n= S(p n ) S(n) S(p ) n n > >, n > >. n= n k= pn n= S(n) k= S(p n ) p

71 70 Wandering in the World of Smarandache umbers Next, we concentrate our attention to equations involving S(n) and other classical functions of Number Theory, such as the divisor function d(n) and the Euler phi function φ(n). These are given in Lemma Lemma To prove these results, we need the results below. Lemma 3.3. : Let n be an integer. Then, α () F( α) n is strictly increasing in α for any n fixed, with F() ; α + () α G( α) n is strictly increasing in α for any n 3 fixed, G(α) is strictly α ( α + ) increasing in α 3 when n = ; α (n + )n (3) H( α) α ( α + ) is strictly increasing in α for any n 3 fixed, with F() > ; H(α) is strictly increasing in α 3 for n =, with F() = = F(3) when n =. Proof : The results corresponding to n = in cases () and (3) can easily be verified. () Note that α+ α F( α + ) n > n F( α) α + α + if and only if n(α + ) > α +, which is true for any α and any n. () In this case, α α G( α + ) n > n G( α) ( α + )( α + ) α( α + ) if and only if nα > α + which is true for any α and any n 3. (3) Since α α (n + )n (n + )n H( α + ) > H( α) ( α + )( α + ) α( α + ) if and only if n α > α +, the result follows for any α and any n 3. Hence the proof is complete. Lemma 3.3. : Let n be an integer. Then, () for n, (a) () for n 3, (a) (3) for n 5, (a) (4) for n 7, (a) (5) for n, (a) α n for α 3, (b) α + α n 3 for α, (b) α + α n > for α, (b) α + α n > 3 for α, (b) α + (n + )n α( α + ) α for α, (b) α n > 3 for α 4, α + (c) α n > for α 3, (c) 4α + α n > for α 3, α( α + ) α n > for α, α( α + ) n(n + ) α( α + ) α for α. α n > for α 4, 3α + α n > for α 4, α( α + )

72 7 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function We now consider the problem of finding the solutions of the equation S(n) + d(n) = n. The problem has been considered by Asbacher [5]. Here, we follow a different approach. Lemma : The only solutions of the equation S(n) + d(n) = n. are n = 8, 9. Proof : Let, for some integer n, S(n) + d(n) = n. (3.3.) Clearly, n is not a prime, since for any prime p, S(n) + d(n) = p + > p = n. Thus, n must be a composite number. First, we consider some particular cases. Case : When n is of the form n = p α, p is a prime, (α ). In this case, the equation (3.3.), together with Corollary 3.3(), becomes p α = S(p α ) + d(p α ) pα + (α + ) = α(p + ) +. () When α =, from (), p p + 3, which is true only if p 3. Now, let p =, so that () takes the form α 3α +, which is absurd if α 4 (by Lemma 3.3.). Next, let p = 3 in () to get 3 α 4α +, which is impossible if α 3 (by Lemma 3.3.). Thus, in this case, the only possible candidates for (3.3.) to hold true are n =, 3, 3. Checking with these values of n, we see that S(4) + d(4) = = 7 > 4 = n, S(8) + d(8) = = 8 = n, S(9) + d(9) = = 9 = n. Thus, in this case, n = 8 and n = 9 are two solutions of the equation (3.3.). Case : When n is of the form n = p α, p 3 is a prime. In this case, by parts () and (3) of Lemma 3.3., p α > p α + (α + ) S(p α ) + d(p α ) if p = 3 and α, or if p 5 and α. Now, checking with n = 3 = 6, we see that S(6) + d(6) = = 7 > 6 = n. Thus, any n of the form n = p α cannot be a solution of the equation (3.3.) Now, to deal with the general case, let α α α α β n = p q p p... p k k be the representation of n in terms of its distinct prime factors p, q, p, p,, p k. Let, for definiteness, S(n) = S(p α ). (3.3.)

73 7 Wandering in the World of Smarandache umbers Now, by Theorem 0. (in Chapter 0), d(n) = (α + )(β + )(α + )(α + ) (α k + ). () Dividing throughout of (3.3.) by d(n), using () together with Lemma 3.7 (that S(p α ) p α ), and the fact that β +, α i + for all i k, we get the following chain of inequalities : α α α α β k α α p q p p p k p p , α + β + α + α + α k + α + ( β + )( α + )...( α k + ) α + so that, after rearrangement of terms, we get α α αk α β p q p p p k α + β + α + α + α k + Case 3 : When n is of the form n = p α q β, p and q are distinct primes. In this case, (3) takes the form α β p q. α + β + Here also, we consider some particular cases. First, let p =, q 3. α β Since q for any α (by Lemma 3.3.()), (4) gives, which is impossible if α + β + q = 3 and β, or if q 5 and β (by parts () and (3) of Lemma 3.3.). For α 4, (4) with Lemma 3.3.() give Next, let p = 3, q 3. When α =, so that if q 5 (any β ). α 3 = 3 >, α + (4) gives β q 3 <, β + β q <, β + (3) (4) which is impossible if q 3 and β. which is impossible if q = and β 3, or α β Since 3 q 3 for α (by Lemma 3.3.()), (4) gives 3, α + β + which is impossible if q = and β (with strict equality sign when q =, β = and α = ), or if q 5 and β. Thirdly, let p 5, q 5. α β p q In this case, since > for any α (by Lemma 3.3.(3)), (4) gives, α + < β + which is impossible if q = and β, or if q 3 and β. Thus, in this case, the only possible candidates for (3.3.) to hold true are n = 3., 3. 3, 3., p α. The case n = p α has been verified in Case. Checking with the other values of n, we see that

74 73 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function S() + d() = = 0 < = n, S(4) + d(4) = = < 4 = n, S(36) + d(36) = = 5 < 36 = n. Hence, there is no solution of (3.3.) of the form n = p α q β. Now, we confine our attention to (3) with k. From (3), we get α α α β k q p p p k..... <, β + α + α + α + k where, without loss of generality, we may assume that p i 5 for all i k. But then, by Lemma 3.3., we have a contradiction. In the proof of Lemma above, we proved more. They are summarized below. Corollary : Let d(n) be the divisor function. Then, () the equation S(n) + d(n) = n has the only solutions n = 8, 9; () the inequality S(n) + d(n) > n has the solutions n =, 4, 6, p; (3) S(n) + d(n) < n for any integer n, 4, 6, 8, 9, p. The following lemma is due to Jingping [7]. Lemma : The only solutions of the equation S(n) = φ(n) (3.3.3) are n =, 8, 9,. Proof : Clearly, the equation (3.3.3) admits no prime solution, since for any prime p, S(p) = p > p = φ(p). Thus, any solution n of (3.3.3) must be a composite number. So, let α β α α αk n = p q p p... p (3.3.) k be the representation of n in terms of its distinct prime factors p, q, p, p,, p k. Let, for definiteness, S(n) = S(p α ). Now, by Theorem 0.5 (in Chapter 0), φ(n) is multiplicative, so that (3.3.3) can be written as α α α p (q ) p α β k α (p ) φ φ(p p... p k ) α α αk p φ β (q ) φ (p p... p k ), p which is impossible if k, since in such a case α α α β k φ(q ) φ(p p... p ). k Thus, any solution of the equation (3.3.3) can be either of the two forms n = p α and n = p α q β. ()

75 74 Wandering in the World of Smarandache umbers We now consider these two possibilities below. Case : When n is of the form n = p α (α ). In this case, note that, if p =, then φ( α ) α > α S( α ) if α 5, and if p 3, α 3, then φ(p α ) p α (p ) p α > αp S(p α ). Thus, in this case, the only possible candidates for (3.3.3) to hold true are n =, 3, 4, 3. Checking with these values of n, we see that S(4) = 4 > = φ(4), S(8) = 4 = φ(8), S(6) = 6 < 8 = φ(8), S(9) = 6 = φ(9). Thus, in this case, there are two solutions, namely, n = 4, 9. Case : When n is of the form n = p α q β. In this case, from (), we get β p β φ(q ) φ (q ) = β =, p =, q = 3. p With n = 3. α, (3.3.3) takes the form φ(3. α ) α = S(3. α ) = max{s( α ), S(3)} = S(3. α ), with the solution α =. Hence, n = is the only solution in this case, with S() = 4 = φ(). All these establish the lemma. The following three lemmas are due to Yi Yuan [8]. Lemma : The only solutions of the equation S(n ) = φ(n) are n =, 4, 50. Lemma : The only solutions of the equation S(n 3 ) = φ(n) are n =, 48, 98. Lemma : The only solution of the equation S(n 4 ) = φ(n) is n =. The following result has been established by Maohua Le [9]. Lemma : The only solution of the equation S(n + n + + n n ) = φ(n)s()s() S(n) is n =. Weiguo Duan and Yanrong Xue [0] have established the following result. Lemma : The only solutions of the equation S() + S() + + S(n) = φ( are n =, 0. n(n + ) )

76 75 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function Lemma : A necessary and sufficient condition that the equation [S(n)] + S(n) = kn (3.3.4) possesses a solution for some integers n and k, is that the following equation has a solution : r(r + ) S( ) = r for some integers r and k. k Proof : Solving the equation [S(n)] + S(n) kn = 0 for positive S(n), we get S(n) = ( + 4kn ). (3.3.5) Thus, a necessary condition such that the equation (3.3.4) has a solution is that, k and n must be such that + 4kn is a perfect square. Let + 4kn = x for some integer x. Then, 4kn = x = (x )(x + ). () Since x and x + has the same parity, from (), both are even. Letting x = r, from (), 4kn = (r).(r + ) r(r + ) n =. k Finally, we get from (3.3.), r(r + ) S( ) = r. (3.3.6) k r(r + ) Conversely, n = satisfies the equation (3.3.4), as is shown below : k [S(n)] + S(n) = r + r = r(r + ) = kn. All these complete the proof of the lemma. Lemma 3.3. : For any integer k fixed, the equation n(n + ) S( ) = n k has an infinite number of solutions. Proof : Given the integer k, let p be a prime of the form p = ka, a. () p + p + By Lemma 3.7, S( ) < p, so that by Lemma 3..3, k k p(p + ) p + S( ) = max{s(p), S( )} = S(p) = p. k k This shows that n = p is a solution of the given equation. Now, since there are an infinite number of primes of the form (), the lemma is proved.

77 76 Wandering in the World of Smarandache umbers Remark 3.3. : Equation (3.3.6) allows us to find the solutions of the equation (3.3.4) for any particular value of k, as given in Lemma () When k =, the solutions of (3.3.4) are n = p(p + ), where p 5 is a prime. When p 5, p + 6 is a composite number, and so by Lemma 3.9, p + S(p + ) < p. Now, since (p, p + ) =, by Lemma 3..3, S(p(p + )) = max{ S(p), S(p + ) } = S(p) = p. () When k =, besides the trivial solution n =, other solutions of (3.3.4) are p(p + ) n =, where p 3 is a prime. In particular, any even perfect number n = a ( a+ ) is also a solution of (3.3.4), where a is an integer such that a+ is prime : In such a case S( a ) a < a+ S(n) = max{ S( a ), S( a+ ) } = a+. (3) When k = 5, r = 4 is a solution of (3.3.6), and n = 4 is a solution of (3.3.4), besides the solutions given in Lemma Following a different approach, the above results have also been found by Rongi Chen and Maohua Le [] The Smarandache perfect number and the completely Smarandache perfect number are defined as follows. Definition 3.3. : Given an integer n, () n is called Smarandache perfect if and only if k n = S(d i ). i= where d, d,, d k are the proper divisors of n. () n is called completely Smarandache perfect if and only if n = d n S(d), where the sum is over all divisors d of n (including n). Remark : The proof of Maohua Le [], with the modification that S() = 0, shows that the only Smarandache perfect number is n =. Gronas [3] has shown that, under the condition S() = 0, all completely Smarandache perfect numbers are n = p, 9, 6, 4. However, the situation is different if S() =. The Smarandache perfect numbers and the completely Smarandache perfect numbers are given in Lemma 3.3. and Lemma respectively.

78 77 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function Lemma 3.3. : The only Smarandache perfect numbers are n =, 6. Proof : Let n be an Smarandache perfect number, so that, by Definition 3.3., n = S(d). (3.3.7) d n d< n Clearly, no prime is a solution of (3.3.7). Also, note that, if n is of the form n = p α (α ), then the proper divisors of n are, p, p,, p α, with S() =, so that the r.h.s. of (3.3.7) is not divisible by p, while p divides the l.h.s. Thus, there is no solution of the form n = p α. By Lemma 3.8, for any divisor d of n, S(d) S(n), with strict inequality sign for at least one divisor of n. Therefore, from (3.3.7), we get n = S(d) < [d(n) ]S(n) n < d(n)s(n). () d n d< n Thus, any solution n of (3.3.7) must satisfy the inequalities in (). We first consider some particular cases. Case : When n is of the form n = pq, p and q are primes with q > p. In this case, d(n) = 4, and so, from () we get n = pq < 3q p =. Now, if p =, then d n d< n S(d) = S() + S(p) + S(q) = + p + q = 3 + q = n = q q = 3. Thus, in this case, n = 3 = 6 is the only solution of the equation (3.3.7). Case : When n is of the form n = α q, α, q 3 is a prime. First, let S(n) = S( α ). Since d(n) = (α + ), from (), we get α α α n = q < ( α + )S( ) α( α + ) q < α( α + ), which is impossible if (i) q = 3 and α 6, or if (ii) q = 5 and α 4, or if (iii) q 7 and α. Thus, the possible candidates for the equation (3.3.7) to hold true are n = 3. α, α 5; (n = 5. 3 is to be excluded since S(5. 3 ) S( 3 )). But, when n = q, when n = 3 q, for n = 4 q, S(d) = S() + S() + S( ) + S(q) + S(q) = q + q = 7 + q; d n d< n 3 S(d) = S() + S() + S( ) + S( ) + S(q) + S(q) + S( q) = 5 + q; d n d< n d n d< n S(d) = S() + S() + S( ) + S( ) + S( ) + S(q) + S(q) + S( q) + S( q) = 5 + q; for n = 5 q, d n d< n S(d) = S() + S() + S( ) + S( ) + S( ) + S( ) + S(q) + S(q) + S( q) + S( q) + S( q) = 39 + q; so that, in each case, the term on the right of (3.3.7) is odd, while n is even.

79 78 Wandering in the World of Smarandache umbers Next, let S(n) = q. Then, from (), we get n α α = q < ( α + )q < α +, which is possible only for α =, and so, n = q, which is not a solution of (3.3.7) for any q 3. Thus, there is no solution of the equation (3.3.7) of the form n = α q. Case 3 : When n is of the form n = p α q β (α, β, αβ ). Let S(n) = S(p α ). Now, since d(n) = (α + )(β + ), () takes the form α β α β α p q n = p q < ( α + )( β + )S(p ) ( α + )( β + ) αp. <. α( α + ) β + α p Now, since > for p 7 and α, () is impossible for p 7, q p, α and β α( α + ) (by part (4) of Lemma 3.3.). Thus, it is sufficient to check for the primes p =, 3, 5 only. First, let p = (so that q 3, α β ). Then, () reads as α β q. <. α( α + ) β + β q α Since 3 if q 3 and β, (3) gives 3. <, which is impossible for any α β + α( α + ) (by Lemma 3.3.(5)). Thus, in this case, (3.3.7) has no solution (by Case above). Next, let p = 3, so that from () α β q 3. <. α( α + ) β + If q 5 and β (so that β q > ), (4) gives β + (by Lemma 3.3.(5)); if q = and β 3 (so that impossible for any α. Moreover, from (4), α.3 <, α( α + ) β ), from (4), β + α 3 <, α( α + ) () (3) (4) which is impossible for any α α.3 <, α( α + ) which is impossible for α 4. Thus, the only possible candidates for (3.3.7) to hold true are n =.3 α,.3 α with α 3. α When n =.3, which is α α S(d) = S() + S() + S(3) S(3 ) + S(.3) + S(.3 ) S(.3 ) d n d< n α α (3k) (3k) 3( ), = = α + k= k= and the equation 3(α + ) =.3 α has no solution for α. And when n =.3 α, d n d< n α α α S(d) = S() + S() + S( ) + S(3) S(3 ) + S(.3) S(.3 ) + S(.3) S(.3 ) α α = (3k) (3k), which is not divisible by 3. k= k=

80 79 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function Finally, let p = 5, so that () takes the form α β q 5. <. α( α + ) β + If q 7 and β, then since α ; if q = and β 3 (so that β q > 3, β + (5) gives α 3.5 <, α( α + ) β ), then from (5), β + for any α ; if q = 3 and β, then since β 3 3, β + (5) which is impossible for any α.5 <, α( α + ) from (5), which is impossible α 3.5 <, α( α + ) which is α impossible for any α. Also, from (5), 5 <, which is impossible for α 3. α( α + ) Thus, in this case, the only possible candidates for (3.3.7) to hold true are n = 3.5,.5,.5. Note, however, that n = 3.5 violates the inequality (5), since > Now, when n =.5, S(d) = S() + S() + S(5) + S(5 ) + S(.5) = = 3; for n =.5, d n d< n d n d< n S(d) = S() + S() + S( ) + S(5) + S(5 ) + S(.5) + S(.5) + S(.5 ) = 4. Thus, there is no solution of (3.3.7) of the form n = p α q β. Now, we consider the general case. So, let α β α α αk n = p q p p... p k be the representation of n in terms of its distinct prime factors p, q, p, p, p,, p k (k ). Let, for definiteness, S(n) = S(p α ). Then, from (), we get α n < d(n)s(n) d(n)s(p ) α α α α β k p q p p p k α < S(p ) αp. α + β + α + α + α + k Now, if p =, without loss of generality, q 5, and so (6) takes the form α α α α k p p p k..... <, α( α + ) α + α + α + k which is impossible. On the other hand, if p = 3, with q 5, (6) takes the form α α α α k.3 p p p k..... <, α( α + ) α + α + α k + which is also impossible. All these show that n =, 6 are the only solutions of (3.3.7), which we intended to prove. (6)

81 80 Wandering in the World of Smarandache umbers Lemma : The only completely Smarandache perfect numbers are n =, 8. Proof : Let n be a completely Smarandache perfect number, so that, by Definition 3.3., n = S(d). (3.3.8) d n d n Clearly, no prime is a solution of (3.3.8). Thus, any solution of (3.3.8) must be a composite number. In this case, any solution of (3.3.8) must satisfy the following condition : n = S(d) < d(n)s(n). () d n d n We proceed like that in the proof of Lemma 3.3., considering the following cases : Case : When n is of the form n = p α (α ). In this case, however, there is no solution of the equation (3.3.8). Case : When n is of the form n = pq with q p. In this case, S(d) = S() + S(p) + S(q) + S(pq) = + p + q + q = + p + q, d n d n which is odd only if p =. But, then d n d n S(d) = 3 + q>q. Thus, there is no solution in this case. Case 3 : When n is of the form n = α q, q 3, α. Assuming that S(n) = S( α ), the possible candidates for the equation (3.3.8) to hold true are n = 3., But, when n = q, the divisors are,,, q, q and q, so that S(d) = S() + S() + S( ) + S(q) + S(q) + S( q) = q + q + 4 = + q, d n d n which is odd. Again, when n = 3 q, the divisors are,,, 3, q, q, q and 3 q, so that 3 3 S(d) = S() + S() + S( ) + S( ) + S(q) + S(q) + S( q) + S( q) d n d n = q + q = 9 + q, which is odd. Next, assuming that S(n) = q, so that α, we get from (), n α α = q < ( α + )q < α +, which is possible only for α =, and so, n = q. Now, since

82 8 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function we see that d n d n S(d) = S() + S() + S( ) + S(q) + S(q) + S( q) = q + q + q = 7 + 3q, S(d) = 7 + 3q = q q = 7. d n d n Thus, in this case, n =.7 = 8 is the only solution of the equation (3.3.8). Case 4 : When n is of the form n = p α q β. Letting S(n) = S(p α ) with p = 3, the only possible candidates for (3.3.8) to hold true are n =.3 α,.3 α with α 3. When n =.3 α, α α S(d) = S() + S() + S(3) + S(3 ) S(3 ) + S(.3) + S(.3 ) S(.3 ) d n d< n α = + + (3k) k= α( α + ) = = α + α + 3( ), and the equation 3(α + α + ) =.3 α has no solution for α. Again, when n =.3 α, d n d< n which is not divisible by 3. α S(d) = S() + S() + S( ) + S(3) + S(3 ) S(3 ) α α S(.3) S(.3 )... S(.3 ) S(.3) S(.3 )... S(.3 ) α α = (3k) (3k), k= k= When p = 5, the only possible candidates for () to hold true are n = 5.,.5,.5. But, when n =.5, and when n =.5, S(d) = S() + S() + S(5) + S(5 ) + S(.5) + S(.5 ) = 33; d n d n S(d) = S() + S() + S( ) + S(5) + S(5 ) + S(.5) + S(.5) + S(.5 ) + S(.5 ) = 5. d n d< n α β α α αk In the general case when n = p q p p... p k with k, an analysis similar to that in the proof of Lemma 3.3. shows that the equation (3.3.8) has no solution. All these complete the proof of the lemma. The following result is due to Ashbacher []. Here, we give a simplified proof.

83 8 Wandering in the World of Smarandache umbers Lemma : Given the successive values of the Smarandache function S() =, S() =, S(3) = 3, Z(4) = 4, S(5) = 4, S(6) = 3, S(7) = 7, S(8) = 4,, the number r is constructed by concatenating the values in the following way : r = Then, the number r is irrational. Proof : The proof is by contradiction. Let, on the contrary, r be a rational number. Then, from some point on, a group of consecutive digits must repeat infinite number of times, that is, r has the form r = d d d k d d d k. Let N be the integer such that S(N) = d d d i, i k. By Lemma 3., S((d d d i )!) = d d d i, and (d d d i )! is the maximum such number (so that, for any integer n > (d d d i )!, S(n) d d d i ). Thus, the digits d d d i in r cannot repeat infinitely often. This leads to a contradiction, and consequently, r is irrational. The following lemma, giving an exact formula for the number of primes less than or equal to given integer N 4, is due to Seagull [4]. Lemma : Given any integer N 4, the number of primes less than or equal to N, denoted by π(n), is N S(n) π (N) =, n= n where [x] denotes the greatest integer less than or equal to x. Proof : By Lemma and Lemma 3.7, S(n), if n = 4 or if n is prime n = 0, otherwise Thus, in the above sum, a prime contributes, increasing the count of prime by. Since n = 4 is not prime, its contribution to the sum is to be deducted. The idea of Lemma can be extended to find an exact formula for the number of twin primes less than or equal to a given integer N 3. Recall that, p and p + are called twin primes if both p and p + are primes. Some of the twin prime pairs are (3, 5), (5, 7), (, 3), (7, 9), (9, 3) and (4, 43). A formula has been put forward by Mehendale [5] to find the number of twin primes less than or equal to a given integer N 3. In the following lemma, we reproduce the formula in a modified form. Lemma : Given any integer N 3, the number of twin primes less than or equal to N, denoted by π (N), is S(n) S(n + ) (N). ([x] denotes the greatest integer not exceeding x). N π = n= 3 n n +

84 83 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function Proof : By Lemma and Lemma 3.7, S(n) S(n ), if both n and n + are primes +. n n = + 0, otherwise Thus, in the sum, a contribution of a count of comes only from a twin prime pair. Since it is known that, the first twin prime pair in the sequence of twin primes is (3, 5), it is reasonable to start searching from n = 3; and this automatically excludes the case n = 4. The idea is extended further in the following lemma, due to Mehendale [5]. Lemma : Given any integer N 3, the number of prime pairs (p, p + k) (where k is a pre specified integer) less than or equal to N, denoted by π k (N), is N k S(n) S(n + k) π k (N) =., n= 3 n n + k (where [x] denotes the greatest integer less than or equal to x). And we can extend still further to find a formula of the number of prime triplets less than or equal to a given integer N. A prime triplet is of the form (p, p +, p + 6) where p, p + and p + 6 are all primes, or of the form (p, p + 4, p + 6) with p, p + 4 and p + 6 all primes. The triplets (5, 7, ) and (, 3, 7) are examples of prime triplets of the first kind, while two prime triplets of the second kind are (7,, 3) and (3, 7, 9). It is conjectured that, like the twin-primes, the twin triplets (of both kinds) are infinite in number (Hardy and Wright [6], pp. 5), but the conjectures still remain open. Lemma : Let N 5 be any given integer. () The number of prime triplets of the form (p, p +, p + 6) less than or equal to N, denoted by π 3 (N), is S(n) S(n + ) S(n + 6) (N).. ; N 6 π 3 = n= 5 n n + n + 6 () The number of prime triplets of the form (p, p + 4, p + 6) less than or equal to N, denoted by π 3 (N), is N 6 S(n) S(n + 4) S(n + 6) π 3 (N) =.. ; n= 7 n n + 4 n + 6 (where [x] denotes the greatest integer less than or equal to x). Lemma Lemma show how the Smarandache function comes into role in the formulas for determining the exact number of primes, twin primes or prime triplets upto a given integer N. However, it should be borne in mind that, from the computational point of view, these formulas are not that useful, because the calculation of S(n) for large values of n involves much memory.

85 84 Wandering in the World of Smarandache umbers 3.4 Some Observations and Remarks In this section, some observations about some of the properties of the Smarandache function are given. Some questions are raised in the relevant cases. Remark 3.4. : The proof of Lemma 3.3. shows that S(n + ) S(n) is both unbounded above and unbounded below; in fact, the proof also shows that lim sup S(n + ) S(n) = +. n The minimum value of S(n + ) S(n) is clearly, and some instances are given below : S() S() =, S(3) S() =, S(4) S(3) =, S(5) S(4) =, S(9) S(0) =, S(6) S(5) =, S(35) S(36) =, S(64) S(63) =, S(99) S(00) =, S(76) S(75) =, S(96) S(95) =. The question is : Is there infinitely many n with S(n + ) S(n) =? (see Jason Earls [6]). Remark 3.4. : Since for any prime p 5, p S(p ) < p = S(p), it follows that, the inequality S(n + ) > S(n) holds for infinitely many n. Again, for any p 5, p + S(p + ) < p = S(p), which shows that the inequality S(n + ) < S(n) holds for infinitely many n. Lemma proves that the equation S(n) S(n + ) = n has no solution. However, for the equation, S(n) S(n + ) = kn, k, the situation is quite different : If n is a prime, say n = p, then the equation has a solution corresponding to k = S(p + ). Besides this trivial case, there are solutions of the equation for particular values of k. We have examples corresponding to k =, 4, 5, 6, 7, 9, 0,, 3, 7, 9, 37, which shows that the equation S(n)S(n + ) = kn has more than one solutions for these values of k. For these values of k, there are composite n as well. A limited search upto n = 4800 reveals that, there are solutions for other values of k as well. These are given below : k = 8 : n = 3, 7, 9, 3, 383, k = 4 : n = 97, 93, k = : n = 4, k = 3 : n = 37, 9, 367, k = 6 : n = 337, k = 4 : n = 63, 409, k = 43 : n = 57, k = 47 : n = 8, k = 53 : n =, 37, k = 59 : n = 353, k = 67 : n = 40, k = 7 : n = 83, k = 79 : n = 57, k = 83 : n = 33, k = 97 : n = 93, k = 39 : n = 77, k = 57 : n = 33, k = 99 : n = 397. Note that, in each case, n is prime. The question is : Is there a solution of the equation S(n)S(n + ) = kn for any k?

86 85 Chapter Chapter 4 : The Pseudo 3 : The Smarandache Function Remark : In more detailed study of the properties of the Smarandache function S(n), the inequalities given in Lemma 3..6 and Corollary 3..3 may prove very useful. Sandor [3] derives a number of equations and inequalities involving S(n). The derivations of some of them become simpler if Lemma 3..6 is exploited. For example, from Lemma 3..6, it follows immediately that S(n n ) S(n ) S(n ) n S(n ) for any integers n, n. As another example, let us consider the problem of finding the solution of S(ab) = a k S(b). If a = = b, the equation becomes trivial. So, let a, b >. Then, by Lemma 3..6, k k S(ab) = a S(b) S(a) S(b) a S(a) a k =, S(a) = a. Therefore, S(ab) = S(a) S(b), and hence, by Corollary 3..3, we must have S(ab) = S(a) + S(b). Now, the proof of Lemma 3..7 shows that, we must have a = = b. Remark : As has been proved by Sandor [3], n n n n n + { } S(n) S(n) S(n) lim inf = 0, lim sup = = max : n Z. The proof is simple : In the first case, the proof follows from Lemma In the second case, the proof follows from the fact that, for any prime p, S(p). p = Remark : The following results have been established by Sandor [3] : n n n n n + { } SS(n)) S(S(n)) S(S(n)) lim inf = 0, lim sup = = max : n Z. To prove the first part, note that S(S(p!)) S(p) p 0 < = = = 0 as p. p! p! p! (p )! The proof of the remaining part is as follows : For any prime p, S(S(p)) S(p) = =. p p Remark : The proof of Lemma shows that, the inequality S(n) + d(n) > n holds true for infinitely many n (namely, when n = p is a prime), and the inequality S(n) + d(n) < n also holds true for infinitely many n (namely, when n is of the form n=p α, where p 3 is a prime). Remark : In [3], Sandor conjectures that S * ((n + )! (n + 3)!) = q n, where q n is the first prime following n + 3. From the values of S * ((n + )! (n + 3)!), given in Remark 3..., we see that the case n = is a violation to the above conjecture. However, for values up to n = 0, given in Sandor [3], n = is the only exception to the conjecture. The question is : Is the conjecture true for n? In this respect, we have only the result given in Lemma

87 86 Wandering in the World of Smarandache umbers Remark : Several researchers have considered the function m T p(m) = k, k= p which counts the exponent of the prime p in m! (where [x] is the greatest integer less than or equal to x). Since for any prime p and any integer α, α α S(p ) = min{m : p m!}, it follows that S(p α ) = min{m : T (m) α }. p Using T p (m), Farris and Mitchell [7] have proved the following result. k p k Lemma 3.4. : ( ) k S for k. p = (p )p Remark : From Lemma 3.3.5, S(n) < n for any composite number n 4. To find the k S(n) S(p ) maximum value, max { : n 4, p n }, it is sufficient to consider max : p, k, k p where the maximum is over all primes p and all k. By Lemma 3.3. and Lemma 3..4, the sequence Since k k k S(p ) k p k= is strictly decreasing in k for any prime p fixed, with S(p k ) = kp = k if k p. k k k p p p p is strictly deceasing in both k and p, to find S(n) max { : n 4, p} n consider only the cases when p = with k = 3 and p = 3 with k =. Since 3 S(3 ) S( ) 3 = 6 = > =, we have the following lemma. Lemma 3.4. : For any composite number n > 4, S(n), n 3 (with the equality sign for n = 9 only). Ashbacher [] raises the following question : Question 3.4. : For what rational numbers r there exists an n such that S(n) r? n =, it is sufficient to From Lemma 3.4., we see that Question 3.4. is valid only when r (0, ]. Ashbacher [] 3 also proved that, if r is of the form r =, where p is a prime, then n = 8, 7, and these are the p only solutions.

88 87 Chapter 4 Chapter : The Pseudo 3 : The Smarandache Function A simpler proof is as follows : In order that S(n) =, n must be of the form n = p α. From n p Lemma 3..4, Lemma 3..5 and Lemma 3.4., it is clear that α p +. Now, when α p, then α S(p ) αp = = α = α α α p p p p α = p, α = p = 3 = α, and when α = p +, p+ S(p ) p = = = p = p =. p+ p+ p p p p p We conclude this chapter with the following conjecture related to S(n). Conjecture 3.4. : For any integer n, S(n + ) S(n). Let, on the contrary, S(n + ) = S(n) = m 0 for some integers n and m 0. () Clearly, neither n nor n + can be prime. So, let α α α α β β β βt n = p p p... p, n + = q q q... q r r t be the representations of n and n + in terms of their prime factors, where the primes p, p, p,, p r and the primes q, q, q,, q t are all distinct. Moreover, let for definiteness α β S(n) = S(p ), S(n + ) = S(q ). () Thus, the problem of finding the solution of () reduces to finding the primes p and q, and the integers α and β such that α β S(p ) = S(q ). (3) Without any loss of generality, we may assume that p > q, so that α β. From (3), it follows that p must divide S(q β ) and q must divide S(p α ). From Lemma 3..4 and Lemma 3..5, we see that, since p > q, we must have α p, β q +, so that S(p α ) = kp for some integer k p (p does not divide k), a S(q β ) = q (q + b) for some integers a and b (q does not divide b). Again, since n m 0!, (n + ) m 0!, and since (n, n + ) =, it follows that n(n + ) m 0! S(n(n + )) m o S(n(n + )) = m 0, since, by Lemma 3.8, S(n(n + )) S(n) m 0. Therefore, S(n + ) = S(n) S(n + ) = S(n) = S(n(n + )).

89 88 Wandering in the World of Smarandache umbers References [] Smarandache, F. A Function in the Number Theory, An. Univ. Timisoara, Ser. St. Mat., 8(), 79 88, 980. [] Ashbacher, Charles. An Introduction to the Smarandache Function, Erhus University Press, Vail, USA [3] Sandor, Jozsef. Geometric Theorems, Diophantine Equations, and Arithmetic Functions, American Research Press, Rehoboth, 00. [4] Stuparu, A.Valcea and Sharpe, D.W. Problem of Number Theory (5), Smarandache Function Book Series, Vol. 4 5 (Ed. C. Dumitrescu and V. Seleacu), Number Theory Publishing Co., pp. 4, 994. [5] Ashbacher, Charles. Pluckings from the Tree of Smarandache Sequences and Functions, American Research Press, Lupton, AZ, USA [6] Hardy, G.H. and Wright, E.M. An Introduction to the Theory of Numbers, Oxford Science Publications, Clarendon Press, 5 th Edition, 00. [7] Jingping, M. An Equation Involving the Smarandache Function, Scientia Magna, (), 89 90, 005. [8] Yi Yuan. An Equation Involving the Euler Function and the Smarandache Function, Scientia Magna, (), 73 75, 005. [9] Maohua Le. An Equation Concerning the Smarandache Function, Scientia Magna, (), 78 80, 005. [0] Weiguo Duan and Yanrong Xue. On the Solvability of an Equation Involving the Smarandache Function and Euler Function, Scientia Magna, 4(), 9 33, 008. [] Chen, R. and Le, M. On the Functional Equation S(n) + S(n) = kn, Smarandache otions Journal,, 73 76, 000. [] Maohua Le. On Smarandache Perfect Numbers, Scientia Magna, (), 4 6, 006. [3] Gronas, Pal. The Solution of the Diophantine Equation σ η (n) = n, Smarandache otions Journal, 4 5, 4 6, 994. [4] Seagull, A. The Smarandache Function and the Number of Primes Less Than x, Mathematical Spectrum, 8(3), pp. 53, 995/6. [5] Mehendale, D.P. The Classical Smarandache Function and a Formula for Twin Primes, Scientia Magna, (), 4 5, 005. [6] Jason Earls, On Consecutive Values of the Smarandache Function, Scientia Magna, (), 9 30, 005. [7] Farris, M. and Mitchell, P. Bounding the Smarandache Function, Smarandache otions Journal, 3, 37 4, 00.

90 89 Chapter 4 : The Pseudo Smarandache Function Chapter 4 The Pseudo Smarandache Function Chapter 3 gives some properties of the Smarandache function S(n). The other function, widely-studied by the Smarandache number theorists, is the pseudo Smarandache function. In this chapter, the pseudo Smarandache function, denoted by Z(n), is considered. The pseudo Smarandache function, Z(n), introduced by Kashihara [], is as follows : Definition 4. : For any integer n, the pseudo Smarandache function Z(n) is the m(m + ) smallest positive integer m such that m is divisible by n. Thus, Z(n) = min{ m : m Z + m(m + ), n }; n, + (where Z is the set of all positive integers) : Let T(m) be m th the triangular number m(m + ) T(m), m Z +. Then, T(m) is strictly increasing in m. This implies that the function Z(n) is well-defined. By Definition 4., Z(n) is the smallest positive integer m such that the corresponding triangular number T(m) is divisible by n. Thus, Z(n) = m 0 if and only if the following two conditions are satisfied : m 0 (m0 + ) () n divides, m(m + ) () n does not divide Let, for any n Z + fixed, Z(n) = m 0, so that m 0 (m0 + ) T(m 0 ) for any m with m m 0. = k n for some k Z +. Then, m 0 can be found by solving the above equation for the positive root. Thus, m 0 = ( 8kn ) +, where k is such that + 8 k n is a perfect square; moreover, the smallest m 0 is found by choosing the smallest such k. Conversely, if k is the smallest integer such that + 8 k n is a perfect square, m 0 (m0 + ) then m 0 is the smallest integer such that n, and hence, Z(n) = m 0. We thus arrive at the following equivalent definition of Z(n), as has been pointed out by Ibstedt [].

91 90 Wandering in the World of Smarandache umbers Definition 4. : Z(n) min { k ( 8kn ) : k + = + Z, + 8kn is a perfect square}. Kashihara [], Ibstedt [] and Ashbacher [3] studied some of the elementary properties satisfied by Z(n). Their findings are summarized in the following Lemma 4. Lemma 4.6. The proofs of some of these results depend on the following lemma which is almost trivial. Lemma 4. : Let p be a prime. Let an integer m ( > p) be divisible by p k for some integer k ( ). Then, p k does not divide m + (and m ). Lemma 4. : For any n Z +, Z(n). Moreover, () Z(n) = if and only if n =, () Z(n) = if and only if n = 3. Proof : In either case, the proof of the if part is trivial. So, we prove the only if part. () Let Z(n) = for some n Z +. Then, by definition, n, and hence, n =. () Let Z(n) = for some n Z +. Then, by definition, n 3, and hence, n = 3. The following lemma gives lower and upper bounds of Z(n). Lemma 4.3 : 3 Z(n) n for all n 4. m(m + ) Proof : T(m) and only if n = 3. Therefore, for n 4, Z(n) T() = 3. Again, since n T(n ), it follows that Z(n) cannot be greater than n. is strictly increasing in m (m Z + ) with T() = 3. Thus, Z(n) = if The upper bound of Z(n) in Lemma 4.3 can be improved in the case when n is odd. In fact, we can prove the following result. Lemma 4.4 : Z(n) n for any odd integer n 3. Proof : If n 3 is an odd integer, then n is even, and hence, n T(n ) n(n ) The lower bound of Z(n) can also be improved. For example, since T(4) = 0, it follows that Z(n) 5 for all n. A better lower bound of Z(n) is given in Lemma 4.5 below for the case when n is a composite number. Lemma 4.5 : For any composite number n 4, Z(n) max{ Z(d) : d n }. Proof : Let Z(n) = m 0 for some m 0 Z +. m 0(m0 + ) Then, n T(m 0 ) (and m 0 is the smallest integer with this property). Clearly, any divisor d of n also divides T(m 0 ), so that Z(d) m 0. Lemma 4.6 : For any integer n 4, Z(n) ( + 8n ) > n. Proof : From Definition 4., since the minimum admissible value of k =, Z(n) ( + 8n ) > n..

92 9 Chapter 4 : The Pseudo Smarandache Function This chapter gives some results related to the pseudo Smarandache function Z(n). In 4., we give some preliminary results which would be needed in the following section. 4. gives some simple explicit expressions for Z(n). An expression for Z(pq) is given in Theorem 4.., whose proof shows that the solution of Z(pq) involves the solution of two Diophantine equations in p and q. Some particular cases of Theorem 4.. are given in Corollaries Some generalizations of the pseudo Smarandache function are given in is devoted to some miscellaneous topics. We conclude this chapter with some remarks about the properties of Z(n) in the final Some Preliminary Results In this section, we give some preliminary results that would be required in the next section to find explicit expressions of Z(n) in some cases. Lemma 4.. : 6 m(m + )(m + ) for any m Z + ; and 6 (p ) for any prime p 5. Proof : The first part is well-known : Since the product of two consecutive integers is divisible by and the product of three consecutive integers is divisible by 3, and since (, 3) =, the result follows. In particular, for any prime p 5, 6 (p )p(p + ). But since p ( 5) is not divisible by 6, it follows that 6 (p )(p + ). Lemma 4.. : For any integer k 0, () 4 divides 3 k, () 4 divides 3 k+ +. Proof : In either case, the proof is trivial for k = 0. So, let k. () Writing 3 k = (3 k )(3 k + ), the result follows immediately. () Since 3 k+ + = 3(3 k ) + 4, the result follows by virtue of part (). Lemma 4..3 : For any integer k 0, () 3 divides k, () 3 divides k+ +. Proof : In either case, the proof is trivial for k = 0. So, let k. () By Lemma 4.., 3 divides ( k ) k ( k + ). Since 3 does not divide k, 3 must divide ( k )( k + ) = k. () Noting that k+ + = ( k ) + 3, the result follows by virtue of part (). Lemma 4..4 : For any integer k 0, 5 divides 4k. Proof : The proof is by induction on k. The result is clearly true for k =. So, we assume its validity for some integer k. Now, since 4(k+) = 6( 4k ) + 5, it follows that the result is true for k + as well, completing induction. Lemma 4..5 : For any integer k 0, 7 divides 3k. Proof : The result is clearly true for k =. To proceed by induction, we assume the validity of the result for some integer k. Now, 3(k+) = 8( 3k ) + 7, showing that the result is true for k + as well, which we intended to prove.

93 9 Wandering in the World of Smarandache umbers Lemma 4..6 : For any integer k 0, () divides 5(k+) +, () divides 0k. Proof : In either case, the result is true when k = 0. So, let k. () The result can easily be checked to be true for k =. Now, assuming its validity for some integer k, by the induction hypothesis, together with the fact that 5(k+3) + = 0 [ 5(k+) + ] ( 5 + )( 5 ), it follows that the result is true for k + as well, completing induction. () Since 0k = ( 5k )( 5k + ), the result follows by virtue of part (). Lemma 4..7 : For any integer k 0, () 3 divides 6(k+) +, () 3 divides k. Proof : In either case, the result is trivially true when k = 0. So, let k. () The result is true for k =. To proceed by induction, we assume that it is true for some integer k. Now, 6(k+3) + = ([ 6(k+) + ] ( 6 + )( 6 ), which shows the validity of the result for k +. This completes induction. () Noting that k = ( 6k )( 6k + ), the result follows by part () above. Lemma 4..8 : For any integer k 0, () 7 divides 4(k+) +, () 7 divides 8k. Proof : In either case, the result is true when k = 0. So, let k. () The result is clearly true for k =. To proceed by induction, we assume its validity for some integer k. Then, since 4(k+3) + = 8 ([ 4(k+) + ] ( 4 + )( 4 ), we see that the result is true for k + as well. This completes the proof by induction. () Since 8k = ( 4k )( 4k + ), the result follows by virtue of part (). Lemma 4..9 : For any integer k 0, () 9 divides 9(k+) +, () 9 divides 8k. Proof : In either case, the result is true when k = 0. So, let k. () The result is clearly true for k =. To proceed by induction, we assume its validity for some integer k. Then, 4(k+3) + = 8 ([ 4(k+) + ] ( 4 + )( 4 ), showing that the result is true for k + also. This completes the proof by induction. () Writing 8k = ( 4k )( 4k + ), the result follows by part () above. Lemma 4..0 : For any integer k 0, 3 divides 5k. Proof : The proof is by induction on k. The result is clearly true for k =. So, we assume its validity for some integer k. Now, since 5(k+) = 3( 4k ) + 3, it follows that the result is true for k + as well, completing induction.

94 93 Chapter 4 : The Pseudo Smarandache Function 4. Some Explicit Expressions for Z(n) k(k + ) In this section, we first give the explicit expressions for Z( ), Z( k ) and Z(p k ) in Lemmas (see Kashihara [] and Ashbacher [3]). Lemmas give the expressions for Z(np) and z(np ) for some particular n. We also derive the explicit forms of Z(3. k ), Z(5. k ), Z(7. k ), Z(. k ), Z(3. k ), Z(7. k ), Z(9. k ), Z(3. k ), Z(p k ) and Z(3p k ), as well as the expressions for Z(4p), Z(5p), Z(6p), Z(7p), Z(8p), Z(9p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(3p), Z(4p), Z(48) and Z(96p) in Lemmas respectively. Theorems give two alternative expressions for Z(pq). k(k + ) Lemma 4.. : Z( ) = k for any k Z +. Proof : Noting that k(k + ) = m(m + ) if and only if k = m, the result follows. Lemma 4.. : For any k Z +, Z( k ) = k+. Proof : By definition, Z( k ) = min{ m : k m(m + ) } = min{m : k+ m(m + ) }. () Now, note that, only one of m and m + is even, and as such, k+ divides exactly one of m and m +. Hence, the minimum m in () is k+. This proves the lemma. Lemma 4..3 : For any prime p 3 and any k Z +, Z(p k ) = p k. Proof : By definition, Z(p k ) = min{m : p k m(m + ) } = min{m :.p k m(m + )}. () If p k divides m(m + ), then p k must divide either m or m +, but not both (by Lemma 4.). Again, note that, if p k divides one of m and m +, then divides the other. Thus, the minimum m in () may be taken as p k, which establishes the lemma. Lemma 4..4 and Lemma 4..5 give respectively the expressions for Z(np) and Z(np ) (p is a prime) for some particular n. Then, the expressions for Z(p) and Z(3p) follow from Lemma 4..4, and are given in Corollary 4... The expressions for Z(p ) and Z(3p ), which follow from Lemma 4..5, are given in Corollary 4... Lemma 4..4 : If p 3 is a prime and n is an integer (not divisible by p), then { p, if n (p ) Z(np) = p, if n (p + ) Proof : By Definition 4., m(m + ) Z(np) = min{m : np } = min{m : np m(m + )}. (3) Then p must divide either m or m +, but not both (by Lemma 4.). Thus, the minimum m in (3) is p or p depending on whether p or p + respectively is divisible by n.

95 94 Wandering in the World of Smarandache umbers Corollary 4.. : Let p be an odd prime. Then, () for p 3, { p, if 4 (p ) Z(p) = p, if 4 (p + ) () for p 5, { p, if 3 (p ) Z(3p) = p, if 3 (p + ) Proof : The above results follow immediately from Lemma 4..4, in Case () since any prime p 3 is either of the form 4a + or 4a + 3 (for some integer a ), and in Case () since any prime p is of the form 3a + or 3a + (for some integer a ), and since both p and p + are divisible by. Lemma 4..5 : If p 3 is a prime and n is an integer not divisible by p, then Z(np ) = p, if n (p + ) Proof : Using the definition, p, if n (p ) Z(np ) = min{m : np m(m + ) } = min{m : np m(m + )}. (4) Therefore, p must divide exactly one of m and m + (by Lemma 4.), and then n must divide the other. Now, if n divides p, then the minimum m in (4) may be taken as p ; on the other hand, if p + is divisible by n, the minimum m is p +. Corollary 4.. : Let p be an odd prime. Then, () Z(p ) = p, for p 3; () Z(3p ) = p, for p 5. Proof : In both the cases, the results follow from Lemma () Noting that 4 divides (p )(p + ) = p, the result follows immediately. () By Lemma 4.., 6 divides p. The result then follows. Remark 4.. : Since Z(p) 3(p ) = Z() Z(p) for any odd prime p, and Z(3p ) = p Z(p ) + Z(p ) for any prime p 5, it follows that Z(n) is neither multiplicative nor additive. The expressions for Z(3. k ) and Z(5. k ), due to Ashbacher [3], are given in Lemma 4..6 and Lemma 4..7 respectively. Lemma 4..8 gives an expression for Z(7. k ). This result also appears in Ibstedt [4], but we follow a different approach to the proof. Note that, for any prime p 3 fixed, the values of Z(p. k ) depends on the form of k.

96 95 Chapter 4 : The Pseudo Smarandache Function Lemma 4..6 : For any integer k, k+ k, if k is odd Z(3. ) = k+, if k is even Proof : Using the definition, Z(3. k ) = min{m : 3. k m(m + ) } = min{m : 3. k+ m(m + )}. (5) Then, k+ must divide one of m and m + and 3 must divide the other. Now, we consider the two possibilities separately : Case () : When k is odd, say, k = a +, for some integer a 0. In this case, by Lemma 4..3(), 3 divides a+ + = k +. By Lemma 4..3(), 3 k+ ( ) 3. k+ k+ k+ ( ), and hence, the minimum m in (5) may be taken as k+. Case () : When k is even, say, k = a, for some integer a. By Lemma 4..3(), 3 divides a = k. Also, 3 k+ ( + ) 3. k+ k+ k+ ( + ), and hence, in this case, the minimum m in (5) may be taken as k+. Lemma 4..7 : For any integer k, k+, if 4 k k+ k, if 4 (k ) Z(5. ) = k+, if 4 (k ) k+, if 4 (k 3) Proof : Using the definition, Z(5. k ) = min{m : 5. k m(m + ) } = min{m : 5. k+ m(m + )}. (6) Here, k+ must divide one of m and m +, and 5 must divide the other. We have to consider separately the four possible cases : Case () : When k is of the form k = 4a for some integer a. In this case, by Lemma 4..4, 5 divides 4a = k. Now, since k+ + = 4( k ) + 5, we see that 5 k+ ( + ) 5. k+ k+ k+ ( + ), and hence, the minimum m in (6) may be taken as k+. Case () : When k is of the form k = 4a + for some integer a 0. By Lemma 4..4, 5 divides 4a = k. Now, since k+ + = 4( k ) + 5 = 4( 4a ) + 5, we see that 5 k+ ( + ) 5. k+ k+ k+ ( + ), and hence, in this case, the minimum m in (6) may be taken as k+.

97 96 Wandering in the World of Smarandache umbers Case (3) : When k is of the form k = 4a + for some integer a 0. In this case, writing k+ = 6( k ) + 5, and noting that, by Lemma 4..4, 5 divides 4a = k, we see that 5 k+ ( ) 5. k+ k+ k+ ( ). Hence, in this case, the minimum m in (6) may be taken as k+. Case (4) : When k is of the form k = 4a + 3 for some integer a 0. Here, since k+ = 6( k 3 ) + 5, and since 5 divides 4a = k 3 (by Lemma 4..4), we see that 5 k+ ( ) 5. k+ k+ k+ ( ). Hence, in this case, Z(5. k ) = k+. All these complete the proof of the lemma. Lemma 4..8 : For any integer k, k+ 3., if 3 k k k+ Z(7. ) =, if 3 (k ) k+, if 3 (k ) Proof : By definition, Z(7. k ) = min{m : 7. k m(m + ) } = min{m : 7. k+ m(m + )}. (7) Here, k+ must divide one of m and m +, and 7 must divide the other. We now consider the three possible cases below : Case () : When k is of the form k = 3a for some integer a. By Lemma 4..5, 7 divides 3a = k. Now, 3. k+ + = 6( k ) + 7, which shows that 3. k+ + is divisible by 7. Hence, 7. k+ divides 3. k+ (3. k+ + ), and consequently, the minimum m in (7) can be taken as 3. k+. Case () : When k is of the form k = 3a + for some integer a 0. Here, by virtue of Lemma 4..5, 7 k+ k 3a ( ) = 8( ) + 7 = 8( ) k+ k+ k+ ( ), and hence, in this case, the minimum m in (7) may be taken as k+. Case (3) : When k is of the form k = 3a + for some integer a 0. In this case, by virtue of Lemma 4..5, 7 k+ k 3a ( ) = 8( ) + 7 = 8( ) k+ k+ k+ ( ), and hence, Z(7. k ) = k+. Hence, the lemma is established. In Lemmas respectively, the expressions for Z(. k ), Z(3. k ), Z(7. k ), Z(9. k ) and Z(3. k ) are given. These results would be needed later in 4.4.

98 97 Chapter 4 : The Pseudo Smarandache Function Lemma 4..9 : For any integer k, k+ 5., if 0 k k+ 3., if 0 (k ) k+ 3, if 0 (k ) k+, if 0 (k 3) k+ k, if 0 (k 4) Z(. ) = k+ 5., if 0 (k 5) k+ 3., if 0 (k 6) k+ 3, if 0 (k 7) k+, if 0 (k 8) k+, if 0 (k 9) Proof : By definition, Z(. k ) = min{m :. k m(m + ) } = min{m :. k+ m(m + )}. (8) Here, k+ must divide one of m and m +, and must divide the other. We now consider all the possible cases below : Case () : When k is of the form k = 0a for some integer a. By Lemma 4..6(), divides 0a = k. Now, (5. k+ + ) = 0( k ) +. k+ (5. k+ ).(5. k+ + ). Therefore, the minimum m in (8) can be taken as 5. k+. Case () : When k is of the form k = 0a + for some integer a 0. Here, by virtue of Lemma 4..6(), (3. k+ ) = ( k ) +. k+ (3. k+ ).(3. k+ ), and hence, in this case, the minimum m in (8) may be taken as 3. k+. Case (3) : When k is of the form k = 0a + for some integer a 0. In this case, writing k+3 = 0a+5 in the form k+3 = 0a+5 = ( 0a )( 5 + ) + ( 5 + ) ( 0a ), we see, by virtue of Lemma 4..6, that ( k+3 + ). k+ k+3 ( k+3 + ), and hence, Z(. k ) = k+3. Case (4) : When k is of the form k = 0a + 3 for some integer a 0. Here, since k+ = 0a+5, it follows that ( k+ + ). k+ k+ ( k+ + ) Z(. k ) = k+. Case (5) : When k is of the form k = 0a + 4 for some integer a 0. Here, since k+ = 0a+5, it follows that ( k+ + ). k+ k+ ( k+ + ) Z(. k ) = k+.

99 98 Wandering in the World of Smarandache umbers Case (6) : When k is of the form k = 0a + 5 for some integer a 0. In this case, 5. k+ = 0( 0a+5 + ) (5. k+ ), and hence, Z(. k ) = 5. k+. Case (7) : When k is of the form k = 0a + 6 for some integer a 0. Here, the result follows from the following chain of implications : 3. k+ + = ( 0a+5 + ) (3. k+ + ) Z(. k ) = 3. k+. Case (8) : When k is of the form k = 0a + 7 for some integer a 0. In this case, k+3 = 0(a+) ( k+3 ) Z(. k ) = k+ 3. Case (9) : When k is of the form k = 0a + 8 for some integer a 0. Here, the desired result is obtained as follows : k+ = 0(a+) ( k+ ) Z(. k ) = k+. Case (0) : When k is of the form k = 0a + 9 for some integer a 0. In this case, the result follows, since k+ = 0(a+) ( k+ ) Z(. k ) = k+. All these complete the proof. Lemma 4..0 : For any integer k, k+ 3., if k k+ 3., if (k ) k+ 5., if (k ) k+ 3, if (k 3) k+, if (k 4) k+ k, if (k 5) Z(3. ) = 3. k+, if (k 6) k+ 3., if (k 7) k+ 5., if (k 8) k+ 3, if (k 9) k+, if (k 0) k+, if (k ) Proof : By definition, Z(3. k ) = min{m : 3. k m(m + ) } = min{m : 3. k+ m(m + )}. (9) Here, k+ must divide one of m and m +, and 3 must divide the other. We now consider all the possible cases, twelve in number : Case () : When k is of the form k = a for some integer a. By Lemma 4..7(), 3 divides a = k. Now, 3. k+ + = ( k ) (3. k+ + ), so that, 3. k+ divides 3. k+ (3. k+ + ). Hence, the minimum m in (9) is 3. k+.

100 99 Chapter 4 : The Pseudo Smarandache Function Case () : When k is of the form k = a + for some integer a 0. Here, by virtue of Lemma 4..7(), 3 (3. k+ + ) = ( k ) k+ (3. k+ ).(3. k+ + ), and hence, in this case, the minimum m in (9) may be taken as 3. k+. Case (3) : When k is of the form k = a + for some integer a 0. In this case, 5. k+ = 40( k ) k+ (5. k+ ).(5. k+ ), and hence, Z(3. k ) = 5. k+. Case (4) : When k is of the form k = a + 3 for some integer a 0. Here, k+3 = a+6. Expressing this as follows : k+3 = a+6 = ( a )( 6 + ) + ( 6 + ) ( a ), we see, by Lemma 4..7, that 3 ( k+3 + ) 3. k+ k+3 ( k+3 + ) Z(3. k ) = k+3. Case (5) : When k is of the form k = a + 4 for some integer a 0. Here, since k+ = a+6, it follows that 3 ( k+ + ) 3. k+ k+ ( k+ + ) Z(3. k ) = k+. Case (6) : When k is of the form k = a + 5 for some integer a. In this case, k+ + = a ( k+ + ), and hence, Z(3. k ) = k+. Case (7) : When k is of the form k = a + 6 for some integer a 0. Here, the result follows from the chain of implications below : 3. k+ = ( a+6 + ) 3 3 (3. k+ ) Z(3. k ) = 3. k+. Case (8) : When k is of the form k = a + 7 for some integer a 0. In this case, the result is obtained as follows : 3. k+ = ( a+6 + ) 3 3 (3. k+ ) Z(3. k ) = 3. k+. Case (9) : When k is of the form k = a + 8 for some integer a 0. Here, 5. k+ + = 40( a+6 + ) 39 3 (5. k+ + ) Z(3. k ) = 5. k+. Case (0) : When k is of the form k = a + 9 for some integer a 0. In this case, by Lemma 4..7(), k+3 = (a+) 3 ( k+3 ) Z(3. k ) = k+3. Case () : When k is of the form k = a + 0 for some integer a 0. In this case, k+ = (a+) 3 ( k+ ) Z(. k ) = k+. Case () : When k is of the form k = a + for some integer a 0. In this case, the result follows by virtue of the fact that 3 ( k+ ). Hence, the lemma is established.

101 00 Wandering in the World of Smarandache umbers Lemma 4.. : For any integer k, k+ 4, if 8 k k+ 3, if 8 (k ) k+, if 8 (k ) k+ k, if 8 (k 3) Z(7. ) = k+ 4, if 8 (k 4) k+ 3, if 8 (k 5) k+, if 8 (k 6) k+, if 8 (k 7) Proof : By definition, Z(7. k ) = min{m : 7. k m(m + ) } = min{m : 7. k+ m(m + )}. (0) Here, k+ must divide one of m and m +, and 7 must divide the other. We now consider all the possibilities cases below : Case () : When k is of the form k = 8a for some integer a. By Lemma 4..8(), 7 divides 8a = k. Now, k+4 + = 6( k ) ( k+4 + ), so that 7. k+4 divides k+4 ( k+4 + ), and the minimum m in (0) can be taken as k+4. Case () : When k is of the form k = 8a + for some integer a 0. Here, by virtue of Lemma 4..8(), 7 ( k+3 + ) = 6( 8a ) + 7 = 6( k ) + 7, and hence, in this case, the minimum m in (0) may be taken as k+3. Case (3) : When k is of the form k = 8a + for some integer a 0. In this case, k+ = 8a+3. Then, since k+ = 8a+4 = ( 8a )( 4 + ) + ( 4 + ) ( 8a ), by virtue of Lemma 4..8, it follows that 7 ( k+ + ) 7. k+ k+ ( k+ + ) Z(7. k ) = k+. Case (4) : When k is of the form k = 8a + 3 for some integer a 0. Here, since k+ = 8a+4, it follows that, 7 divides k+ +. Hence, Z(7. k ) = k+. Case (5) : When k is of the form k = 8a + 4 for some integer a 0. In this case, k+ = 8a+5, so that k+4 = 6( 8a+4 + ) 7 is divisible by 7. Hence, Z(7. k ) = k+4. Case (6) : When k is of the form k = 8a + 5 for some integer a 0. In this case, 7 divides k+3 = 6( 8a+4 + ) 7,and hence, Z(7. k ) = k+3. Case (7) : When k is of the form k = 8a + 6 for some integer a 0. Here, k+ = 8(a+), which is divisible by 7. Thus, Z(7. k ) = k+. Case (8) : When k is of the form k = 8a + 7 for some integer a 0. In this case, the result follows, since 7 divides k+ = 8(a+).

102 0 Chapter 4 : The Pseudo Smarandache Function The same procedure can be followed to find explicit expressions for Z(p. k ), where p is an odd prime. But, in some cases, it would be complicated as the number of possible cases to be considered becomes larger and larger. For example, to find Z(9. k ), we have to consider 8 possible cases. In the following two lemmas, we give the expressions for Z(9. k ) and Z(3. k ), with outlines of the proofs. Lemma 4.. : For any integer k, k+ 9., if 9 k k+ 5., if 9 (k ) k+ 3 3., if 9 (k ) k+ 3., if 9 (k 3) k+ 3., if 9 (k 4) k+ 4, if 9 (k 5) k+ 3, if 9 (k 6) k+, if 9 (k 7) k+ k Z(9. ) =, if 9 (k 8) 9. k+, if 9 (k 9) k+ 5., if 9 (k 0) k+ 3 3., if 9 (k ) k+ 3., if 9 (k ) k+ 3., if 9 (k 3) k+ 4, if 9 (k 4) k+ 3, if 9 (k 5) Proof : By definition, k+, if 9 (k 6) k+, if 9 (k 7) Z(9. k ) = min{m : 9. k m(m + ) } = min{m : 9. k+ m(m + )}. The eighteen possible cases are given below. Case () : When k is of the form k = 8a for some integer a. Here, by Lemma 4..9(), 9 (9. k+ + ) = 8( 8a ) + 9. Thus, Z(9. k ) = 9. k+. Case () : When k is of the form k = 8a + for some integer a 0. Since 9 divides 5. k+ = 0( 8a ) + 9, we have, Z(9. k ) = 5. k+. Case (3) : When k is of the form k = 8a + for some integer a 0. In this case, 9 divides. k+ = 96( 8a ) Hence, Z(9. k ) = 3. k+3. Case (4) : When k is of the form k = 8a + 3 for some integer a 0. Here, 9 divides 6. k+ = 96( 8a ) Consequently, Z(9. k ) = 3. k+. Case (5) : When k is of the form k = 8a + 4 for some integer a 0. In this case, 9 divides 3. k+ = 96( 8a ) Therefore, Z(9. k ) = 3. k+. Case (6) : When k is of the form k = 8a + 5 for some integer a 0. In this case, 9 divides 8. k+ + = 5( 8a ) + 53, and so, Z(9. k ) = k+4.

103 0 Wandering in the World of Smarandache umbers Case (7) : When k is of the form k = 8a + 6 for some integer a 0. Here, 9 divides 4. k+ + = 5( 8a ) + 53, and hence, Z(9. k ) = k+3. Case (8) : When k is of the form k = 8a + 7 for some integer a 0. In this case, 9 divides. k+ + = 5( 8a ) As such, Z(9. k ) = k+. Case (9) : When k is of the form k = 8a + 8 for some integer a 0. Writing k+ = 8a+9 = ( 8a )( 9 + ) + ( 9 + ) ( 8a ), it follows, by Lemma 4..9(), that 9 divides k+ +. Consequently, Z(9. k ) = k+. Case (0) : When k is of the form k = 8a + 9 for some integer a 0. Here, 9 divides 9. k+ = 8( 8a+9 + ) 9, so that, Z(9. k ) = 9. k+. Case () : When k is of the form k = 8a + 0 for some integer a 0. In this case, 9 divides 5. k+ + = 0( 8a+9 + ) 9, and hence, Z(9. k ) = 5. k+. Case () : When k is of the form k = 8a + for some integer a 0. Here, 9 divides. k+ + = 96( 8a+9 + ) 95. Then, Z(9. k ) = 3. k+3. Case (3) : When k is of the form k = 8a + for some integer a 0. In this case, 9 divides 6. k+ + = 96( 8a+9 + ) 95. Thus, Z(9. k ) = 3. k+. Case (4) : When k is of the form k = 8a + 3 for some integer a 0. Here, 9 divides 3. k+ + = 96( 8a+9 + ) 95. Therefore, Z(9. k ) = 3. k+. Case (5) : When k is of the form k = 8a + 4 for some integer a 0. In this case, 9 divides k+4 = 8(a+). And consequently, Z(9. k ) = k+4. Case (6) : When k is of the form k = 8a + 5 for some integer a 0. Here, 9 divides k+3 = 8(a+), so that, Z(9. k ) = k+3. Case (7) : When k is of the form k = 8a + 6 for some integer a 0. In this case, 9 divides k+ = 8(a+). Therefore, Z(9. k ) = k+. Case (8) : When k is of the form k = 8a + 7 for some integer a 0. Noting that, k+ = 8(a+) is divisible by 9, it follows that Z(9. k ) = k+. Hence, the lemma is established. Lemma 4..3 : For any integer k, k+ 5., if 5 k k+ 4, if 5 (k ) k Z(3. ) = k+ 3, if 5 (k ) k+, if 5 (k 3) Proof : Using the definition, Z(3. k ) = min{m : 3. k k+, if 5 (k 3) m(m + ) } = min{m : 3. k+ m(m + )}. We now consider the five possible cases separately below : Case () : When k is of the form k = 5a for some integer a. By Lemma 4..0, 3 divides 5. k+ + = 30( 5a ) + 3. Thus, Z(3. k ) = 5. k+. Case () : When k is of the form k = 5a + for some integer a 0. Here, since 3 divides 8. k+ = 3( 5a ) + 3, it follows that Z(3. k ) = k+4.

104 03 Chapter 4 : The Pseudo Smarandache Function Case (3) : When k is of the form k = 5a + for some integer a 0. In this case, k+ = 8( 5a ) + 8, so that 3 divides 4. k+. Thus, Z(3. k ) = k+3. Case (4) : When k is of the form k = 5a + 3 for some integer a 0. Here,. k+ = 3( 5a ) + 3 is divisible by 3. Therefore, Z(3. k ) = k+. Case (5) : When k is of the form k = 5a + 4 for some integer a 0. In this case, 3 divides k+ = 5(a+). And so, Z(3. k ) = k+. All these complete the proof of the lemma. The expressions for Z(p k ) and Z(3p k ) (k 3) are given in Lemmas Lemma 4..4 : If p 3 is a prime and k 3 is an integer, then k k p, if 4 (p + ) and k is odd Z(p ) = k p, otherwise Proof : By definition, Z(p k ) = min{m : p k m(m + ) } = min{m : 4p k m(m + )}. We consider the two possibilities : Case : p is of the form 4a + for some integer a. In this case, p k = (4a + ) k = (4a) k + kc (4a) k + + kc (k ) (4a) +, showing that 4 divides p k. Hence, in this case, Z(p k ) = p k. Case : p is of the form 4a + 3 for some integer a. Here, p k = (4a + 3) k = (4a) k + kc (4a) k (3) + + kc (k ) (4a)3 k + 3 k. () If k is even, then by Lemma 4.., 4 (3 k ), so that 4 (p k ). Thus, Z(p k ) = p k. () If k 3 is odd, then by Lemma 4.., 4 (3 k + ), and so 4 (p k + ). Hence, Z(p k ) = p k. All these complete the proof of the theorem. Lemma 4..5 : If p 5 is a prime and k 3 is an integer, then k k p, if 3 (p + ) and k is odd Z(3p ) = k p, otherwise Proof : Using the definition, Z(3p k ) = min{m : 3p k m(m + ) } = min{m : 6p k m(m + )}. We now consider the following two possible cases : Case : p is of the form 3a + for some integer a. In this case, since p k = (3a + ) k = (3a) k + kc (3a) k + + kc (k ) (3a) +, it follows that 3 (p k ). Thus, in this case, Z(3p k ) = p k. Case : p is of the form 3a + for some integer a. Then, p k = (3a + ) k = (3a) k + kc (3a) k () + + kc (k ) (3a) k + k. () If k is even, then by Lemma 4..3, 3 ( k ), and so 3 (p k ). Thus, Z(3p k ) = p k. () If k 3 is odd, then by Lemma 4..3, 3 ( k + ), so that 3 (p k + ). Hence, Z(3p k ) = p k.

105 04 Wandering in the World of Smarandache umbers Lemma 4..6 Lemma 4..5 give the expressions for Z(4p), Z(5p), Z(6p), Z(7p), Z(8p), Z(9p), Z(0p), Z(p), Z(p) and Z(3p) respectively, where p is a prime. Lemma 4..6 : If p 3 is a prime, then p, if 8 (p ) p, if 8 (p + ) Z(4p) = 3p, if 8 (3p ) 3p, if 8 (3p + ) Proof : By definition, m(m + ) Z(4p) = min{m : 4p } = min{m : 8p m(m + )}. Here, p must divide only one of m and m +. We now consider separately the following four cases that may arise depending on p : Case () : p is of the form p = 8a + for some integer a. In this case, 8 divides p, and hence, by Lemma 4..4, Z(4p) = p. Case () : p is of the form p = 8a + 7 for some integer a 0. Here, 8 divides p +, and hence, Z(4p) = p (by Lemma 4..4). Case (3) : p is of the form p = 8a + 3 for some integer a 0. In this case, 8 divides 3p = 8(3a + ), and hence, Z(4p) = 3p. Case (4) : p is of the form p = 8a + 5 for some integer a 0. Here, 8 divides 3p + = 8(3a + ), and hence, Z(4p) = 3p. All these complete the proof of the theorem. Lemma 4..7 : If p = 3 or p 7 is a prime, then p, if 0 (p ) p, if 0 (p + ) Z(5p) = p, if 5 (p ) p, if 5 (p + ) Proof : Using definition, m(m + ) Z(5p) = min{m : 5p } = min{m : 0p m(m + )}. () Now, p must divide one of m and m +, and 5 must divide the other. By Lemma 4..4, the minimum m in () may be taken as p or p if 0 divides p or p + respectively. Here, we have to consider the following possible four cases : Case () : p is a prime whose last digit is. Here, 0 divides p, and hence, Z(5p) = p. Case () : p is a prime whose last digit is 9. In this case, 0 divides p +, and so, Z(5p) = p. Case (3) : p is a prime whose last digit is 3. Here, 5 (p ), so that 0p p(p ). Thus, the minimum m in () may be taken as p. Hence, Z(5p) = p. Case (4) : p is a prime whose last digit is 7. Here, 5 divides p +, and hence, Z(5p) = p. Hence, the proof is complete.

106 05 Chapter 4 : The Pseudo Smarandache Function Lemma 4..8 : If p 5 is a prime, then p, if (p ) p, if (p + ) Z(6p) = 3p, if (p 5) 3p, if (p 7) Proof : By definition, m(m + ) Z(6p) = min{m : 6p } = min{m : p m(m + )}. Here, p must divide exactly one of m and m +. Now, the following four cases may arise depending on the nature of the prime p : Case () : p is of the form p = a + for some integer a. In this case, divides p, and hence, by Lemma 4..4, Z(6p) = p. Case () : p is of the form p = a + for some integer a 0. Here, divides p +, and hence, Z(6p) = p (by Lemma 4..4). Case (3) : p is of the form p = a + 5 for some integer a 0. In this case, 4 divides 3p + = 4(9a + 4), so that p 3p(3p + ). Thus, Z(6p) = 3p. Case (4) : p is of the form p = a + 7 for some integer a 0. Here, 4 divides 3p = 4(9a + 5), and hence, Z(6p) = 3p. All these establish the lemma. Lemma 4..9 : If p 3 with p 7 is a prime, then p, if 7 (p ) p, if 7 (p + ) 3p, if 7 (3p + ) Z(7p) = 3p, if 7 (3p ) p, if 7 (p + ) p, if 7 (p ) Proof : Since m(m + ) Z(7p) = min{m : 7p } = min{m : 4p m(m + )}, p must divide exactly one of m or m +, and 7 must divide the other. We consider the six cases that may arise. Case () : p is of the form p = 7a + for some integer a. In this case, 7 divides p, and consequently, by Lemma 4..4, Z(7p) = p. Case () : p is of the form p = 7a + 6 for some integer a. Here, 7 divides p +, and so, Z(7p) = p (by Lemma 4..4). Case (3) : p is of the form p = 7a + for some integer a. In this case, 7 divides 3p + = 7(3a + ), so that 7p 3p(3p + ). Therefore, Z(7p) = 3p.

107 06 Wandering in the World of Smarandache umbers Case (4) : p is of the form p = 7a + 5 for some integer a 0. Here, 7 (3p ). Clearly, (3p ) as well. Thus, 4p 3p(3p ). Hence, Z(7p) = 3p. Case (5) : p is of the form p = 7a + 3 for some integer a 0. In this case, 7 divides p + = 7(a + ), and hence, Z(7p) = p. Case (6) : p is of the form p = 7a + 4 for some integer a. Here, 7 divides p = 7(a + ), and hence, Z(7p) = p. Hence, the lemma is established. Lemma 4..0 : For any prime p 3, p, if 6 (p ) p, if 6 (p + ) 5p, if 6 (5p + ) 5p, if 6 (5p ) Z(8p) = 3p, if 6 (3p + ) 3p, if 6 (3p ) 7p, if 6 (7p ) 7p, if 6 (7p + ) Proof : By definition, m(m + ) Z(8p) = min{m : 8p } = min{m : 6p m(m + )}. Then, p must divide exactly one of m or m +. Here, we have to consider the following eight possibilities : Case () : p is of the form p = 6a + for some integer a. In this case, 6 divides p, and so, by Lemma 4..4, Z(8p) = p. Case () : p is of the form p = 6a + 5 for some integer a. Here, 6 divides p +, and hence, Z(8p) = p (by Lemma 4..4). Case (3) : p is of the form p = 6a + 3 for some integer a 0. In this case, 6 divides 5p +, so that 6p divides 5p(5p + ). Hence, Z(8p) = 5p. Case (4) : p is of the form p = 6a + 3 for some integer a 0. Here, 6 divides 5p = 6(5a + 4), and hence, Z(8p) = 5p. Case (5) : p is of the form p = 6a + 5 for some integer a 0. In this case, 6 divides 3p + = 6(3a + ), and hence, Z(8p) = 3p. Case (6) : p is of the form p = 6a + for some integer a 0. Here, 6 divides 3p = 6(3a + ), and hence, Z(8p) = 3p. Case (7) : p is of the form p = 6a + 7 for some integer a 0. In this case, 6 divides 7p = 6(7a + 3), and hence, Z(8p) = 7p. Case (8) : p is of the form p = 6a + 9 for some integer a. Here, 6 divides 7p + = 6(7a + 4), and hence, Z(8p) = 7p. Thus, the proof is complete.

108 07 Chapter 4 : The Pseudo Smarandache Function Lemma 4.. : For any prime p 5, p, if 8 (p ) p, if 8 (p + ) p, if 9 (p ) Z(9p) = p, if 9 (p + ) 4p, if 9 (4p ) 4p, if 9 (4p + ) Proof : By definition, m(m + ) Z(9p) = min{m : 9p } = min{m : 8p m(m + )}. The six possible cases are considered separately below : Case () : p is of the form p = 8a + for some integer a. In this case, 8 divides p, and so, by Lemma 4..4, Z(9p) = p. Case () : p is of the form p = 8a + 7 for some integer a 0. In this case, 8 divides p +, and hence, Z(9p) = p (by Lemma 4..4). Case (3) : p is of the form p = 8a + 5 for some integer a 0. Here, p = 9(4a + ), so that 8p divides p(p ). Thus, Z(9p) = p. Case (4) : p is of the form p = 8a + 3 for some integer a 0. In this case, 9 divides p + = 9(4a + 3), and consequently, Z(9p) = p. Case (5) : p is of the form p = 8a + 7 for some integer a 0. Here, 9 divides 4p = 9(8a + 3), so that 8p divides 4p(4p ). Thus, Z(9p) = 4p. Case (6) : p is of the form p = 8a + for some integer a 0. In this case, since 9 divides 4p + = 9(8a + 5), it follows that, Z(9p) = 4p. Lemma 4.. : For any prime p 3 with p 5, p, if 0 (p ) p, if 0 (p + ) 5p, if 0 (p 3) 5p, if 0 (p + 3) Z(0p) = 3p, if 0 (p 7) 3p, if 0 (p + 7) 4p, if 0 (p 9) 4p, if 0 (p + 9) Proof : By definition, m(m + ) Z(0p) = min{m : 0p } = min{m : 0p m(m + )}. The eight possible cases are considered below : Case () : p is of the form p = 0a + for some integer a. In this case, 0 divides p, and so, by Lemma 4..4, Z(0p) = p. Case () : p is of the form p = 0a + 9 for some integer a 0. Here, 0 divides p +, and hence, Z(0p) = p (by Lemma 4..4).

109 08 Wandering in the World of Smarandache umbers Case (3) : p is of the form p = 0a + 3 for some integer a 0. Here, 4 divides 5p + = 4(5a + 4), so that 0p divides 5p(5p + ). Thus, Z(0p) = 5p. Case (4) : p is of the form p = 0a + 7 for some integer a 0. In this case, 4 divides 5p = 4(5a + ), and hence, Z(0p) = 5p. Case (5) : p is of the form p = 0a + 7 for some integer a 0. Here, 0 divides 3p = 0(3a + ), and consequently, Z(0p) = 3p. Case (6) : p is of the form p = 0a + 3 for some integer a 0. In this case, 0 divides 3p + = 0(3a + ), and hence, Z(0p) = 3p. Case (7) : p is of the form p = 0a + 9 for some integer a. Here, 5 divides 4p = 5(6a + 7), so that 0p divides 4p(4p ). Thus, Z(0p) = 4p. Case (8) : p is of the form p = 0a + for some integer a 0. In this case, since 5 divides 4p + = 5(6a + 9), it follows that, Z(0p) = 4p. Lemma 4..3 : For any prime p 3 with p, p, if (p ) p, if (p + ) 5p, if (5p + ) 5p, if (5p ) 4p, if (4p ) Z(p) = 4p, if (4p + ) 3p, if (3p ) 3p, if (3p + ) p, if (p + ) p, if (p ) Proof : By definition, Z(p) = min{m : p m(m + ) } = min{m : p m(m + )}. Here, p must divide exactly one of m or m +, and must divide the other. We have to consider ten possible cases. Case () : p is of the form p = a + for some integer a. In this case, divides p, and so, by Lemma 4..4, Z(p) = p. Case () : p is of the form p = a + 0 for some integer a. Here, divides p +, and consequently (by Lemma 4..4), Z(p) = p. Case (3) : p is of the form p = a + for some integer a. In this case, divides 5p + = (5a + ), and hence, Z(p) = 5p. Case (4) : p is of the form p = a + 9 for some integer a. Here, divides 5p = (5a + 4), and therefore, Z(p) = 5p. Case (5) : p is of the form p = a + 3 for some integer a 0. In this case, divides 4p = (4a + ), and hence, Z(p) = 4p.

110 09 Chapter 4 : The Pseudo Smarandache Function Case (6) : p is of the form p = a + 8 for some integer a. Here, divides 4p + = (4a + 3), and consequently, Z(p) = 4p. Case (7) : p is of the form p = a + 4 for some integer a. In this case, divides 3p = (3a + ), and hence, Z(p) = 3p. Case (8) : p is of the form p = a + 7 for some integer a 0. Here, divides 3p + = (3a + ), so that, Z(p) = 3p. Case (9) : p is of the form p = a + 5 for some integer a 0. In this case, divides p + = (a + ), and hence, Z(p) = p. Case (0) : p is of the form p = a + 6 for some integer a. Here, since divides p = (a + ), it follows that, Z(p) = p. Lemma 4..4 : If p 3 is a prime, then p, if 4 (p ) p, if 4 (p + ) 3p, if 8 (3p + ) Z(p) = 3p, if 8 (3p ) 7p, if 4 (7p ) 7p, if 4 (7p + ) Proof : By definition, m(m + ) Z(p) = min{m : p } = min{m : 4p m(m + )}. Here, p must divide exactly one of m or m +. We consider the following eight cases that may arise depending on p : Case () : p is of the form p = 4a + for some integer a. In this case, 4 divides p. Therefore, by Lemma 4..4, Z(p) = p. Case () : p is of the form p = 4a + 3 for some integer a 0. Here, 4 divides p +, and so, Z(p) = p (by Lemma 4..4). Case (3) : p is of the form p = 4a + 5 for some integer a 0. In this case, 8 divides 3p + = 8(9a + ), so that 4p divides 3p(3p + ). Hence, Z(p) = 3p. Case (4) : p is of the form p = 4a + 9 for some integer a 0. Here, 8 divides 3p = 8(9a + 7), and consequently, Z(p) = 3p. Case (5) : p is of the form p = 4a + 7 for some integer a 0. In this case, 4 divides 7p = 4(7a + ), and therefore, Z(p) = 7p. Case (6) : p is of the form p = 4a + 7 for some integer a 0. Here, 4 divides 7p + = 4(7a + 5), and hence, Z(p) = 7p. Case (7) : p is of the form p = 4a + for some integer a 0. In this case, 8 divides 3p = 8(9a + 4), and so, Z(p) = 3p. Case (8) : p is of the form p = 4a + 3 for some integer a 0. Here, 8 divides 3p + = 8(9a + 5), and hence, Z(p) = 3p. To complete the proof of the lemma, note that Z(36) = 8.

111 0 Wandering in the World of Smarandache umbers Lemma 4..5 : For any prime p, p, if 3 (p ) p, if 3 (p + ) 6p, if 3 (6p + ) 6p, if 3 (6p ) 4p, if 3 (4p + ) 4p, if 3 (4p ) Z(3p) = 3p, if 3 (3p + ) 3p, if 3 (3p ) 5p, if 3 (5p + ) 5p, if 3 (5p ) p, if 3 (p + ) p, if 3 (p ) Proof : By definition, m(m + ) Z(3p) = min{m : 3p } = min{m : 6p m(m + )}. Here, p must divide exactly one of m or m +, and 3 must divide the other. In this case, we have to consider the twelve possible cases that may arise : Case () : p is of the form p = 3a + for some integer a. In this case, 3 divides p, and so, by Lemma 4..4, Z(3p) = p. Case (): p is of the form p = 3a + for some integer a. Here, 3 divides p +, and hence, Z(3p) = p (by Lemma 4..4). Case (3) : p is of the form p = 3a + for some integer a 0. In this case, 3 divides 6p + = 3(6a + ), and hence, Z(3p) = 6p. Case (4) : p is of the form p = 3a + for some integer a 0. Here, 3 divides 6p = 3(6a + 5), and hence, Z(3p) = 6p. Case (5) : p is of the form p = 3a + 3 for some integer a 0. In this case, 3 divides 4p + = 3(4a + ), and hence, Z(3p) = 4p. Case (6) : p is of the form p = 3a + 0 for some integer a. Here, 3 divides 4p = 3(4a + 3), and hence, Z(3p) = 4p. Case (7) : p is of the form p = 3a + 4 for some integer a. In this case, 3 divides 3p + = 3(3a + ), and hence, Z(3p) = 3p. Case (8) : p is of the form p = 3a + 9 for some integer a. Here, 3 divides 3p = 3(3a + ), and hence, Z(3p) = 3p. Case (9) : p is of the form p = 3a + 5 for some integer a 0. In this case, 3 divides 5p + = 3(5a + ), and hence, Z(3p) = 5p. Case 0 : p is of the form p = 3a + 8 for some integer a. Here, Z(3p) = 5p. Case : p is of the form p = 3a + 6 for some integer a. In this case, 3 divides p + = 3(a + ), and hence, Z(3p) = p. Case : p is of the form p = 3a + 7 for some integer a 0. Here, Z(3p) = p.

112 Chapter 4 : The Pseudo Smarandache Function In Lemmas , we give the expressions for Z(6p) and Z(3p). Lemma 4..6 : For any prime p 3, p, if 3 (p ) p, if 3 (p + ) p, if 3 (p 3) p, if 3 (p 9) 3p, if 3 (p 5) 3p, if 3 (p 7) 9p, if 3 (p 7) Z(6p) = 9p, if 3 (p 5) 7p, if 3 (p 9) 7p, if 3 (p 3) 3p, if 3 (p ) 3p, if 3 (p ) 5p, if 3 (p 3) 5p, if 3 (p 9) 5p, if 3 (p 5) 5p, if 3 (p 7) Proof : By definition, m(m + ) Z(6p) = min{m : 6p } = min{m : 3p m(m + )}. Here, p must divide exactly one of m or m +. In this case, we have to consider the sixteen possible cases that may arise : Case () : p is of the form p = 3a + for some integer a. In this case, 3 divides p, and so, by Lemma 4..4, Z(6p) = p. Case () : p is of the form p = 3a + 3 for some integer a 0. Here, 3 divides p +, and hence, Z(6p) = p (by Lemma 4..4). Case (3) : p is of the form p = 3a + 3 for some integer a 0. Since p = 3(a + ), it follows that 3 divides p. Hence, Z(6p) = p. Case (4) : p is of the form p = 3a + 9 for some integer a 0. Here, Z(6p) = p. Case (5) : p is of the form p = 3a + 5 for some integer a 0. In this case, 3 divides 3p = 3(3a + ), and hence, Z(6p) = 3p. Case (6) : p is of the form p = 3a + 7 for some integer a. Here, Z(6p) = 3p. Case (7) : p is of the form p = 3a + 7 for some integer a 0. In this case, 9p + = 3(9a + ). Therefore, 3 divides 9p +, and hence, Z(6p) = 9p. Case (8) : p is of the form p = 3a + 5 for some integer a. Here, Z(6p) = 9p. Case (9) : p is of the form p = 3a + 9 for some integer a. In this case, 3 divides 7p + = 3(7a + ), and consequently, Z(6p) = 7p. Case (0) : p is of the form p = 3a + 3 for some integer a 0. Here, Z(6p) = 7p.

113 Wandering in the World of Smarandache umbers Case () : p is of the form p = 3a + for some integer a 0. In this case, 3 divides 3p = 3(3a + ), and hence, Z(6p) = 3p. Case () : p is of the form p = 3a + for some integer a. Here, Z(6p) = 3p. Case (3) : p is of the form p = 3a + 3 for some integer a 0. Since 3 divides 5p = 3(5a + ), it follows that, Z(6p) = 5p. Case (4) : p is of the form p = 3a + 9 for some integer a 0. Here, Z(6p) = 5p. Case (5) : p is of the form p = 3a + 5 for some integer a. In this case, 3 divides 5p = 3(5a + 7), and so, Z(6p) = 5p. Case (6) : p is of the form p = 3a + 7 for some integer a 0. Here, Z(6p) = 5p. Lemma 4..7 : For any prime p 3, p, if 64 (p ) p, if 64 (p + ) p, if 64 (p 3) p, if 64 (p 6) 3p, if 64 (p 5) 3p, if 64 (p 59) 9p, if 64 (p 7) 9p, if 64 (p 57) 7p, if 64 (p 9) 7p, if 64 (p 55) 9p, if 64 (p ) 9p, if 64 (p 53) 5p, if 64 (p 3) 5p, if 64 (p 5) Z(3p) = 7p, if 64 (p 5) 7p, if 64 (p 49) 5p, if 64 (p 7) 5p, if 64 (p 47) 7p, if 64 (p 9) 7p, if 64 (p 45) 3p, if 64 (p ) 3p, if 64 (p 43) 5p, if 64 (p 3) 5p, if 64 (p 4) 3p, if 64 (p 5) 3p, if 64 (p 39) 9p, if 64 (p 7) 9p, if 64 (p 37) p, if 64 (p 9) p, if 64 (p 35) 3p, if 64 (p 3) 3p if 64 (p 33)

114 3 Chapter 4 : The Pseudo Smarandache Function Proof : By definition, m(m + ) Z(3p) = min{m : 3p } = min{m : 64p m(m + )}. Here, p must divide exactly one of m or m +. In this case, we have to consider the 3 possible cases that may arise : Case () : p is of the form p = 64a + for some integer a. In this case, 64 divides p, and so, by Lemma 4..4, Z(3p) = p. Case () : p is of the form p = 64a + 63 for some integer a. Here, 64 divides p +, and hence, Z(3p) = p (by Lemma 4..4). Case (3) : p is of the form p = 64a + 3 for some integer a 0. In this case, p + = 64(a + ), so that 64 divides p +. Hence, Z(3p) = p. Case (4) : p is of the form p = 64a + 6 for some integer a 0. Here, p = 64(a + 0). Thus, 3 (p + ), and consequently, Z(3p) = p. Case (5) : p is of the form p = 64a + 5 for some integer a 0. Since 64 divides 3p = 64(3a + ), it follows that, Z(3p) = 3p. Case (6) : p is of the form p = 64a + 59 for some integer a 0. Here, 64 divides 3p + = 64(3a + ). Therefore, Z(3p) = 3p. Case (7) : p is of the form p = 64a + 7 for some integer a 0. In this case, 9p + = 64(9a + ). Thus, 64 divides 9p +, and hence, Z(3p) = 9p. Case (8) : p is of the form p = 64a + 57 for some integer a. Here, 64 divides 9p = 64(9a + 8), and as such, Z(3p) = 9p. Case (9) : p is of the form p = 64a + 9 for some integer a. Since 64 divides 7p + = 64(7a + ), it follows that, Z(3p) = 7p. Case (0) : p is of the form p = 64a + 55 for some integer a. In this case, 64 divides 7p = 64(7a + 6), and hence, Z(3p) = 7p. Case () : p is of the form p = 64a + for some integer a 0. Here, 9p + = 64(9a + 5), so that 64 divides 9p +. Thus, Z(3p) = 9p. Case () : p is of the form p = 64a + 53 for some integer a 0. In this case, 64 divides 9p = 64(9a + 4), and so, Z(3p) = 9p. Case (3) : p is of the form p = 64a + 3 for some integer a 0. Here, 64 divides 5p = 64(5a + ), and consequently, Z(3p) = 5p. Case (4) : p is of the form p = 64a + 5 for some integer a. Since 64 divides 5p + = 64(5a + 4), it follows that, Z(3p) = 5p. Case (5) : p is of the form p = 64a + 5 for some integer a. In this case, 64 divides 7p + = 64(7a + 4), and hence, Z(3p) = 7p. Case (6) : p is of the form p = 64a + 49 for some integer a. Here, 7p = 64(7a + 3). Thus, 64 divides 7p, and so, Z(3p) = 7p. Case (7) : p is of the form p = 64a + 7 for some integer a 0. In this case, 64 divides 5p + = 64(5a + 4), and consequently, Z(3p) = 5p.

115 4 Wandering in the World of Smarandache umbers Case (8) : p is of the form p = 64a + 47 for some integer a 0. Here, 64 divides 5p = 64(5a + ), and hence, Z(3p) = 5p. Case (9) : p is of the form p = 64a + 9 for some integer a 0. In this case, 7p = 64(7a + 8), so that 64 divides 7p. Hence, Z(3p) = 7p. Case (0) : p is of the form p = 64a + 45 for some integer a. Here, 7p + = 64(7a + 9). Thus, 64 divides 7p +, and so, Z(3p) = 7p. Case () : p is of the form p = 64a + for some integer a. In this case, 3p + = 64(a + ), so that 64p divides 3p(3p + ). Thus, Z(3p) = 3p. Case () : p is of the form p = 64a + 43 for some integer a 0. Here, 3p = 64(3a + ). Thus, 64p divides 3p(3p ), and so, Z(3p) = 3p. Case (3) : p is of the form p = 64a + 3 for some integer a 0. In this case, 5p + = 64(5a + 9). Therefore, 64 divides 5p +, and hence, Z(3p) = 5p. Case (4) : p is of the form p = 64a + 4 for some integer a 0. Here, 64 divides 5p = 64(5a + 6), and hence, Z(3p) = 5p. Case (5) : p is of the form p = 64a + 5 for some integer a. In this case, 64 divides 3p + = 64(3a + 9), and hence, Z(3p) = 3p. Case (6) : p is of the form p = 64a + 39 for some integer a. Here, 64 divides 3p = 64(3a + 4), and hence, Z(3p) = 3p. Case (7) : p is of the form p = 64a + 7 for some integer a. In this case, 9p = 64(9a + 8), so that 64 divides 9p. Thus, Z(3p) = 9p. Case (8) : p is of the form p = 64a + 37 for some integer a 0. Here, 64 divides 9p + = 64(9a + ), and hence, Z(3p) = 9p. Case (9) : p is of the form p = 64a + 9 for some integer a 0. Since p + = 64(a + 5), it follows that 64p divides p(p + ), and so, Z(3p) = p. Case (30) : p is of the form p = 64a + 35 for some integer a. Here, 64 (p ) = 64(a + 6), so that 64p divides p(p ). Thus, Z(3p) = p. Case (3) : p is of the form p = 64a + 3 for some integer a 0. In this case, 64 divides 3p = 64(3a + 5), and hence, Z(3p) = 3p. Case (3) : p is of the form p = 64a + 33 for some integer a. Here, 3p + = 64(3a + 6). Thus, 64 divides 3p +, and consequently, Z(3p) = 3p. All these complete the proof of the lemma. We now give the expressions for Z(4p), Z(48p) and Z(96p) in the following three lemmas. We shall make use of these results in proving Lemma and Lemma 4.4. in 4.4 in connection with equations involving Z(n) and the arithmetic functions like d(n) (the divisor function) and φ(n) (Euler s phi function).

116 5 Chapter 4 : The Pseudo Smarandache Function Lemma 4..8 : For any prime p 5, p, if 48 (p ) p, if 48 (p + ) 3p, if 6 (3p + ) 3p, if 6 (3p ) 7p, if 48 (7p ) 7p, if 48 (7p + ) Z(4p) = p, if 48 (p + ) p, if 48 (p ) 5p, if 48 (5p + ) 5p, if 48 (5p ) 5p, if 8 (5p + ) 5p, if 8 (5p ) 9p, if 6 (9p + ) 9p, if 6 (9p ) Proof : By definition, m(m + ) Z(4p) = min{m : 4p } = min{m : 48p m(m + )}. Here, p must divide exactly one of m or m +. In this case, we have to consider the sixteen possible cases that may arise : Case () : p is of the form p = 48a + for some integer a. In this case, 48 divides p, and so, by Lemma 4..4, Z(4p) = p. Case () : p is of the form p = 48a + 47 for some integer a 0. Here, 48 divides p +, and hence, Z(4p) = p (by Lemma 4..4). Case (3) : p is of the form p = 48a + 5 for some integer a 0. In this case, 3p + = 6(9a + ), so that 48p divides 3p(3p + ). Hence, Z(4p) = 3p. Case (4) : p is of the form p = 48a + 43 for some integer a 0. Here, 3p = 6(9a + 8). Thus, 48p divides 3p(3p ), and consequently, Z(4p) = 3p. Case (5) : p is of the form p = 48a + 7 for some integer a 0. In this case, 48 divides 7p = 48(7a + ), and hence, Z(4p) = 7p. Case (6) : p is of the form p = 48a + 4 for some integer a 0. Here, 48 divides 7p + = 48(7a + 6), and consequently, Z(4p) = 7p. Case (7) : p is of the form p = 48a + for some integer a 0. In this case, 3p = 6(9a + ), and so, 48p divides 3p(3p ). Hence, Z(4p) = 3p. Case (8) : p is of the form p = 48a + 37 for some integer a 0. Here, 6 divides 3p + = 6(9a + 7), so that 48p divides 3p(3p + ). Thus, Z(4p) = 3p. Case (9) : p is of the form p = 48a + 3 for some integer a 0. Since, 48 divides p + = 48(a + 3), we see that, Z(4p) = p. Case (0) : p is of the form p = 48a + 35 for some integer a.

117 6 Wandering in the World of Smarandache umbers In this case, 48 divides p = 48(a + 8), and hence, Z(4p) = p. Case () : p is of the form p = 48a + 7 for some integer a 0. Here, 5p + = 6(45a + 6) so that 48p divides 5p(5p + ), and hence, Z(4p) = 5p. Case () : p is of the form p = 48a + 3 for some integer a 0. Note that, 5p =6(45a + 5). Thus, 48p divides 5p(5p ), and so, Z(4p) = 5p. Case (3) : p is of the form p = 48a + 9 for some integer a 0. In this case, 48 divides 5p + = 48(5a + ), and hence, Z(4p) = 5p. Case (4) : p is of the form p = 48a + 9 for some integer a 0. Here, 48 divides 5p = 48(5a + 3), and hence, Z(4p) = 5p. Case (5) : p is of the form p = 48a + 3 for some integer a 0. Since 9p + = 6(7a + 3), it follows that 48p divides 9p(9p + ), and so, Z(4p) = 9p. Case (6) : p is of the form p = 48a + 5 for some integer a. Here, 9p = 6(7a + 4). Thus, 48p divides 9p(9p ), and hence, Z(4p) = 9p. All these establish the lemma. Lemma 4..9 : For any prime p 5, p, if 96 (p ) p, if 96 (p + ) 9p, if 96 (9p + ) 9p, if 96 (9p ) 9p, if 3 (9p + ) 9p, if 3 (9p ) 3p, if 3 (3p + ) 3p, if 3 (3p ) 7p, if 3 (7p + ) 7p, if 3 (7p ) 5p, if 3 (5p + ) Z(48p) = 5p, if 3 (5p ) 5p, if 96 (5p + ) 5p, if 96 (5p ) 5p, if 96 (5p + ) 5p, if 96 (5p ) p, if 3 (p ) p, if 3 (p + ) 3p, if 96 (3p ) 3p, if 96 (3p + ) p, if 96 (p ) p, if 96 (p + ) 3p, if 96 (3p ) 3p, if 96 (3p + ) 7p, if 3 (7p + ) 7p, if 3 (7p )

118 7 Chapter 4 : The Pseudo Smarandache Function Proof : By definition, Z(48p) = min{m : 96p m(m + )}. Here, p must divide exactly one of m or m +. In this case, we have to consider the 3 possible cases that may arise : Case () : p is of the form p = 96a + for some integer a. In this case, Z(48p) = p. Case () : p is of the form p = 96a + 95 for some integer a. Here, Z(48p) = p. Case (3) : p is of the form p = 96a + 5 for some integer a 0. In this case, 9p + = 96(9a + ), so that 96p divides 9p +. Hence, Z(48p) = 9p. Case (4) : p is of the form p = 96a + 9 for some integer a. Here, Z(48p) = 9p. Case (5) : p is of the form p = 96a + 7 for some integer a 0. In this case, 9p + = 3(7a + ), so that 96p divides 9p(9p + ). Hence, Z(48p) = 9p. Case (6) : p is of the form p = 96a + 89 for some integer a 0. In this case, Z(48p) = 9p. Case (7) : p is of the form p = 96a + for some integer a 0. Here, 3 divides 3p = 3(9a + ), and so, 96p divides 3p(3p ). Thus, Z(48p) = 3p. Case (8) : p is of the form p = 96a + 85 for some integer a. In this case, Z(48p) = 3p. Case (9) : p is of the form p = 96a + 3 for some integer a 0. Since 3 divides 7p + = 3(8a + ), we see that, Z(48p) = 7p. Case (0) : p is of the form p = 96a + 83 for some integer a 0. Here, Z(48p) = 7p. Case () : p is of the form p = 96a + 7 for some integer a 0. Here, 3 divides 5p + = 3(45a + 8), so that 96p divides 5p(5p + ). Thus, Z(48p) = 5p. Case () : p is of the form p = 96a + 79 for some integer a 0. Here, Z(48p) = 5p. Case (3) : p is of the form p = 96a + 9 for some integer a 0. In this case, 96 divides 5p + = 96(5a + ), and hence, Z(48p) = 5p. Case (4) : p is of the form p = 96a + 77 for some integer a. Here, Z(48p) = 5p. Case (5) : p is of the form p = 96a + 3 for some integer a 0. Since 5p + = 96(5a + 6), it follows that 96 divides 5p +, and so, Z(48p) = 5p. Case (6) : p is of the form p = 96a + 73 for some integer a 0. Here, 5p = 96(5a + 9). Thus, 96 divides 5p, and hence, Z(48p) = 5p. Case (7) : p is of the form p = 96a + 5 for some integer a. Here, Z(48p) = 9p. Case (8) : p is of the form p = 96a + 7 for some integer a 0. In this case, Z(48p) = 9p. Case (9) : p is of the form p = 96a + 9 for some integer a 0. Here, 3 divides p = 3(63a + 9), so that 96p divides p(p ). Thus, Z(48p) = p. Case (0) : p is of the form p = 96a + 67 for some integer a 0. In this case, Z(48p) = p. Case () : p is of the form p = 96a + 3 for some integer a 0. In this case, 96 divides 3p = 96(3a + 0), and hence, Z(48p) = 3p. Case () : p is of the form p = 96a + 65 for some integer a. In this case, Z(48p) = 3p. Case (3) : p is of the form p = 96a + 35 for some integer a. Here, 96 divides p = 96(a + 4), and so, Z(48p) = p.

119 8 Wandering in the World of Smarandache umbers Case (4) : p is of the form p = 96a + 6 for some integer a 0. Here, Z(48p) = p. Case (5) : p is of the form p = 96a + 37 for some integer a 0. Here, 96 divides 3p = 96(3a + 5), and so, Z(48p) = 3p. Case (6) : p is of the form p = 96a + 59 for some integer a 0. In this case, Z(48p) = 3p. Case (7) : p is of the form p = 96a + 4 for some integer a 0. Here, 96 divides 7p + = 96(7a + 3), so that, Z(48p) = 7p. Case (8) : p is of the form p = 96a + 55 for some integer a. Here, Z(48p) = 7p. Case (9) : p is of the form p = 96a + 43 for some integer a 0. Here, 3 divides 3p = 3(9a + 4), so that, 96p divides 3p(3p ). Hence, Z(48p) = 3p. Case (30) : p is of the form p = 96a + 53 for some integer a 0. In this case, Z(48p) = 3p. Case (3) : p is of the form p = 96a + 47 for some integer a 0. Here, 3 divides 5p = 3(45a + ), and so, 96p divides 5p(5p ). Thus, Z(48p) = 5p. Case (3) : p is of the form p = 96a + 49 for some integer a 0. Here, Z(48p) = 5p. Hence the lemma is established. Lemma : For any prime p 5, the values of Z(96p) are as follows : Form of p Z(96p) Form of p Z(96p) Form of p Z(96p) 9a + p 9a + 35 p 9a + 67 p 9a + 9 p 9a + 57 p 9a + 5 p 9a + 5 5p 9a p 9a + 7 9p 9a p 9a p 9a + 9p 9a + 7 9p 9a p 9a p 9a p 9a p 9a p 9a + 35p 9a p 9a p 9a p 9a p 9a + 5 5p 9a p 9a p 9a p 9a p 9a p 9a + 3 7p 9a+7 5p 9a p 9a p 9a p 9a p 9a p 9a+9 7p 9a p 9a p 9a p 9a p 9a p 9a + 3 5p 9a p 9a p 9a p 9a p 9a p 9a + 5 3p 9a p 9a + 9 9p 9a+67 3p 9a p 9a + 0 9p 9a p 9a + 6 p 9a p 9a p 9a + 3 p 9a p 9a + 3 3p 9a p 9a + 6 3p 9a p

120 9 Chapter 4 : The Pseudo Smarandache Function Proof : By definition, Z(96p) = min{m : 9p m(m + )}. Here, p must divide exactly one of m or m +. In this case, we have to consider the 64 possible cases that may arise : Case () : p is of the form p = 9a + for some integer a. In this case, Z(96p) = p. Case () : p is of the form p = 9a + 9 for some integer a 0. Here, Z(96p) = p. Case 3 : p is of the form p = 9a + 5 for some integer a 0. In this case, 5p + = 64(53a + 4), so that 9p divides 5p(5p + ). Hence, Z(96p) = 5p. Case (4) : p is of the form p = 9a + 87 for some integer a. Here, 9p divides 5p(5p ) = 9(7p).(53a + 49), and consequently, Z(96p) = 5p. Case (5) : p is of the form p = 9a + 7 for some integer a 0. In this case, 9p + = 64(7a + ), so that 9p divides 9p(9p + ). Hence, Z(96p) = 9p. Case (6) : p is of the form p = 9a + 85 for some integer a. Here, 9 divides 9p(9p ) = 9(3p).(7a + 6), and consequently, Z(96p) = 9p. Case (7) : p is of the form p = 9a + for some integer a 0. In this case, 9 divides 35p = 9(35a + ), and so, Z(96p) = 35p. Case (8) : p is of the form p = 9a + 8 for some integer a 0. Here, 9 divides 35p + = 9(35a + 33). And consequently, Z(96p) = 35p. Case (9) : p is of the form p = 9a + 3 for some integer a 0. Since, 9 divides 59p + = 9(59a + 4), we see that, Z(96p) = 59p. Case (0) : p is of the form p = 9a + 79 for some integer a 0. In this case, 9 divides 59p = 9(59a + 55), and hence, Z(96p) = 59p. Case () : p is of the form p = 9a + 7 for some integer a 0. Here, 64 divides 5p + = 64(45a + 4), so that 9p divides 5p(5p + ). Thus, Z(96p) = 5p. Case () : p is of the form p = 9a + 75 for some integer a. In this case, 9p divides 5p(5p ) = 9(5p).(45a + 4), and so, Z(96p) = 5p. Case (3) : p is of the form p = 9a + 9 for some integer a 0. In this case, 64 divides 7p = 64(8a + 8) and 3p divides 7p, and hence, Z(96p) = 7p. Case (4) : p is of the form p = 9a + 73 for some integer a 0. Here, 64 divides 7p + = 64(8a + 73), and hence, Z(96p) = 7p. Case (5) : p is of the form p = 9a + 3 for some integer a 0. Since 5p + = 9(5a + 3), it follows that 9 divides 5p +, and so, Z(96p) = 5p. Case (6) : p is of the form p = 9a + 69 for some integer a. Here, 5p = 9(5a + ). Thus, 9 divides 5p, and hence, Z(96p) = 5p. Case (7) : p is of the form p = 9a + 5 for some integer a. Here, 3p + = 9(3a + 3), so that, 9 divides 3p +. Therefore, Z(96p) = 3p. Case (8) : p is of the form p = 9a + 67 for some integer a 0. In this case, 3p = 9(3a + 0), and so, 9 divides 3p. Thus, Z(96p) = 3p.

121 0 Wandering in the World of Smarandache umbers Case (9) : p is of the form p = 9a + 9 for some integer a 0. Here, 9 divides 53p = 9(53a + 8), so that, Z(96p) = 53p. Case (0) : p is of the form p = 9a + 63 for some integer a 0. Since, 9 divides 53p + = 9(53a + 45), it follows that, Z(96p) = 53p. Case () : p is of the form p = 9a + 3 for some integer a 0. In this case, 9 divides 3p = 9(3a + 5), and hence, Z(96p) = 3p. Case () : p is of the form p = 9a + 6 for some integer a. Since, 9 divides 3p + = 9(3a + 6), it follows that, Z(96p) = 3p. Case (3) : p is of the form p = 9a + 35 for some integer a. Here, 9 divides p = 9(a + ), and so, Z(96p) = p. Case (4) : p is of the form p = 9a + 57 for some integer a 0. Since, 9 divides p + = 9(a + 9), it follows that, Z(96p) = p. Case (5) : p is of the form p = 9a + 37 for some integer a 0. Here, 64 divides 45p = 64(35a + 6), and so, Z(96p) = 45p. Case (6) : p is of the form p = 9a + 55 for some integer a. In this case, 64 divides 45p + = 64(3a + 09), and hence, Z(96p) = 45p. Case (7) : p is of the form p = 9a + 4 for some integer a 0. Here, 64 divides 39p + = 64(7a + 5), so that, Z(96p) = 39p. Case (8) : p is of the form p = 9a + 5 for some integer a. Since 64 divides 39p = 64(7a + 9), it follows that, Z(96p) = 39p. Case (9) : p is of the form p = 9a + 43 for some integer a 0. Here, 64 divides 3p = 64(9a + ), so that 9p divides 3p(3p ). Hence, Z(96p) = 3p. Case (30) : p is of the form p = 9a + 49 for some integer a 0. Then, Z(96p) = 3p. Case (3) : p is of the form p = 9a + 47 for some integer a 0. Here, 64 divides 5p = 64(45a + ), and 9p divides 5p(5p ). Thus, Z(96p) = 5p. Case (3) : p is of the form p = 9a + 45 for some integer a. In this case, 64 divides 5p + = 64(45a + 34). Consequently, Z(96p) = 5p. Case (33) : p is of the form p = 9a + 49 for some integer a. In this case, 9 divides 47p + = 9(47a + ), and consequently, Z(96p) = 47p. Case (34) : p is of the form p = 9a + 43 for some integer a. Here, Z(96p) = 47p. Case (35) : p is of the form p = 9a + 53 for some integer a 0. In this case, 9 divides 9p = 9(9a + 8), so that, Z(96p) = 9p. Case (36) : p is of the form p = 9a + 39 for some integer a 0. Here, Z(96p) = 9p. Case (37) : p is of the form p = 9a + 55 for some integer a. In this case, 9 divides 7p = 9(7a + ). Hence, Z(96p) = 7p. Case (38) : p is of the form p = 9a + 37 for some integer a 0. Here, 9 divides 7p + = 9(7a + 5), and consequently, Z(96p) = 7p.

122 Chapter 4 : The Pseudo Smarandache Function Case (39) : p is of the form p = 9a + 59 for some integer a 0. In this case, 9 divides 3p + = 9(3a + 4), and so, Z(96p) = 3p. Case (40) : p is of the form p = 9a + 33 for some integer a. Here, 9 divides 3p = 9(3a + 9). And consequently, Z(96p) = 3p. Case (4) : p is of the form p = 9a + 6 for some integer a 0. In this case, 64 divides p = 64(63a + 0). Therefore, Z(96p) = p. Case (4) : p is of the form p = 9a + 3 for some integer a 0. Since, 64 divides p + = 64(63a + 43), it follows that, Z(96p) = p. Case (43) : p is of the form p = 9a + 65 for some integer a. Here, 64 divides 63p + = 64(89a + 64), and so, 9p divides 63p(63p + ). Thus, Z(96p) = 63p. Case (44) : p is of the form p = 9a + 7 for some integer a 0. In this case, 9p divides 63p(63p ), and so, Z(63p) = 63p. Case (45) : p is of the form p = 9a + 67 for some integer a 0. In this case, 64 divides p + = 64(63a + ), and hence, Z(96p) = p. Case (46) : p is of the form p = 9a + 5 for some integer a. Here, 64 divides p = 64(63a + 4), and hence, Z(96p) = p. Case (47) : p is of the form p = 9a + 7 for some integer a 0. Since 9p + = 64(7a + 0), it follows that 9p divides 9p(9p + ), and so, Z(96p) = 9p. Case (48) : p is of the form p = 9a + for some integer a. Here, 9p = 64(7a + 7). Thus, 9p divides 9p(9p ), and hence, Z(96p) = 9p. Case (49) : p is of the form p = 9a + 73 for some integer a 0. In this case, 57p = 64(7a + 65), and 3p divides 57p. Therefore, Z(96p) = 57p. Case (50) : p is of the form p = 9a + 9 for some integer a. Here, 57p + = 64(7a + 06). Therefore, Z(96p) = 57p. Case (5) : p is of the form p = 9a + 77 for some integer a. In this case, 9 divides 5p = 9(5a + ), and hence, Z(96p) = 5p. Case (5) : p is of the form p = 9a + 5 for some integer a. Here, Z(96p) = 5p. Case (53) : p is of the form p = 9a + 79 for some integer a 0. Here, 9 divides 7p + = 9(7a + 7), and so, Z(96p) = 7p. Case (54) : p is of the form p = 9a + 3 for some integer a 0. Since, 9 divides 7p = 9(7a + 0), it follows that, Z(96p) = 7p. Case (55) : p is of the form p = 9a + 83 for some integer a 0. Here, 64 (7p ) = 64(8a + 35), so that 9p divides 7p(7p ). Thus, Z(96p) = 7p. Case (56) : p is of the form p = 9a + 09 for some integer a 0. Here, Z(96p) = 7p. Case (57) : p is of the form p = 9a + 85 for some integer a. Here, 64 divides 3p + = 64(9a + 4), so that 9p divides 3p(3p + ). Thus, Z(96p) = 3p. Case (58) : p is of the form p = 9a + 07 for some integer a 0. Here, Z(96p) = 3p.

123 Wandering in the World of Smarandache umbers Case (59) : p is of the form p = 9a + 89 for some integer a 0. Here, 9 divides 4p = 9(4a + 9), so that, Z(96p) = 4p. Case (60) : p is of the form p = 9a + 03 for some integer a 0. Here, Z(96p) = 4p. Case (6) : p is of the form p = 9a + 9 for some integer a. Here, 9 divides 9p = 9(9a + 9). Hence, Z(96p) = 9p. Case (6) : p is of the form p = 9a + 0 for some integer a 0. Here, Z(96p) = 9p. Case (63) : p is of the form p = 9a + 95 for some integer a. Here, 64 divides 33p + = 64(99a + 49), and so, 9p divides 33p(33p + ). Hence, Z(96p) = 33p. Case (64) : p is of the form p = 9a + 97 for some integer a 0. Here, Z(96p) = 33p. All these complete the proof of the lemma. Let When n is a composite number, we observe the following : Z(n) = m o for some integer m o (so that n divides We now consider the following two cases that may arise : Case : m o is even (so that m o + is odd). m 0 m 0 (m0 + ) ). In this case, n does not divide, for otherwise, m n 0 m 0 (m0 ) n Z(n) m o. Case : m o is odd (so that m o + is even). If n is even, then, n does not divide m o. If n is odd, n does not divide m o, for n m o m 0 (m0 ) n Z(n) m o. Thus, if n is a composite number, n does not divide m o. Now, let the representation of n in terms of its distinct prime factors, p, p,, p s, be α α α α α α n p p... p p... p i i+ s = i i+ s i ( α 0, α 0 for all i s), where p, p,, p s are not necessarily ordered. Then, one of m o and m o + is divisible by α + α α α p p... p i i and the other one is divisible by α α + α p p... p j j s j j+ s for some i < s; for i + j s. We now proceed to find an expression for Z(pq), where p and q are two distinct primes. The following result is due to Ibstedt [].

124 3 Chapter 4 : The Pseudo Smarandache Function Theorem 4.. : Let p and q be two primes with q > p. Let g = q p. Then, p(qk + ) q(pk ) Z(pq) = min{, }, g g where both pk + and pk are divisible by g. m(m + ) Proof : By definition, Z(p q) = min{m : p q }. Now, we consider the three possible cases separately. Case () : When p divides m and q divides (m + ). In this case, m = px, m + = qy for some integers x, y, leading to the equation p(x y) = gy. This shows that p divides gy. Now, gy = pk y = pk +, x = qk + p(qk + ), m =. g g g Case () : When p divides (m + ) and q divides m. In this case, m + = px, m = qy for some integers x, y. Thus, we get the equation p(x y) = gy +. This shows that p divides gy +. Now, gy + = pk y = pk, m = g p(qk + ) g Case (3) : When pq divides (m + ). This case cannot occur (see Case 3 in the proof of Theorem 4..). Hence, the theorem is proved. In Theorem 4.., we give an alternative expression for Z(pq). In this connection, we state the following lemma (see, for example, Theorem.. of Gioia [5]). Lemma 4..3 : Let p and q be two distinct primes. Then, the Diophantine equation q y p x = has an infinite number of solutions. Moreover, if (x o, y o ) is a solution of the Diophantine equation, then any solution is of the form x = x o + qt, y = y o + pt, where t 0 is an integer. Theorem 4.. : Let p and q be two primes with q > p 5. Then, Z(pq) = min{qy o, px o }, where y o = min{y : x, y Z +, q y p x = }, x o = min{x : x, y Z +, p x q y = }. Proof : Since m(m + ) Z(pq) = min{m : pq }, (A) it follows that we have to consider the three cases below that may arise : Case : When p divides m and q divides (m + ). In this case,.

125 4 Wandering in the World of Smarandache umbers m = p x for some integer x, m + = q y for some integer y. From these two equations, we get the Diophantine equation q y p x =. By Lemma 4..3, the above Diophantine equation has infinite number of solutions. Let y o = min{y : x, y Z +, q y p x = }. For this y o, the corresponding x o is given by the equation q o y p o x =. Note that y o and x o cannot be both odd or both even. Then, the minimum m in (A ) is given by m + = q y o m = q y o. Case : When p divides (m + ) and q divides m. Here, m + = p x for some integer x, m = q y for some integer y. These two equations lead to the Diophantine equation. p x q y =. Let x o = min{x : x, y Z +, p x q y = }. px0 For this x o, the corresponding y o is given by y o =. Here also, xo and y o both cannot be q odd or even simultaneously. The minimum m in (A) is given by m + = p x o m = p x o. Case 3 : When pq divides (m + ). In this case, m = p q. But then, by Case and Case above, this does not give the minimum m. Thus, this case cannot occur. The proof of the theorem now follows by virtue of Case and Case. Remark 4.. : Let p and q be two primes with q > p 5. Let q = k p + l for some integers k and l with k and l p. We now consider the two cases given in Theorem 4.. : Case : When p divides m and q divides (m + ). In this case, m = p x for some integer x, m + = q y = (k p + l)y for some integer y. From the above two equations, we get l y (x k y)p =. (4..) Case : When p divides (m + ) and q divides m. Here, m + = p x for some integer x, m = (k p + l)y for some integer y. These two equations lead to (x k y)p l y =. (4..)

126 5 Chapter 4 : The Pseudo Smarandache Function Some particular cases are given in Corollary 4..3 Corollary 4..8 below. Corollary 4..3 : Let p and q be two primes with q > p 5. Let q = k p + for some integer k. Then, Z(pq) = q. Proof : From (4..) with l =, we get y (x k y)p =, the minimum solution of which is y =, x = k y = k. Then, the minimum m in (A) is given by m + = q y = q m = q. Note that, from (4..) with l =, we have (x k y)p y =, with the least possible solution y = p (and x k y = ). Corollary 4..4 : Let p and q be two primes with q > p 5. Let q = (k + )p for some integer k. Then, Z(pq) = q. Proof : From (4..) with l = p, we have, y [(k + )y x]p =, whose minimum solution is y =, x = (k + )y = k +. Then, the minimum m in (A) is given by m = q y = q. Note that, from (4..) with l = p, we have [(k + )y x]p y =, with the least possible solution y = p (and (k + )y x = ). Corollary 4..5 : Let p and q be two primes with q > p 5. Let q = k p + for some integer k. Then, q(p ) Z(p q) =. Proof : With l = in (4..), we have (x k y)p y =, with the minimum solution p y = (and x k y = ). q(p ) This gives m = q y = as one possible solution of (A). Now, (4..) with l = gives y (x k y)p =, with the minimum solution

127 6 Wandering in the World of Smarandache umbers p + y = (and x = k y + ). This gives q(p + ) m = q y = as another possible solution of (A). Now, since q(p + ) q(p ) >, it follows that q(p ) Z(pq) =, which we intended to prove. Corollary 4..6 : Let p and q be two primes with q > p 5. Let q = (k + )p for some integer k. Then, q(p ) Z(p q) =. Proof : From (4..) with l = p, we get, [(k + )y x]p y =, whose minimum solution is p y = (and x = (k + )y ). q(p ) This gives m = q y = as one possible solution of (A). Note that, (4..) with l = p gives y [(k + )y x]p =, with the minimum solution p + y = (and x = (k + )y ). Corresponding to this case, we get q(p + ) m = q y = as another possible solution of (A). But since q(p + ) q(p ) >, it follows that q(p ) Z(pq) =, establishing the desired result.

128 7 Chapter 4 : The Pseudo Smarandache Function Corollary 4..7 : Let p and q be two primes with q > p 7. Let q = k p + 3 for some integer k. Then, q(p ), if 3 (p ) 3 Z(pq) = q(p + ), if 3 (p + ) 3 Proof : From (4..) and (4..) with l = 3, we have respectively 3y (x k y)p =, (i) (x k y)p 3 y =. (ii) We now consider the following two possible cases : Case : When 3 divides (p ). In this case, the minimum solution is obtained from (ii), which is p y = (and x k y = ). 3 Also, p is divisible by as well. Therefore, the minimum m in (A) is q(p ) m = q y =. 3 Case : When 3 divides (p + ). In this case, (i) gives the minimum solution, which is p + y = (and x k y = ). 3 Note that, divides p +. Therefore, the minimum m in (A) is q(p + ) m = q y =. 3 Thus, the result is established. Corollary 4..8 : Let p and q be two primes with q > p 7. Let q = (k + )p 3 for some integer k. Then, q(p + ), if 3 (p + ) 3 Z(pq) = q(p ), if 3 (p ) 3 Proof : The equations (4..) and (4..) with l = p 3 give respectively [(k + )y x]p 3y =, (i) 3y [(k + )y x]p =. (ii) We now consider the following two cases : Case : When 3 divides (p + ). In this case, the minimum solution, obtained from (ii), is

129 8 Wandering in the World of Smarandache umbers p + y = (and x = (k + )y ). 3 Moreover, divides p +. Therefore, the minimum m in (A) is q(p + ) m = q y =. 3 Case : When 3 divides (p ). In this case, the minimum solution, obtained from (i), is p y = (and x = (k + )y ). 3 Moreover, divides p. Therefore, the minimum m in (A) is q(p ) m = q y =. 3 Corollary 4..9 : Let p and q be two primes with q > p 7. Let q = k p + 4 for some integer k. Then, q(p ), if 4 (p ) 4 Z(pq) = q(p + ), if 4 (p + ) 4 Proof : With l = 4, (4..) and (4..) take the respective forms 4y (x k y)p =, (i) (x k y)p 4y =. (ii) Now, for any prime p 7, exactly one of the following two cases can occur : Either p is divisible by 4, or p + is divisible by 4. We consider these two possibilities separately below : Case : When 4 divides (p ). In this case, the minimum solution is obtained from (ii), which is p y = (and x = k y + ). 4 Therefore, the minimum m in (A) is q(p ) m = q y =. 4 Case : When 4 divides (p + ). In this case, (i) gives the minimum solution, which is p + y = (and x = k y + ). 4 Therefore, the minimum m in (A) is q(p + ) m = q y =. 4

130 9 Chapter 4 : The Pseudo Smarandache Function Corollary 4..0 : Let p and q be two primes with q > p 7. Let q = (k + )p 4 for some integer k. Then, q(p + ), if 4 (p + ) 4 Z(pq) = q(p ), if 4 (p + ) 4 Proof : From (4..) and (4..) with l = p 4, we have respectively [(k + )y x]p 4y =, (i) 4y [(k + )y x]p =. (ii) We now consider the following two cases which are the only possibilities (as noted in the proof of Corollary 4..9) : Case : When 4 divides (p + ). In this case, the minimum solution obtained from (ii) is p + y = (and x = (k + )y ). 4 q(p + ) Therefore, the minimum m in (A) is m = q y =. 4 Case : When 4 divides (p ). In this case, the minimum solution, obtained from (i), is p y = (and x = (k + )y ). 4 q(p ) Therefore, the minimum m in (A) is m = qy =. 4 Corollary 4.. : Let p and q be two primes with q > p. Let q = k p + 5 for some integer k. Then, q(p ) q(p ), if 5 (p ), if 5 (p ) 5 5 q(p + ) q(a + ), if p = 5a +, if 5 (p ) 5 Z(pq) = = q(a + ), if p = 5a + 3 q(p ), if 5 (p 3) q(p + ) 5, if 5 (p + ) 5 q(p + ), if 5 (p + ) 5 Proof : The equations (4..) and (4..) with l = 5 are respectively 5y (x k y)p =, (i) (x k y)p 5y =. (ii) Now, for any prime p 7, exactly one of the following four cases occur :

131 30 Wandering in the World of Smarandache umbers Case : When p is of the form p = 5a + for some integer a. In this case, 5 divides (p ). Then, the minimum solution is obtained from (ii), which is p q(p ) y = (and x k y = ). Therefore, the minimum m in (A) is m = q y =. 5 5 Case : When p is of the form p = 5a + for some integer a. In this case, from (i) and (ii), we get respectively = 5y (x k y)(5a + ) = 5[y (x k y)a] (x k y), (iii) = (x k y)(5a + ) 5y = (x k y) 5[y (x k y)a]. (iv) Clearly, the minimum solution is obtained from (iii), which is y (x k y)a =, x k y = y = a + (and x = k(a + ) + ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 3 : When p is of the form p = 5a + 3 for some integer a. From (i) and (ii), we get = 5y (x k y)(5a + 3) = 5[y (x k y)a] 3(x k y), (v) = (x k y)(5a + 3) 5y = 3(x k y) 5[y (x k y)a]. (vi) The minimum solution is obtained from (vi) as follows : y (x k y)a =, x k y = y = a + (and x = k(a + ) + ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 4 : When p is of the form p = 5a + 4 for some integer a. In this case, 5 divides (p + ). Then, the minimum solution, obtained from (i), is y = p + 5 (and x k y = ). Therefore, the minimum m in (A) is m = q y = q(p + ) 5. Corollary 4.. : Let p and q be two primes with q > p. Let q = (k + )p 5 for some integer k. Then, q(p ) q(p ), if 5 (p ), if 5 (p ) 5 5 q(p + ) q(a + ), if p = 5a +, if 5 (p ) 5 Z(pq) = = q(a + ), if p = 5a + 3 q(p ), if 5 (p 3) q(p + ) 5, if 5 (p + ) 5 q(p + ), if 5 (p + ) 5 Proof : From (4..) and (4..) with l = p 5, we have respectively [(k + )y x]p 5y =, (i) 5y [(k + )y x]p =. (ii)

132 3 Chapter 4 : The Pseudo Smarandache Function As in the proof of Corollary 4.., we consider the following four possibilities : Case : When p is of the form p = 5a + for some integer a. In this case, 5 divides (p ). Then, the minimum solution is obtained from (i), which is p y = (and x = (k + )y ). 5 Therefore, the minimum m in (A) is q(p ) m = q y =. 5 Case : When p is of the form p = 5a + for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](5a + ) 5y = [(k + )y x] 5[y a{(k + )y x}], (iii) = 5y [(k + )y x](5a + ) = 5[y a{(k + )y x}] [(k + )y x]. (iv) Clearly, the minimum solution is obtained from (iv), which is y a{(k + )y x} =, (k + )y x = y = a + (and x = (k + )(a + ) ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 3 : When p is of the form p = 5a + 3 for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](5a + 3) 5y = 3[(k + )y x] 5[y a{(k + )y x}], (v) = 5y [(k + )y x](5a + 3) = 5[y a{(k + )y x}] 3[(k + )y x]. (vi) The minimum solution is obtained from (v) as follows : y a{(k + )y x} =, (k + )y x = y = a + (and x = (k + )(a + ) ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 4 : When p is of the form p = 5a + 4 for some integer a. In this case, 5 divides (p + ). Then, the minimum solution is obtained from (ii). The minimum y and the minimum m in (A) are given by y = p + 5 (and x = (k + )y ), m = q y = q(p + ) 5 Corollary 4..3 : Let p and q be two primes with q > p 3. Let q = k p + 6 for some integer k. Then, q(p ), if 6 (p ) 6 Z(pq) = q(p + ), if 6 (p + ) 6 Proof : The equations (4..) and (4..) with l = 6 become respectively 6y (x k y)p =, (i) (x k y)p 6y =. (ii) Now, for any prime p 3, exactly one of the following two cases can occur : Either p is divisible by 6, or p + is divisible by 6. We thus consider the two possibilities separately below :.

133 3 Wandering in the World of Smarandache umbers Case : When 6 divides (p ). In this case, the minimum solution is obtained from (ii), which is p y = (and x = k y + ). 6 Therefore, the minimum m in (A) is q(p ) m = q y =. 6 Case : When 6 divides (p + ). In this case, (i) gives the minimum solution, which is p + y = (and x = k y + ). 6 Therefore, the minimum m in (A) is q(p + ) m = q y =. 6 Corollary 4..4 : Let p and q be two primes with q > p 3. Let q = (k + )p 6 for some integer k. Then, q(p + ), if 6 (p + ) 6 Z(pq) = q(p ), if 6 (p ) 6 Proof : From (4..) and (4..) with l = p 6, we have respectively [(k + )y x]p 6y =, (i) 6y [(k + )y x]p =. (ii) We now consider the following two cases which are the only possibilities (as noted in the proof of Corollary 4..3) : Case : When 6 divides (p + ). In this case, the minimum solution, obtained from (ii), is p + y = (and x = (k + )y ). 6 Therefore, the minimum m in (A) is q(p + ) m = q y =. 6 Case : When 6 divides (p ). Here, the minimum solution is obtained from (i), which is p y = (and x = (k + )y ). 6 Therefore, the minimum m in (A) is q(p ) m = q y =. 6

134 33 Chapter 4 : The Pseudo Smarandache Function Corollary 4..5 : Let p and q be two primes with q > p 3. Let q = k p + 7 for some integer k. Then, q(p ) q(p ), if 7 (p ) 7, if 7 (p ) 7 q(3p+ ) q(3a + ), if p = 7a + 7, if 7 (p ) q(p+ ) q(a + ), if p = 7a + 3 Z(pq) = 7, if 7 (p 3) = q(a + ), if p = 7a + 4 q(p ) 7, if 7 (p 4) q(3a + ), if p = 7a + 5 q(3p ) q(p + ) 7, if 7 (p 5), if 7 (p + ) q(p+ ) 7 7, if 7 (p + ) Proof : The equations (4..) and (4..) with l = 7 are respectively 7y (x k y)p =, (i) (x k y)p 7y =. (ii) Now, for any prime p, exactly one of the following six cases occur : Case : When p is of the form p = 7a + for some integer a. In this case, 7 divides (p ). Then, the minimum solution is obtained from (ii), which is p y = (and x k y = ). Therefore, the minimum m in (A) is 7 q(p ) m = q y =. 7 Case : When p is of the form p = 7a + for some integer a. In this case, from (i) and (ii), we get respectively = 7y (x k y)(7a + ) = 7[y (x k y)a] (x k y), (iii) = (x k y)(7a + ) 7y = (x k y) 7[y (x k y)a]. (iv) Clearly, the minimum solution is obtained from (iii), which is y (x k y)a =, x k y = 3 y = 3a + (and x = k(3a + ) + 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 3 : When p is of the form p = 7a + 3 for some integer a. Here, from (i) and (ii), = 7y (x k y)(7a + 3) = 7[y (x k y)a] 3(x k y), (v) = (x k y)(7a + 3) 7y = 3(x k y) 7[y (x k y)a]. (vi) The minimum solution is obtained from (v) as follows : y (x k y)a =, x k y = y = a + (and x = k(a + ) + ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 4 : When p is of the form p = 7a + 4 for some integer a. In this case, from (i) and (ii), we get respectively = 7y (x k y)(7a + 4) = 7[y (x k y)a] 4(x k y), (vii) = (x k y)(7a + 4) 7y = 4(x k y) 7[y (x k y)a]. (viii)

135 34 Wandering in the World of Smarandache umbers Clearly, the minimum solution is obtained from (viii), which is y (x k y)a =, x k y = y = a + (and x = k(a + ) + ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 5 : When p is of the form p = 7a + 5 for some integer a. From (i) and (ii), we have = 7y (x k y)(7a + 5) = 7[y (x k y)a] 5(x k y), (ix) = (x k y)(7a + 3) 7y = 5(x k y) 7[y (x k y)a]. (x) The minimum solution is obtained from (x), which is y (x k y)a =, x k y = 3 y = 3a + (and x = k(3a + ) + 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 6 : When p is of the form p = 7a + 6 for some integer a. In this case, 7 divides (p + ). Then, the minimum solution is obtained from (i), which is p + y = (and x k y = ). 7 Therefore, the minimum m in (A) is q(p + ) m = q y =. 7 Corollary 4..6 : Let p and q be two primes with q > p 3. Let q = (k + )p 7 for some integer k. Then, q(p ) q(p ) 7, if 7 (p ), if 7 (p ) 7 q(3p+ ) 7, if 7 (p ) q(3a + ), if p = 7a + q(p+ ) q(a + ), if p = 7a + 3 7, if 7 (p 3) Z(pq) = = q(a + ), if p = 7a + 4 q(p ) 7, if 7 (p 4) q(3a + ), if p = 7a + 5 q(3p ) q(p + ) 7, if 7 (p 5), if 7 (p + ) 7 q(p+ ) 7, if 7 (p + ) Proof : From (4..) and (4..) with l = p 7, we have respectively [(k + )y x]p 7y =, (i) 7y [(k + )y x]p =. (ii) We now consider the following six possibilities : Case : When p is of the form p = 7a + for some integer a. In this case, 7 divides (p ). Then, the minimum solution is obtained from (i), which is p y = (and x = (k + )y ). 7

136 35 Chapter 4 : The Pseudo Smarandache Function Therefore, the minimum m in (A) is q(p ) m = q y =. 7 Case : When p is of the form p = 7a + for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](7a + ) 7y = [(k + )y x] 7[y a{(k + )y x}], (iii) = 7y [(k + )y x](7a + ) = 7[y a{(k + )y x}] [(k + )y x]. (iv) Clearly, the minimum solution is obtained from (iv), which is y a{(k + )y x} =, (k + )y x = 3 y = 3a + (and x = (k + )(3a + ) 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 3 : When p is of the form p = 7a + 3 for some integer a. Here, from (i) and (ii), = [(k + )y x](7a + 3) 7y = 3[(k + )y x] 7[y a{(k + )y x}], (v) = 7y [(k + )y x](7a + 3) = 7[y a{(k + )y x}] 3[(k + )y x]. (vi) Then, (vi) gives the minimum solution, which is y a{(k + )y x} =, (k + )y x = y = a + (and x = (k + )(a + ) ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 4 : When p is of the form p = 7a + 4 for some integer a. Here, from (i) and (ii), = [(k + )y x](7a + 4) 7y = 4[(k + )y x] 7[y a{(k + )y x}], (vii) = 7y [(k + )y x](7a + 4) = 7[y a{(k + )y x}] 4[(k + )y x]. (viii) Clearly, the minimum solution is obtained from (vii) as follows : y a{(k + )y x} =, (k + )y x = y = a + (and x = (k + )(a + ) ). Hence, in this case, the minimum m in (A) is m = q y = q(a + ). Case 5 : When p is of the form p = 7a + 5 for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](7a + 5) 7y = 5[(k + )y x] 7[y a{(k + )y x}], (ix) = 7y [(k + )y x](7a + 5) = 7[y a{(k + )y x}] 5[(k + )y x]. (x) Then, (ix) gives the following minimum solution : y a{(k + )y x} =, (k + )y x = 3 y = 3a + (and x = (k + )(3a + ) 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 6 : When p is of the form p = 7a + 6 for some integer a. In this case, 7 divides (p + ). Then, the minimum solution is obtained from (ii), which is p + y = 7 (and x = (k + )y ). Therefore, the minimum m in (A) is q(p + ) m = q y =. 7

137 36 Wandering in the World of Smarandache umbers Corollary 4..7 : Let p and q be two primes with q > p 3. Let q = k p + 8 for some integer k. Then, q(p ) q(p ), if 8 (p ) 8, if 8 (p ) 8 q(3a + ), if p = 8a + 3 q(3p ) 8, if 8 (p 3) Z(pq) = = q(3a + ), if p = 8a + 5 q(3p+ ) 8, if 8 (p 5) q(p + ), if 8 (p + ) q(p+ ) 8 8, if 8 (p + ) Proof : With l = 8 in (4..) and (4..), we get respectively 8y (x k y)p =, (i) (x k y)p 8y =. (ii) Now, for any prime p 3, exactly one of the following four cases occur : Case : When p is of the form p = 8a + for some integer a. In this case, 8 divides (p ). Then, the minimum solution is obtained from (ii), which is y = p 8 (and x k y = ). Therefore, the minimum m in (A) is m = q y = Case : When p is of the form p = 8a + 3 for some integer a. In this case, from (i) and (ii), we get respectively q(p ) 8 = 8y (x k y)(8a + 3) = 8[y (x ky)a] 3(x k y), (iii) = (x k y)(8a + 3) 8y = 3(x k y) 8[y (x k y)a]. (iv) Clearly, the minimum solution is obtained from (iv), which is y (x k y)a =, x k y = 3 y = 3a + (and x = k(3a + ) + 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 3 : When p is of the form p = 8a + 5 for some integer a. From (i) and (ii), we get = 8y (x k y)(8a + 5) = 8[y (x k y)a] 5(x k y), (iv) = (x k y)(8a + 5) 8y = 5(x k y) 8[y (x k y)a]. (v) The minimum solution is obtained from (iv) as follows : y (x k y)a =, x k y = 3 y = 3a + (and x = k(3a + ) + 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 4 : When p is of the form p = 8a + 7 for some integer a. In this case, 8 divides (p + ). Then, the minimum solution is obtained from (i), which is p + y = 8 (and x k y = ). Therefore, the minimum m in (A) is q(p + ) m = q y =. 8.

138 37 Chapter 4 : The Pseudo Smarandache Function Corollary 4..8 : Let p and q be two primes with q > p 3. Let q = (k + )p 8 for some integer k. Then, q(p ) q(p ), if 8 (p ) 8, if 8 (p ) 8 q(3p ) q(3a + ), if p = 8a + 3 8, if 8 (p 3) Z(pq) = = q(3a + ), if p = 8a + 5 q(3p+ ) 8, if 8 (p 5) q(p + ), if 8 (p + ) q(p+ ) 8 8, if 8 (p + ) Proof : From (4..) and (4..) with l = p 8, we have respectively [(k + )y x]p 8y =, (i) 8y [(k + )y x]p =. (ii) We consider the four possibilities that may arise : Case : When p is of the form p = 8a + for some integer a. In this case, 8 divides (p ). Then, the minimum solution is obtained from (i), which is p y = (and x = (k + )y ). 8 q(p ) Therefore, the minimum m in (A) is m = q y =. 8 Case : When p is of the form p = 8a + 3 for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](8a + 3) 8y = 3[(k + )y x] 8[y a{(k + )y x}], (iii) = 8y [(k + )y x](8a + 3) = 8[y a{(k + )y x}] 3[(k + )y x]. (iv) Clearly, the minimum solution is obtained from (iii), which is y a{(k + )y x} =, (k + )y x = 3 y = 3a + (and x = (k + )(3a + ) 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 3 : When p is of the form p = 8a + 5 for some integer a. In this case, from (i) and (ii), we get respectively = [(k + )y x](8a + 5) 8y = 5[(k + )y x] 8[y a{(k + )y x}], (v) = 8y [(k + )y x](8a + 5) = 8[y a{(k + )y x}] 5[(k + )y x]. (vi) The minimum solution is obtained from (vi) as follows : y a{(k + )y x} =, (k + )y x = 3 y = 3a + (and x = (k + )(3a + ) 3). Hence, in this case, the minimum m in (A) is m = q y = q(3a + ). Case 4 : When p is of the form p = 8a + 7 for some integer a. In this case, 8 divides (p + ). Then, the minimum solution is obtained from (ii), which is p + y = (and x = (k + )y ). 8 Therefore, the minimum m in (A) is q(p + ) m = q y =. 8

139 38 Wandering in the World of Smarandache umbers 4.3 Some Generalizations Since its introduction by Kashihara [], the pseudo Smarandache function has seen several variations. This section is devoted to some of these generalizations, given in the following two subsections The Pseudo Smarandache Dual Function Z * (n) The pseudo Smarandache dual function, denoted by Z * (n), introduced by Sandor [6], is defined as follows. Definition : For any integer n, the pseudo Smarandache dual function, Z * (n), is + m(m + ) Z * (n) = max m : m Z, n. Let Z * (n) = m 0 for some integer m 0. Then, by Definition 4.3.., the following two conditions are satisfied : () m (m ) divides n, () n is not divisible by m(m + ) for any m > m 0. Lemma : For any integer n, Z *(n) ( + 8n ). Proof : Let Z * (n) = m 0 for some integer m 0, so that, by definition, m 0(m0 + ) n = k. for some integer k. Then, m 0(m0 + ) n, which gives the r.h.s. inequality. The other part of the inequality is obvious. k(k + ) Lemma : Z *( ) = k for any integer k. Proof : By definition, k(k + ) m(m + ) k( k + ) Z * ( ) = max m :. () Now, since m(m + ) = k(k + ) if and only if m = k, the maximum m in () is k.

140 39 Chapter 4 : The Pseudo Smarandache Function Corollary : Let p and q = p be two primes. Then, Z * (pq) = q. Proof : Since pq = q + ( ).q, the result follows from Lemma Lemma : For any integers a, b, Z * (ab) max{ Z * (a), Z * (b)}. Proof : Let Z * (a) = m 0 for some integer m 0. Then, by definition, m 0 (m0 + ) m 0 (m0 + ) a ab Z * (ab) m 0 = Z * (a). Lemma : Let p 3 be a prime. Then, for any integer k, k, if p = 3 Z *(p ) =, if p 3 k + m(m + ) k Proof : By definition, Z * (p ) = max m : m Z, p. If one of m and m + is p, then the other one must be. Now, m + = p, m = p = 3, m + =, m = p p =. Thus, in the second case when p 3, we must have m =. Lemma : For any integer n, Z(n) Z * (n). Proof : Let Z(n) = m 0, Z * (n) = m for some integers m 0 and m. Then, by definitions of Z * (n) and Z(n) respectively, m(m + ) m 0 (m0 ) m(m ) m 0 (m0 ) n, n Thus, m 0 m, which we intended to prove. Lemma : Any solution of the equation Z(n) = Z * (n) is of the form k(k + ) n =, k. Proof : Let Z(n) = k = Z * (n) for some integer k. Then, k(k + ) k(k + ) k(k + ) n, n n =. Corollary : There is an infinite number of n satisfying the equation Z(n) = Z * (n).

141 40 Wandering in the World of Smarandache umbers 4.3. The Additive Analogues of Z(n) and Z * (n) The function, called the additive analogue of Z(n), to be denoted by Z S (n), has been introduced by Sandor [7]. Definition : For any real number x (0, ), + m(m + ) Z S (x) = min m : m Z, x. The dual of Z S (x), to be denoted by Z S* (x), is defined as follows (Sandor [7]). Definition : For any real number x [, ), + m(m + ) Z S* (x) = max m : m Z, x. From Definition and Definition 4.3.., it is clear that, for any integer k, Z S (x) = k if and only if (k )k k(k + ) x (, ], k(k + ) (k + )(k + ) Z S* (x) = k if and only if x [, ). We then have the following result. Lemma : For any real number x, and any integer k, k(k+) (k+ )(k + ) Z S* (x) +, if x (, ) Z S (x) = k(k+ ) Z S* (x), if x = Lemma : For any fixed real number x, ( 8x + 3) < Z S* (x) ( 8x + ). (4.3..) Proof : Let, for some real number x fixed, Z S* (x) = k for some integer k. Then, by definition, k(k + ) (k + )(k + ) x <, () where k is the largest integer satisfying the above inequality. Now, () is satisfied if and only if k + k x 0 < k + 3k (x ) if and only if ( 8x + 3) < k ( 8x + ), which now gives the desired result. Corollary : For any fixed real number x, S* Z (x) = [ ( 8x + ] ([y] denotes the integer part of y).

142 4 Chapter 4 : The Pseudo Smarandache Function Proof : The result follows from Lemma 4.3.., noting that ( 8x + 3) ( 8x + ) =, k is an integer, and further k is the maximum integer satisfying the inequality (4.3..). Corollary : Z S* (x) ( 8x + ) as x. Proof : follows from Corollary above. If the domain is restricted to the set of positive integers, The first few terms of Z S (n) are,,, 3, 3, 3, 4, 4, 4, 4, 5,, + Z, we get the function Z S (n). k(k + ) (k )k where the value k ( ) is repeated = k times. Similarly, we get the + function Z S* (n), with domain Z, which assumes the successive values,,,,, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5,, where the value k ( ) is repeated (k + )(k + ) k(k + ) = k + times. Lemma : For any integer n, Z(n) Z S* (n). Proof : Let Z(n) = m 0, Z S* (n) = m for some integers m 0 and m. Then, by definitions of Z(n) and Z S* (n), m 0 (m0 + ) m(m + ) n, n Thus, m 0 m, which we intended to prove. m 0 (m0 + ) m( m + ) n. Lemma : Any solution of the equation Z(n) = Z S* (n) is of the form k(k + ) n =, k. Proof : Let Z(n) = k = Z S* (n) for some integer k. Then, k(k ) k(k ) n +, n + k(k + ) n =. Corollary : Any solution of the equation Z(n) = Z S (n) is of the form k(k + ) n =, k. Proof : follows from Lemma and Lemma

143 4 Wandering in the World of Smarandache umbers 4.4 Miscellaneous Topics In this section, we give some properties of the pseudo Smarandache function Z(n). One major difference in the properties of S(n) and Z(n) is that, though the equation S(n) = n has an infinite number of solutions, the equation Z(n) = n does not permit any solution, as is proved in Lemma The next lemma, Lemma 4.4., is due to Ashbacher [3]. Lemma 4.4. : The equation Z(n) = n has no solution. Proof : Let, for some integer n, Z(n) = n, so that, n(n + ) (n )n n n Z(n) n, which is a contradiction to the assumption. Lemma 4.4. : Z(n + ) Z(n) is unbounded. Proof : Let n + = k for some integer k. Then, by Lemma 4.. and Lemma 4.4, Z(n + ) = Z( k ) = k+, Z(n) n = k. Therefore, Z(n + ) Z(n) = Z(n + ) Z(n) ( k+ ) ( k ) = k +, which can be made arbitrarily large (by choosing k large and larger). The following lemma gives a set of conditions such that Z(n + ) divides Z(n). Lemma : Let p be a prime of the form p = 4a +, a is an integer. Let q be another prime such that q = p = 8a +. Then, Z(p) divides Z(q) = Z(p ). Proof : By Corollary 4..(),Z(p) = p. By construction, q is a prime, so that, by Lemma 4..3, Z(q) = Z(p ) = (p ). Hence, Z(p) = p divides Z(q) = Z(p ) = (p ). Lemma : Given any integer k, the equation Z(kn) = n has an infinite number of solutions. Proof : Given the integer k, we choose a prime p such that k (p + ). Then, Z(kp) = p (by Lemma 4..4). Now, given k, there is an infinite number of primes of the desired form. All these complete the proof of the lemma. The following three results, given in Lemma Lemma 4.4.7, are due to Pinch [8], though the proofs given here are modified ones. Lemma : For any integer k fixed, the equation n = kz(n) has an infinite number of solutions. Proof : The equation n = k Z(n) is equivalent to Z(kn) = n.

144 43 Chapter 4 : The Pseudo Smarandache Function Let Now, Lemma : The ratio Z(n) is unbounded above. Z(n) Proof : Given the integer k, we choose a prime p such that k (p + ). Let n = p(p + ) Z(n) = m. T(p) (so that Z(n) = p). k m(m+ ) k m(m+ ) k p p(p + ) = n, n p p m(m + ). We now consider the following two cases : Case () : When p divides m and k+ divides (m + ). In this case, m = pt for some integer t >. Note that, t, for otherwise t = m = p T(m) = T(p) = n n = T(m) n. Now, k (p + ) k (m + t), and since k (m + ), it follows that k (t ), so that t k +. Hence, Z(n) m t k, Z(n) = = > p which can be made arbitrarily large (by choosing k sufficiently large). Case () : When k+ divides m and p divides (m + ). In this case, m = pt for some integer t >. In this case, t, for otherwise t = m = p T(m) = T(p ) n = T(p) > T(p ) n. Now, k (p + ) k (m + t + ), and since k m, it follows that k (t + ), so that t k. Hence, Z(n) m k = = t, Z(n) p p p which can be made arbitrarily large (by choosing k sufficiently large). Example 4.4. : To find n such that the ratio Z(n) Z(n) a prime p such that 3 = k (p + ). Choosing p = 3, so that n = Z(n) Z(99) = = 960 = > 30. Z(n) Z(496) 3 is greater than 30, we have to choose 3 3 = 496, we get

145 44 Wandering in the World of Smarandache umbers Lemma : Given any real number L > 0, the inequalities Z(n + ) > L, Z(n) Z(n ) > L Z(n) each has infinite number of solutions. Proof : Given the number L > 0, we choose an integer k as follows : k = max{ 4, [L]} ([x] denotes the least integer not less than x). Let k(k + ) n = ( + kt) T(k)( + kt), where t is an integer such that n + = T(k)( + kt) + () is prime. Let m = k[ + (k + )t]. Note that m(m + ) = k(k + )( + kt)[ + (k + )t] = n[ + (k + )t], so that Therefore, Now, if and only if m(m + ) n Z(n) m. Z(n + ) n T(k)( + kt) =. Z(n) m k + tt(k) T(k)( + kt) T(k)[ + k(t + )] f (t) > f (t + ) k + tt(k) k + T(k)(t + ) ( + kt)[k + T(k)(t + )] > [ + k(t + )][k + tt(k)], that is, if and only if T(k) > k, which is true. T(k)( + kt) This shows that the function f (t) k + tt(k) lim f (t) k, t it follows that Z(n + ) > k L. Z(n) is strictly decreasing in t. Now, since Next, for k 4, let n = T(k)( + kt), where t is an integer such that n = T(k)( + kt) () is prime. Then,

146 45 Chapter 4 : The Pseudo Smarandache Function Z(n ) n T(k)( kt) = +. Z(n) m k + tt(k) Now, T(k)( + kt) T(k)[ + k(t + )] g(t) > g(t + ) k + tt(k) k + T(k)(t + ) if and only if [T(k)( + kt) ][k + T(k)(t + )] > [T(k){ + k(t + )} ][k + tt(k)] that is, if and only if [T(k)] 4T(k) > k T(k) that is, if and only if k(k + ) 4 > k, which is true. T(k)( + kt) This shows that the function g(t) is strictly decreasing in t. Now, since k + tt(k) lim g(t) k, t it follows that Z(n ) > k L. (3) Z(n) Since there is an infinite number of primes of the forms () or (), the result follows. Remark 4.4. : In Lemma 4.4.7, k can be any integer greater than. But we choose k 4 simply for the convenience in proving the inequality (3). Z(n + ) Example 4.4. : To find the sequences Z(n) and Z(n ), each term of which Z(n) is greater than L =, we choose k = 4. Then, T(k) = 0. () We have to choose the integer t 0 such that n + = T(k)( + kt) + = 0( + 4t) + = + 40t is prime. We then get successively the primes, 3,, 5, 33, 49, with Z() 0 Z(3) = =.5, = 30 = 3.33, Z(0) 4 Z(30) 39 Z() = 0 = 0.5, Z(0) 0 Z(33) = 330 = 7.5, Z(330) 44 Z(5) = 50 =.06, Z(50) 4 Z(49) = 490 =.5,. Z(490) 95

147 46 Wandering in the World of Smarandache umbers () We have to choose t 0 such that n = 0( + 4t) = t is prime. The successive primes are 89, 409, 449, 769, with Z(89) 88 Z(409) = =.5, = 408 = 0., Z(90) 35 Z(40) 40 Z(449) = 448 = 4.5, Z(450) 99 Z(769) = 768 = 3.96,. Z(770) 55 Lemma : Given any even integer k, the equation Z(n + ) = k Z(n) has solutions. Proof : Given the even integer k, we choose a prime p such that k (p + ). Let q be another prime of the form q = kp +. Now, letting n = kp (so that n + = q), by virtue of Lemma 4..4, Z(kp) = p, so that Z(n + ) Z(q) Z(kp + ) kp = = = = k. Z(n) Z(kp) Z(kp) p Example : Corresponding to k = in Lemma 4.4.8, some of the (p, q) pairs are (p, q) = (3, 7), (, 3), (3, 47), with Z(7) 6 Z(3) Z(47) = =, = =, = 46 =. Z(6) 3 Z() Z(46) 3 Corresponding to k = 4, we have (p, q) = (7, 9), (79, 37), (7, 509), with Z(9) 8 Z(37) 36 Z(509) = = 4, = =, = 508 = 4. Z(8) 7 Z(36) 79 Z(508) 7 Remark 4.4. : Since Z(n) n <, it follows that the equation n n Z(n) = k n has no solution if k. The case k = is forbidden by Lemma 4.4..

148 47 Chapter 4 : The Pseudo Smarandache Function The problem of convergence of (infinite) series involving Z(n) and its different variants has been treated by several researchers. The following two lemmas are due to Kashihara []. Lemma : The series is divergent. n= Z(n) Proof : The proof follows by observing that > > >, Z(n) Z(n) n n n= n=,3,... n= n= where we have used the fact that the harmonic series is divergent. Lemma : The series Z(n) is divergent. n n= + k Proof : First note that the series is divergent, since k k= Therefore, Z(n) k+ > >. k n n= k= k+ lim 0. k k Pinch [8] has proved the more general result, given below. Lemma 4.4. : The series is convergent if and only if s >. s n= [Z(n)] The following results are due to Sandor [9]. Lemma 4.4. : Z (n) Z(n) Z (n) Z(n) lim inf * = 0, lim sup * =. n n Proof : With n = p, together with Lemma 4..3 and Lemma , we have, if p = 3 Z * (n) Z * (p) p = = Z(n) Z(p), if p 3 p from which the first limiting value is obtained. k(k + ) Next, let n = for any integer k. Since by Lemma 4.. and Lemma 4.3.., k(k + ) k(k + ) Z( ) = k = Z * ( ), the other part of the lemma is established. Corollary 4.4. : The series Z (n) * is divergent. n= Z(n)

149 48 Wandering in the World of Smarandache umbers Lemma : Z (n) n Z (n) n lim inf * = 0, lim sup * =. n n Proof : With n = p, together with Lemma 4..3 and Lemma , we get the first limiting value (as in the proof of Lemma 4.4. above). To get the remaining part, first note that, for any integer k, k(k + ) Z * ( ) lim = lim k = lim k =. k k(k + ) k k(k + ) k k + Now, from Lemma 4.3.., Z * (n) n +. 8n 8n Corollary 4.4. : The series Z (n) * is divergent. n= n Lemma : Each of the series irrational number. Z (n) * and n= n! Z (n) n+ * ( ) converges to an n= n! The next three lemmas are concerned with the convergence of series involving Z S (n) and Z S* (n). The first one is due to Sandor [7], and the rest are due to Jiangli Gao [0]. Lemma : The series is convergent for s > 0. s n= n[z S* (n)] Lemma : For any s >, () = ζ(s ), () s s [Z (n)] [Z (n)] n= S n= S* where ζ(s) is the Riemann zeta function. Proof : Let (k )k k(k + ) k(k + ) (k + )(k + ) I k (, ], J k [, ). () From the definition, Z S (n) = k if n I k. Therefore, s s s n= S k= n I k k S = = ζ(s ) + ζ(s). = = k = ζ(s ). [Z (n)] [Z (n)] k () Since Z S* (n) = k if n J k, we get, = = k + = ζ(s ) + ζ(s). [Z (n)] [Z (n)] k s s s n= k= n J k= S* k S*

150 49 Chapter 4 : The Pseudo Smarandache Function Lemma : For any s >, () () n= n ( ) s s s S = ( ) ζ(s) ζ(s, ), 4 [Z (n)] 4 n ( ) s s n= [Z S* (n)] = ( ) ζ(s). where ζ(s, α ) is the Hurwitz zeta function. Proof : Let I k and J k be as in the proof of Lemma () First note that n I k ( ) S n [Z (n)] s s, if k =,5,9,... k = 0, if k is even s, if k = 3,7,,... k In the above sum, the sign of the first element of I k is Therefore, [ (k )] + ( ). n ( ) = + = s s s s s n= [Z S (n)] k=,5,... k k= 3,7,... k k=,3,5,... k k=,5,... k, which now gives the desired result, since () In this case, = ( ) ζ (s), = = ζ(s, ). 4 k k (4k + ) 4 s s s s s k=,3,5,... k=,5,... k= 0 n J k ( ) S n [Z (n)] s s, if k =, 6,0,... k = 0, if k is odd s, if k = 4,8,,... k with the first element of J k having the sign Therefore, k ( ). n ( ) = s + s = s s + s s s n= [Z S* (n)] k=,6,... k k= 4,8,... k k=,3,5,... k 4 k=,,3,... k = ( ) + k 4 k which now gives the desired result after simplification., s s s s s k=,,3,... k=,,3,...

151 50 Wandering in the World of Smarandache umbers We now consider the problem of finding the solutions of the equation Z(n) = d(n). The problem has been considered by Zhong Li [], and later by Maohua Le []. Lemma gives the result in respect to the above equation. We follow the same line of approach as that adopted by Maohua Le, filling in the gaps in his proof. Lemma : Z(n) > d(n) for all integers n > 0. Proof : If n = p, where p 5 is a prime, then Z(p) = p > = d(n), so that, the result is true if n is a prime. Thus, it is sufficient to consider the case when n is a composite number. Let there be an integer n such that Z(n) d(n). () Let n have the prime factor representation of the form α α αi αi+ αs n = p p... p i p i+... p (4.4.) s so that (by Theorem 0. in Chapter 0) d(n) = (α + )(α + ) (α k + ). Let the function f(p α ) be defined as follows : α / p f (p α ) =. () α + Then, from (4.4.) and (), Now, α α f (p )f (p )...f (p α ) k k α/ α/ αk / p p pk =.... α + α + α + k = n d(n) >, if p = and α 6, if p = 3 and α α f (p ) 5, if p = 5 and α 5 >, if p 7 and α To verify (4.4.), first note that n Z(n) <. (3) (4.4.)

152 5 Chapter 4 : The Pseudo Smarandache Function α α / p, if and only if p ( ) α α + f (p ) (4.4.3) α + 5 α 5, if and only if p 4 ( α + ) from which the last two inequalities of (4.4.) follow immediately. Again, since 5 < 6 but 6 > 7, and 3 < but 3 3, the first two inequalities in (4.4.) are established. But, (4.4.) contradicts (3) except for the following five cases : () When n = α, α 5, () When n = 3p, where p 5 is a prime, (3) When n = α p, α 5 and p 3 is a prime, (4.4.4) (4) When n = α.3 p, α 5 and p 5 is a prime, (5) When n = α 5, α 5. Thus, except for the numbers of the forms listed above, the result in Lemma is true. To complete the proof, we have to consider the case when n is of the forms () (5) in (4.4.4). This is done below. In Case (), when n = α, by Lemma 4.., Z(n) = Z( α ) = α+ > α + = d( α ) = d(n) for α 5. In Case (), when n = 3p, by Corollary 4..(), Z(n) = Z(3p) p 4 = d(3p) = d(n) for any prime p 5. In Case (3), when n = α p, α 5, d(n) = (α + ). Now, from Corollary 4..(), Z(p) p 4 = d(p) for p 5, (with Z( 3) = 3 < 4 = d(6), Z( 5) = 4 = d(0)), from Lemma 4..6, Z( p) p 6 = d( p) for p 7 (with strict inequality when p = 7), Z( p) 3p > 6 = d( p) for p = 3, 5, from Lemma 4..0, Z( 3 p) p > 8 = d( 3 p) for p, Z( 3 p) 3p > 8 = d( 3 p) for p = 3, 5, 7, from Lemma 4..6, Z( 4 p) p 0 = d( 4 p) for p (with strict inequality when p = ), Z( 4 p) 9p 0 = d( 4 p) for p = 3, 5, 7, from Lemma 4..7, Z( 5 p) p = d( 5 p) for p 3 (with strict inequality sign for p = 3), Z( 5 p) 9p > = d( 5 p) for p = 3, 5, 7,.

153 5 Wandering in the World of Smarandache umbers In Case (4), when n = α.3 p, α 5 and p 5 is a prime, d(n) = 4(α + ). Now, from Lemma 4..8, Z(6p) p > 8 = d(6p) for p, Z(6p) 3p > 8 = d(6p) for p = 5, 7, (with Z(6 3) = 8 > 6 = d(8)), from Lemma 4..4, Z(p) p = d(p) for p 3 (with strict inequality when p = 3), Z(p) 3p > = d(p) for p = 5, 7,, (with Z( 3) = 8 < 9 = d(36)), from Lemma 4..8, Z(4p) p 6 = d(4p) for p 7 (with strict inequality when p = 7), Z(4p) 3p > 6 = d(4p) for p = 7,, 3, (with Z(4 3) = 63 > = d(7)). from Lemma 4..9, Z(48p) p 0 = d(48p) for p 3 (with strict inequality when p = 3), Z(48p) 3p 0 = d(48p) for p = 7,, 3, 7, 9, (with strict inequality for p = 7; Z(48 3) = 63 > 5 = d(44), Z(48 5) = 95 > 5 = d(40)). To check for Z(96p), we shall make use of Lemma 4..9 and Lemma 4.5. Now, since Z(96p) Z(48p), we see that Z(96p) p > 4 = d(96p) for p 9, Z(96p) 3p > 4 = d(96p) for p =, 3, 7, 9, 3, (with Z(96 3) = 63 > 8 = d(88), Z(96 5) = 55 > 4 = d(480), Z(96 7) = 63 > 4 = d(67)). In Case (5), when n = α 5, α 5, note that d( α 5 ) = d( α 5 ) = 3(α + ). But, by Lemma 4.5, Z( α 5 ) Z(5 ) = 4 > 3(α + ) = d( α 5 ) for all α 5. All these complete the proof of the lemma. In the proof of Lemma above, we proved more. They are summarized below. Corollary : Let d(n) be the divisor function. Then, (4) the equation Z(n) = d(n) has the only solutions n =, 3, 0; (5) the inequality Z(n) < d(n) has the only solutions n = 6, 36, 0; (6) Z(n) > d(n) for any integer n, 3, 6, 0, 36, 0. It is interesting to note here that, all the numbers for which the result fails to hold true (that is, the numbers, 3, 6, 0, 36 and 0) are triangular numbers.

154 53 Chapter 4 : The Pseudo Smarandache Function Corollary : Z(n) + φ(n) > d(n) for any integer n. Proof : From Corollary 4.4.3, the result follows, except possibly for the numbers 6, 36 and 0. Now, by Theorem 0. (in Chapter 0), φ(6) = φ( 3) =, φ(36) = φ(.3 ) =, φ(0) = φ( 3.3.5) = 3. Checking with these numbers, we get the desired result. An immediate consequence of Corollary is the following corollary. The same problem has also been treated by Zhong Li and Maohua Le [3]. Corollary : The equation Z(n) + φ(n) = d(n) has no solution. The following two results are due to Ashbacher [3]. Lemma : The equation Z(n) = φ(n) has an infinite number of solutions. Proof : The proof is simple : For any prime p 3, Z(p) = p = φ(p). Lemma : The equation Z(n) + φ(n) = n has an infinite number of solutions. Proof : Let n = 3. k, where k is an integer. Then, from Lemma 4..6 and Theorem 0. (in Chapter 0), Z(n) = Z(3. k ) = k+, φ(n) = φ(3. k k ) = (3. )( )( ) = k. For such an n, 3 Z(n) + φ(n) = Z(3. k ) + φ(3. k ) = k+ + k = 3. k = n. Another related problem of interest is the following, due to Zhong Li []. Lemma 4.4. : The equation Z(n) + d(n) = n has the only solution n = 56. Proof : Let, for some integer n, Z(n) + d(n) = n. () By definition, Z(n)[Z(n) + )] [n d(n)][n {d(n) }] n = n(n + ) + d(n)[d(n) ] = nd(n). () Now, if n is odd, from (), we see that d(n)[d(n) ] n n d(n)[d(n) ]. On the other hand, if n is even, from (), d(n)[d(n) ] = (k )n for some integer k.

155 54 Wandering in the World of Smarandache umbers Thus, in either case, d(n)[d(n) ] n [d(n)] > n <. (3) d(n) But, this contradicts (4.4.) except for the five cases given in (4.4.4). Thus, the equation () has no solution, except for the case of numbers of the forms of (4.4.4). To complete the proof, we now consider the five cases in (4.4.4) separately. In Case (), when n = α, by Lemma 4.., Z(n) + d(n) = Z( α ) + d( α ) = α+ + α > α = n for any α. In Case (), when n = 3p, by Corollary 4..(), Z(n) + d(n) = Z(3p) + d(3p) p + 3 < 3p = n for any prime p 5. In Case (3), when n = α p, α 5, d(n) = (α + ). Now, from Corollary 4..(), Z(p) + d(p) p + 3 < p for p 5 (with Z( 3) + d( 3) = > 6), from Lemma 4..6, Z( p) + d( p) 3p + 5 < 4p for p 7, (Z(.3) + d(.3) = >, Z(.5) + d(.5) = > 0), from Lemma 4..0, Z( 3 p) + d( 3 p) 7 p + 7 < 8p for p, (Z(4) + d(4) = < 4, Z(40) + d(40) = < 40, Z(56) + d(56) = = 56), from Lemma 4..6, Z( 4 p) + d( 4 p) 5p + 9 < 6p for p, (Z(48) + d(48) = < 48, Z(80) + d(80) = < 80, Z() + d() = < ), from Lemma 4..7, Z( 5 p) + d( 5 p) 3p + < 3p for p, (with strict inequality for p =, and Z(96) + d(96) = 63 + < 96, Z(60) + d(60) = 64 + < 60, Z(4) + d(4) = 63 + < 4). In Case (4), when n = α.3 p, α 5 and p 5 is a prime, d(n) = 4(α + ). Now, from Lemma 4..8, Z(6p) + d(6p) 3p + 7 < 6p for p 5, (with Z(6 3) + d(6 3) = = 4 < 8), from Lemma 4..4, Z(p) + d(p) 7p + < p for p 5, (with Z( 3) + d( 3) = = 7 < 36), from Lemma 4..8, n

156 55 Chapter 4 : The Pseudo Smarandache Function Z(4p) + d(4p) 5p + 5 < 4p for p 5, (with Z(4 3) + d(4 3) = 63 + = 75 > 7), from Lemma 4..9, Z(48p) + d(48p) 3p + 9 < 48p for p 3, (with Z(48 3) + d(48 3) = = 78 < 44), from Lemma 4..30, Z(96p) + d(96p) 59p + 3 < 96p for p 3. (with Z(96 3) + d(96 3) = = 8 < 88). In Case (5), when n = α 5, α 5, we verify the five cases directly : Z(50) + d(50) = < 50, Z(00) + d(00) = < 00, Z(00) + d(00) = 75 + < 00, Z(400) + d(400) = < 400, Z(800) + d(800) = < 800. All these complete the proof of the lemma. The proof of Lemma 4.4. shows that n = 56 is the only solution of the equation Z(n) + d(n) = n, and in other cases, either Z(n) + d(n) > n or Z(n) + d(n) < n. The following result is due to Maohua Le [], whose proof relies on a result of Ibstedt [4] which is not complete. See, however, Conjecture 4.5. in 4.5. Lemma 4.4. : The only solutions of the equation Z(n) = σ(n) are n = k, k. It is straight forward to verify that Z( k ) = k+ = σ( k ). Definition 4.4. : An integer n is called pseudo Smarandache perfect if and only if n = k i= Z(d ). i where d, d,, d k are the proper divisors of n. In [4], Ashbacher reports that the only pseudo Smarandache perfect numbers less than 0 6 are n = 4, 6, However, since n = is of the form n = 8p with p = (a prime), its only proper divisors are,, 4, 8, p, p and 4p. Since 8 divides p + = 58944, it follows from Lemma 4..4 that Z(p) = p, Z(p) = p, Z(4p) = p, so that n = > k i= Z(d ). i Thus, n = is not pseudo Smarandache perfect. Note that, the number 6 is also perfect.

157 56 Wandering in the World of Smarandache umbers Lemma : Any solution of the equation [Z(n)] + Z(n) = n () k(k + ) is of the form n =, k. Proof : Solving the equation [Z(n)] + Z(n) n = 0 for positive Z(n), we get Z(n) = ( 8n + ). This shows that () has an integer solution only if 8n + is a perfect square. Let Then, 8n + = x for some integer x 3. 8n = (x )(x + ). () Since x and x + have the same parity, from (), we see that both must be even. Let x = k for some integer k. Then, from (), 8n = k(k + ) n = k(k + ). Hence, the lemma. We now look at some equations and inequalities involving both S(n) and Z(n). Lemma : The solution of the equation S(n) = Z(n) is n = tp, where t is such that t (p )! and Z(tp) = p. Proof : Let, for some integer n, S(n) = Z(n) = m for some integer m. Furthermore, let α α α j α j+ αk n = q q... q q...q j j+ k be the representation of n in terms of its prime factors p, p,, p k (not necessarily ordered). m(m + ) Now, m cannot be even, for otherwise, (since n ) for some j k, α α α j m α j+ α j+ αk q q... q j and q j+ q j+... q k (m + ). Again, since n m!, it follows that ( m ) But then α α α j α! j+ α j+ αk q q... q j and q j+ q j+... q k (m )!. α α α α α n = q q... q q...q (m )! S(n) m, j j+ k j j+ k

158 57 Chapter 4 : The Pseudo Smarandache Function and we reach to a contradiction. Thus, m must be odd, and α α α j q q... q m, j α j+ α j+ αk α m j+ α j+ αk q j q j... q k and q j+ q j+... q k (m )!. Now, we show that m must be a prime. The proof is by contradiction. So, let β β β βr m = p p p... p r with β r { } p β = max p, p β,..., p β. But then, for some j k, r α α α j α j+ α j+ αk q q... q p! and q q... q (m )! β ( ) j j+ j+ k α α α α α n = q q... q q... q (m )!, j j+ k j j+ k a contradiction. Again, if m = p β with β, then n does not divide (m )!, but β β β β n (p )! =.... (p)... p... (p )... (p )... (p )... p, β β + so that the power of p in n is more than β(β ). But then, n does not divide p (p ). Thus, m must be a prime, say, m = p. Therefore, the problem of finding the solution of S(n) = Z(n) reduces to the problem of finding the solution to the following equations : S(n) = p, Z(n) = p. The lemma now follows by virtue of Lemma (in Chapter 3). Example : Let p be such that (p + ) is divisible by 4. Then, Z(p) = p = S(p). Hence, n = p, where p is a prime such that 4 (p + ), is a solution of S(n) = Z(n). This shows that the equation S(n) = Z(n) possesses an infinite number of solutions. Example : Let p 5 be a prime such that 3 (p + ). Then, S(3p) = p = Z(3p). Example : An even perfect number n = k ( k+ ) is a solution of S(n) = Z(n), since, by Lemma 4.. and Remark 3.3. (in Chapter 3), k+ k+ k ( ) k+ k Z( p) = Z( ) = = S( p). Since Z(p) = p < p = S(p), it follows that Z(n) < S(n) infinitely often. Again, since Z( k ) = k+ > k S( k ), Z(n) > S(n) infinitely often.

159 58 Wandering in the World of Smarandache umbers Lemma : The equation Z(S(n)) = Z(n) has an infinite number of solutions. Proof : Clearly, any solution of S(n) = n is a solution of Z(S(n)) = Z(n). Besides the trivial solution n = p (where p 3 is a prime), another set of solution is (p )p n =, p 3 is prime, since, in such a case, p < p, and by Lemma 4.., (p )p (p )p Z( ) = p, S( ) = p, Hence, for such an n, (p )p (p )p Z(S(n)) = Z(S( )) = Z(p) = p = S( ) = S(n). Lemma : The equation S(Z(n)) = S(n) has an infinite number of solutions. Proof : Let n = p, where p 3 is a prime such that 4 divides (p + ). Then, S(Z(n)) = S(Z(p)) = S(p) = p = S(p) = S(n). Lemma : The equation S(Z(n)) = Z(n) has an infinite number of solutions. Proof : Let n = p, where p 3 is a prime such that 4 divides (p + ). Then, S(Z(n)) = S(p) = p = Z(p) = Z(n). Lemma : The equation Z(S(n)) = S(n) has no solution. Proof : follows from Lemma The following two results are due to Maohua Le [5]. Lemma : There exists infinitely many n satisfying the inequality S(Z(n)) > Z(S(n)). Proof : Let n = p, where p 3 is a prime such that 4 divides (p + ). Then, S(Z(n)) = S(p) = p > p = Z(p) = Z(S(p)). Lemma : There exists infinitely many n satisfying the inequality S(Z(n)) < Z(S(n)). Proof : Let n = p, where p 3 is a prime. For such an n, S(Z(n)) = S(Z(p)) = S(p ) < p = Z(p) = Z(S(p)). Ashbacher [3] conjectures that, there is an infinite number of integers n such that Z(n) S(n) (Z(n) S(n)) is a perfect square. The following lemma confirms the conjecture. Lemma : Let p be a prime of the form p = 8a + 3 () for some integer a, such that 3a + = x is a perfect square for some integer x. Let q be another prime of the form q = kp + 8 for some integer k. () Then, Z(pq) S(pq) is a perfect square.

160 59 Chapter 4 : The Pseudo Smarandache Function Proof : Let p and q be two primes of the forms () and () respectively. Then, by Corollary 4..7, Z(pq) = qx. Now, since S(pq) = q, it follows that Z(pq) S(pq) = (qx ).(q) = (qx). Corollary : There exists an infinite number of n such that Z(n) S(n) but Z(n)S(n) is a perfect square. Proof : Let q be a prime of the form q = a + 8 for some integer a. (*) Then, by Corollary 4..7, Z(q) = 4q, so that, Z(q) S(q) = (4q).(q) = (q). Now, since there is an infinite number of primes of the form (*), the result is proved. Example : Using Corollary 4.4.6, we get successively the primes q = 9, 4, 07, 5, 73, 39, 83, 349,, with Z( 9) S( 9) = ( 9), Z( 4) S( 4) = ( 4), Z( 07) S( 07) = ( 07), Z( 5) S( 5) = ( 5), Z( 73) S( 73) = ( 73), Z( 39) S( 39) = ( 39), Z( 83) S( 83) = ( 83), Z( 349) S( 349) = ( 349). Lemma : lim inf n = 0. n Z(n) Proof : Let p and q be two primes with q = kp for some integer k. Let n = pq. Then, by Corollary 4..4, n pq pq = = =, Z(n) Z(pq) q p which can be made arbitrarily small for infinitely many primes q. The following result is due to Ashbacher [3]. Here, we give a simplified proof. Lemma : Given the successive values of the pseudo Smarandache function Z() =, Z() = 3, Z(3) =, Z(4) = 7, Z(5) = 4,, the number r is constructed by concatenating the values in the following way : r = Then, the number r is irrational. Proof : The proof is by contradiction. Let, on the contrary, r be rational. Then, from some point on, a group of consecutive digits must repeat infinite number of times. Let N = t t t i in the recurring group correspond to n (so that Z(n) = N). But, by Lemma 4.., N(N + ) Z(k) N for any integer k >, so that the group of integers N cannot repeat infinitely often.

161 60 Wandering in the World of Smarandache umbers 4.5 Some Observations and Remarks Some observations about the pseudo Smarandache function are given below : Remark 4.5. : The following questions are raised by Kashihara []) : () Is there any integer n such that Z(n) > Z(n + ) > Z(n + ) > Z(n + 3)? () Is there any integer n such that Z(n) < Z(n + ) < Z(n + ) < Z(n + 3)? The following examples answer the questions in the affirmative : () Z(56) = 5 > 56 = Z(57) > Z(58) = 8 > = Z(59) > Z(60) = 39, () Z(59) = 53 < 64 = Z(60) < Z(6) = 69 < 80 = Z(6) < Z(63) = 6. Ashbacher [3] provides lists of seven consecutive increasing and decreasing terms of the sequence {Z(n)}, with n = 8886 in the first case and n = 756 in the second case. Remark 4.5. : The proof of Lemma 4.4. shows that Z(n + ) Z(n) is both unbounded above and unbounded below. The minimum value of Z(n + ) Z(n) is, and looking at the values of Z(n), it seems safe to conjecture that Z(n + ) Z(n) = if and only if n =, 5. Remark : In [3], Ashbacher raises the question : Is there a value of k where there are only finite number of solutions to the equation Z(kn) = n? Lemma answers to this question in the negative. Remark : Lemma shows that the equation Z(n) = φ(n) has an infinite number of solutions, namely, n = p, where p 3 is a prime. A second set of solution, provided by Ashbacher [3], is n = p, where p is a prime such that 4 divides p. Then, Ashbacher poses the question : Are there any other infinite family of solutions to the equation Z(n) = φ(n)? To answer the above question, we first note that, by Lemma 4..4, Z(kp) = p for any prime p such that k divides p. Now, since (by virtue of the multiplicative property of φ(n)) φ(kp) = φ(k) φ(p), it follows that Z(kp) = φ(kp) if and only if φ(k) =. But φ(k) = if and only if k =,. Thus, for any solution of the equation Z(n) = φ(n) of the form n = kp, we must have k =,. It may be mentioned that, by Corollary 4..(), for any prime p such that 4 divides p, Z(p) = p, so that, by Theorem 0. (in Chapter 0), φ (p) = p( )( ) = p = Z(p). p

162 6 Chapter 4 : The Pseudo Smarandache Function Remark : From Lemma 4.4.6, we see that Z(n) Z(n) can be made arbitrarily large. It can be shown that, the equation Z(n) = has infinite number of solutions. The proof is as Z(n) follows : Let p be a prime of the form p = 4a + for some integer a. Then, choosing n = p k, where k is any integer, by Lemma 4..3 and Lemma 4..4, k k Z(n) Z(p ) p = = = k k. Z(n) Z(p ) p Thus, for p = 5, we get n = 5, 5 3, 5 4, with 3 4 Z(.5 ) 4 Z(.5 ) 4 Z(.5 ) ,,, Z(5 ) = 4 = Z(5 ) = 4 = Z(5 ) = 64 = while, corresponding to p = 3, we get n = 3, 3 3, with 3 Z(.3 ) Z(.3 ) = 68, = = 96 = 3 Z(3 ) Z(3 ) Again, keeping k =, we get n = 5, 3, 7, 9, 37, 4, with Z(.7 ) Z(.9 ) Z(.37 ) Z(.4 ) = 88, 840, 368, = = 840 = = 368 = = 680 = Z(7 ) Z(9 ) Z(37 ) Z(4 ) Another infinite family of solutions to Z(n) = Z(n) can be constructed as follows : Let p be a prime of the form p = 4a for some integer a. Let n = p k, where k is any even integer. Then, by Lemma 4..3 and Lemma 4..4, k k Z(n) Z(p ) p = = = k k. Z(n) Z(p ) p For example, for p = 3, we have and for p = 7, we get 4 6 Z(.3 ) Z(.3 ) Z(.3 ) = 8, 80, 78, 8 = = 80 = = 78 = 4 6 Z(3 ) Z(3 ) Z(3 ) 4 Z(.7 ) Z(.7 ) = 48, = = 400 = 4 Z(7 ) Z(7 ) Remark : The equation Z(3n) = has infinite number of solutions. To prove this, let Z(n) n = p, where p 3 is a prime. Then, by Lemma 4..3 and Corollary 4..(), Z(3n) Z(3p ) p = = =. Z(n) Z(p ) p

163 6 Wandering in the World of Smarandache umbers We now concentrate on some conjectures related to Z(n). Conjecture 4.5. : Z(n) = n if and only if n = k for some integer k 0. Let, for some integer n, m 0 (m0 + ) Z(n) = m o, where m o = n (so that n ). Note that the conjecture is true for n = (with k = 0). Also, note that n must be composite. Let Z(n) = m. We want to show that m = m o +. m0 + Since n, it follows that moreover, (m0 + ) (m0 + ) + n = ; n does not divide (m + ) = (m + ) + for all m m o. Thus, m = m o + = (n ) + = n. All these show that Z(n) = n Z(n) = n. Continuing this argument, we see that Z(n) = n Z( k n) = k+ n for any integer k. Since Z() =, it then follows that Z( k ) = k+. Conjecture 4.5. : If n is not of the form k for some integer k 0, then Z(n) < n. By Lemma 4.4., we can exclude the possibility that Z(n) = n. So, let Z(n) = m o with m o > n. Note that, n must be a composite number, not of the form p k (p 3 is prime, k 0). Let m o = a n + b for some integers a, b n. Then, m o (m o + ) = (a n + b)(a n + b + ) = n(a n + ab + a) + b(b + ). Therefore, n m o (m o + ) if and only if b + = n. But, by Conjecture 4.5., b + = n leads to the case when n is of the form k. Thus, the assertion is proved.

164 63 Chapter 4 : The Pseudo Smarandache Function Conjecture : Z(n) = n if and only if n = p k for some prime p 3, k. Let, for some integer n, Z(n) = m o, where m o = n. Then, m o and n (m o + ); moreover, n does not divide m + for any m m o. Let Z(n ) = m. Since n (m o + ), it follows that n (m o + ) = (m o + m o ) + ; moreover, n does not divide (m + ) = (m + m) + for all m m o. Thus, m = m o + m o = (n ) + (n ) = n, so that (since m o m ) Z(n) = n Z(n ) = n. Continuing this argument, we see that Z(n) = n Z(n k ) = n k for any integer k. Next, let Z(n k+ ) = m for some integer k. Since n (m o + ), it follows that n k+ (m o + ) k+ = [(m o + ) k+ ] + ; moreover, n k+ does not divide (m + ) k+ = [(m + ) k+ ] + for any m m o. Thus, m = (m o + ) k+ = n k+, so that (since m o m ) Z(n) = n Z(n k+ ) = n k+ for any integer k. All these show that Z(n) = n Z(n k ) = n k for any integer k. Finally, since Z(p) = p for any prime p 3, it follows that Z(p k ) = p k. Remark : Kashihara [] proposes to find all the values of n such that Z(n + ) = Z(n). In this respect, we make the following conjecture :

165 64 Wandering in the World of Smarandache umbers Conjecture : For any integer n, Z(n + ) Z(n). Let Z(n + ) = Z(n) = m o for some n Z +, m o. Then, m 0 (m0 + ) m 0 (m0 + ) m 0 (m0 + ) n, (n + ) n(n + ), since (n, n + ) =. Therefore, Z(n(n + )) m o Z(n(n + )) = m 0, since, by Lemma 4.5, Z(n(n + )) Z(n) m 0. Therefore, m 0 (m0 + ) n(n + ) m 0 (m0 + ) n(n + ) n(n + ) Z( ) m o. Thus, by virtue of Lemma 4.., Z( n(n + ) ) = n m o = Z(n). But, the above inequality contradicts Conjecture Remark : Kashihara [] raises the following question : Given any integer m o, how many n are there such that Z(n) = m o? Given any integer m o, let Z (m o ) = {n : n Z +, Z(n) = m o }. (4.5.) By Lemma 4.., Z (m o ) is defined for all integers m 0 Z + ; moreover, m 0 (m0 + ) n max Z (m o ). This shows that the set Z (m o ) is non-empty with n max its largest element. Note that, n Z (m o ) if and only if the following two conditions are satisfied () n divides m 0 (m0 + ), () n does not divide Thus, for examples, m(m + ) for any m with 3 m m o. Z () = {}, Z () = {3}, Z (8) = {8,, 8, 36}. We can look at (4.5.) from another point of view : On the set as follows : + Z, we define the relation R For any n, n Z +, n R n if and only if Z(n ) = Z(n ). (4.5.)

166 65 Chapter 4 : The Pseudo Smarandache Function + It is an easy exercise to verify that R is an equivalence relation on Z. The sets Z (m), m Z +, + are then the equivalence classes induced by the equivalence relation R on Z with the following two properties : () () Z (m) m= = Z +, Z (m ) Z (m ) =, if m m. Thus, for any n Z +, there is one and only one m Z + such that n Z (m). Given any integer m o, let C(m o ) be the number of integers n such that Z(n) = m o, that is, C(m o ) denotes the number of elements of Z (m o ). Then, for m o, m + ) d(m ) d(, if m is odd m C(m ) d( )[d(m ) ], if m is even where, for any integer n, d(n) denotes the number of divisors of n including and n. Note, however, that the above upper bound is not tight. Now, let p 3 be a prime. By Lemma 4..3, Z(p) = p, so that p Z (p ) for all p 3. Let n Z (p ), so that, n divides p(p ). Then, n must divide p, for otherwise p (p )(p ) n n Z(n) p, contradicting the assumption. Thus, any element of Z (p ) is a multiple of p. In particular, p is the minimum element of Z (p ). Thus, if p 5 is a prime, then Z (p ) contains at least p(p ) two elements, namely, p and. m 0 (m0 + ) Next, let p be a prime factor of. Since Z(p) = p, we see that p Z (m o ) if and only if p m o, that is, if and only if p m o +. Z(n) In Lemma 4.4.3, we actually proved that the equation Z(n + ) = has solutions. However, the number of solutions is rather scarce. A limited search upto n 5000 revealed only solutions : n = 73, 93, 673, 53, 0, 3, 657, 753, 07, 37, 40, 473, 593, 37, 333, 336, 348, 3697, 37, 4057, 477.

167 66 Wandering in the World of Smarandache umbers In this connection, the following questions may be posed : Z(n) Question 4.5. : Does the equation k (k ) Z(n + ) = () have an infinite number of solutions when k =? () have solutions for k? Lemma shows that the equation n = k (k ) has an infinite number of solutions. Z(n) When k =, we can find more solutions. For example, letting n = p a, where p is a prime such that 4 divides (p + ), a is an odd integer, we have, by virtue of Lemma 4..4, n a a p p = = =. a a Z(n) Z(p ) p Again, n = 3p a, where p is a prime such that 3 divides (p + ), a is an odd integer, is a solution corresponding to k = 3, since by Lemma 4..5, n a a 3p 3p = = = 3. a a Z(n) Z(3p ) p This suggests the following question : Question 4.5. : Does the equation solutions for 0 < k <. n = k (k ) have other solutions? Does it have Z(n) The following question is related to Lemma Question : Does the equation Z(n + ) = k Z(n) () have solutions when k 3 is odd? () have an infinite number of solutions when k is even? when k is odd? The following question has been raised by Ashbacher [3] : Question : What is the minimum value of s 0 such that the series is convergent for all s > s 0? Ashbacher [3] has proved that the series simple : Let p n be the n th prime. Then, n= n= [S(n) + Z(n)] is divergent. The proof is S(n) + Z(n) n= S(n) + Z(n) n= S(p n ) + Z(p n ) n= p n + (pn ) n= pn > = > >. s

168 67 Chapter 4 : The Pseudo Smarandache Function We close this chapter with the following questions and problems, old but still open. Question : Are there solutions to the following equations? () Z(n + ) = Z(n + ) + Z(n)? () Z(n) = Z(n + ) + Z(n + )? (3) Z(n + ) = Z(n + )Z(n)? (4) Z(n) = Z(n + )Z(n + )? (5) Z(n + ) + Z(n) = Z(n + )? (6) Z(n + ) Z(n) = [Z(n + )]? Question : Are there infinitely many instances of 3 consecutive increasing or decreasing terms in the sequence { Z(n) }? n = Problem 4.5. : Find all solutions of the equation Z(n) + = S(n). Note that, for any prime p 3, Z(p) = p, S(p) = p, so that n = p is a solution of the equation Z(n) + = S(n). So, the question is : Are there other solutions to the equation Z(n) + = S(n)? Question : Given a palindromic number n, what is the largest value of k such that Z r (n) are palindromic for all r k, where Z r (n) is the r fold composition of Z(n) with itself? Recall that, a number n is called palindromic if it reads the same forwards and backwards. Given a palindromic prime, it is possible, in some cases to find palindromic number n such that Z(n) is also a palindrome, using Lemma Thus, starting with the palindromic prime 0, we can get n = 303, 404, 606, 707, 808, 909, such that Z(n) is a palindrome in each case : Z(303) = 0, Z(404) = 303 = Z(606) = Z(808), Z(707) = 0, Z(909) = 404, Z() = 505. As has been pointed out by Ashbacher [3], Z 3 (909) = Z (404) = Z(303) = 0. The prime 3 gives rise to three palindromes : Z(6) = 3 = Z(44) = Z(88). Clearly, the palindromic numbers formed in this way are also palindromic with respect to the Smarandache function S(n) as well, since, for example, S(0n) = 0 for any n 9. Given below are other examples where both n and Z(n) are palindromes : Z(444) = = Z(777) = Z(888) (but is not prime). Z(00) = 77 = Z(3003), Z(00) = 363 (but 00 is not prime).

169 68 Wandering in the World of Smarandache umbers References [] Kashihara, Kenichiro. Comments and Topics on Smarandache Notions and Problems. Erhus University Press, USA [] Ibstedt, Henry. Surfing on the Ocean of Numbers A Few Smarandache Notions and Similar Topics. Erhus University Press, USA [3] Ashbacher, Charles. Pluckings from the Tree of Smarandache Sequences and Functions, American Research Press, Lupton, AZ, USA [4] Ibstedt, Henry. On the Pseudo Smarandache Function and Iteration Problems. Part I. The Euler φ Function, Smarandache otions Journal,, 36 4, 00. [5] Gioia, Anthony A. The Theory of Numbers An Introduction. Dover Publications Inc., NY, USA. 00. [6] Sandor, Jozsef. Geometric Theorems, Diophantine Equations, and Arithmetic Functions, American Research Press, Rehoboth, USA. 00. [7] Sandor, Jozsef. On Additive Analogues of Certain Arithmetic Functions, Smarandache otions Journal, 4, 8 3, 004. [8] Pinch, Richard. Some Properties of the Pseudo Smarandache Function, Scientia Magna, (), 67 7, 005. [9] Sandor, Jozsef. On a Dual of the Pseudo Smarandache Function, Smarandache otions Journal, 3, 8 3, 00. [0] Jiangli Gao. On the Additive Analogues of the Simple Function, Scientia Magna, (3), 88 9, 006. [] Zhong Li. Solution of Two Questions Concerning the Divisor Function and the Pseudo-Smarandache Function, Smarandache otions Journal,, 3 35, 00. [] Maohua Le. Two Functional Equations, Smarandache otions Journal, 4, 80 8, 004. [3] Zhong Li and Maohua Le. On the Functional Equation Z(n) + φ(n) = d(n), Smarandache otions Journal,, 77 78, 000. [4] Ashbacher, Charles. On Numbers that are Pseudo-Smarandache and Smarandache Perfect, Smarandache otions Journal, 4, 40 4, 004. [5] Maohua Le. On the Difference S(Z(n)) Z(S(n)), Smarandache otions Journal,, 6, 000. [6] Majumdar, A.A.K. A Note on the Pseudo Smarandache Function, Scientia Magna, (3), 5, 006.

170 69 Chapter 4 : The Pseudo Smarandache Function Chapter 5 Smarandache umber Related Triangles This chapter is devoted to Smarandache function related and pseudo Smarandache function related triangles, defined formally below. The concept of the Smarandache function related triangles was first introduced by Sastry []. Later, it was extended by Ashbacher [] to include the pseudo Smarandache function related triangles as well. Let ABC be a triangle with sides a = BC, b = AC, c = AB. Following Sastry [], we denote by T(a, b, c) the triangle ABC. Let T(a, b, c ) be a second triangle with sides of lengths a, b, c. Recall that two triangles T(a, b, c) and T(a, b, c ) are similar if and only if their three sides are proportional (in any order). Thus, for example, if a = b = c (or if a = b = c a b c c a b ), then the two triangles T(a, b, c) and T(a, b, c ) are similar. The following two definitions are due to Sastry [] and Ashbacher [] respectively. Definition 5. : Two triangles T(a, b, c) and T(a, b, c ) (where a, b, c and a, b, c are all positive integers) are said to be Smarandache function related (or, S related) if S(a) = S(a ), S(b) = S(b ), S(c) = S(c ), (where S(.) is the Smarandache function, defined in Chapter 3). Definition 5. : Two triangles T(a, b, c) and T(a, b, c ) are said to be pseudo Smarandache function related (or, Z related) if Z(a) = Z(a ), Z(b) = Z(b ), Z(c) = Z(c ), (where Z(.) is the pseudo Smarandache function, treated in Chapter 4). A different way of relating two triangles has been proposed by Sastry [] : The triangle ABC, with angles α, β and γ (α, β and γ being positive integers), can be denoted by T(α, β, γ). Then, we have the following definition, due to Sastry [] and Ashbacher []. Definition 5.3 : Given two triangles, T(α, β, γ) and T(α, β, γ ), with α + β + γ = 80 = α + β + γ, (5.) () they are said to be Smarandache function related (or, S related) if S(α) = S(α ), S(β) = S(β ), S(γ) = S(γ ); () they are said to be pseudo Smarandache function related (or, Z related) if Z(α) = Z(α ), Z(β) = Z(β ), Z(γ) = Z(γ ). In Definition 5. Definition 5., the sides of the pair of triangles are S related / Z related, while their angles, measured in degrees, are S related / Z related in Definition 5.3. Note that the condition (5.) merely states the fact the sum of the three angles of a triangle is 80 degrees. 5. gives some preliminary results, which are used in 5. and 5.3 for the 60 degrees and 0 degrees triangles respectively. The Pythagorean case is considered in treats the case given in Definition 5.3. Some remarks are given in the final 5.6.

171 70 Wandering in the World of Smarandache umbers 5. Some Preliminary Results Let T(a, b, c) be a triangle with sides a, b and c, and angles A, B and C, as shown in the figure. Then, C a = b = c, (5..) sin A sin B sin C where b a sinc = sin(80 (A + B)) = sin(a + B) = sina cosb + cosa sinb. A B c Lemma 5.. : Let T(a, b, c) be a triangle with sides a, b and c, whose A = 60 0 (as shown in the figure). Then, 4a = (c b) + 3b. (5..) Proof : If A = 60 0, then sinc = C ( 3 cosb + sinb). Therefore, (5..) reads as a = b = c. () 3 sin B 3 cos B + sin B From the RHS of (), 3 b cosb = (c b) sinb. Squaring both sides, we get 3b ( sin B) = (c b) sin B, that is, [(c b) + 3b ]sin B = 3b. () Again, from the LHS of (), sinb = 3b a. (3) Now, eliminating sinb from () and (3), we finally get, 3b 3b = a +, (c b) 3b which gives the desired result after simplifications. Lemma 5.. : Let T(a, b, c) be a triangle with sides a, b and c, whose A = 0 0 (as shown in the figure). Then, 4a = (c + b) + 3b. (5..3) Proof : If A = 0 0, then sinc = ( 3 cosb sinb). Therefore, (5..) takes the form a = b = c. (i) 3 sin B 3 cos B sin B A b 60 0 c a B

172 7 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function The LHS of (i) gives C sinb = 3b a, (ii) while the RHS of (i) gives b 0 a b c 0 = =. (iii) 3 sin B 3 cos B sin B A Now, eliminating sinb from (iii), using (ii), we get the desired result. a c B Note that, when A = 60 0, then either b a c or else, c a b (that is, a is in between the smallest and the largest sides of the triangle T(a, b, c)); and if A = 0 0, then a is the largest side of the triangle. Also, note that, for our problem, a, b, and c are positive integer-valued. Lemma 5..3 : If (a 0, b 0, c 0 ) is a non trivial solution of the Diophantine equation 4a = (c b) + 3b, (5..) then one of b 0 and c 0 is greater than a 0 and the other one is less than a 0. Proof : We write (5..) in the equivalent form a = b bc + c. (5..a) Let (a 0, b 0, c 0 ) be a solution of (5..), so that a 0 = b 0 b 0 c 0 + c 0. () We now consider the following two cases : Case () : Let b 0 > a 0, c 0 > a 0. Let b 0 = a 0 + m, c 0 = a 0 + n for some integers m > 0, n > 0, with m n. Then, from (), we get a 0 = (a 0 + m) (a 0 + m)(a 0 + n) + (a 0 + n) that is, 0 = a 0 (m + n) + m + n mn that is, mn = a 0 (m + n) + m + n. () Now, since m + n > mn, from (), we get mn > m + n > mn, leading to a contradiction. Case () : Let b 0 < a 0, c 0 < a 0. Without loss of generality, we may assume that c 0 > b 0. Let a 0 = b 0 + c, c 0 = b 0 + d for some integers c > 0, d > 0, where c > d. Then, from (), (b 0 + c) = b 0 b 0 (b 0 + d) + (b 0 + d) that is, c(b 0 + c) = d(b 0 + d). (3) But, this leads to a contradiction, since c > d c(b 0 + c) > d(b 0 + c) > d(b 0 + d) > d(b 0 + d). Thus, neither Case () nor Case () can happen. All these establish the lemma.

173 7 Wandering in the World of Smarandache umbers By Lemma 5..3, if (a 0, b 0, c 0 ) is a (non trivial) solution of the Diophantine equation 4a = (c b) + 3b, then, without loss of generality, b 0 < a 0 < c 0. (5..4) If one solution of the Diophantine equation (5..) is known, then we can find a second independent solution. This is given in the following lemma. Lemma 5..4 : If (a 0, b 0, c 0 ) is a solution of the Diophantine equation 4a = (c b) + 3b, (5..) then (a 0, c 0 b 0, c 0 ) is also a solution of (5..). Proof : If (a 0, b 0, c 0 ) is a solution of (5..), then (since (5..) is equivalent to (5..a)) a 0 = b 0 b 0 c 0 + c 0. () Rewriting () in the equivalent form a 0 = (c 0 b 0 ) (c 0 b 0 )c 0 + c 0, we see that (a 0, c 0 b 0, c 0 ) is also a solution of a = b bc + c, or equivalently, of (5..). Lemma 5..4 shows that the Diophantine equation 4a = (c b) + 3b possesses (positive integer) solutions in pairs, namely, (a 0, b 0, c 0 ) and (a 0, c 0 b 0, c 0 ), which are independent. Note that, by symmetry, (a 0, c 0, b 0 ) and (a 0, c 0, c 0 b 0 ) are also solutions of the Diophantine equation. The following table gives the solutions of 4a = (c b) + 3b when a 00. Table 5.. : Solutions of 4a = (c b) + 3b for a 00 a b c a b c a b c The limited search for the solutions of the Diophantine equation 4a = (c b) + 3b for a 00 shows that, it possesses solutions only for certain primes (and their multiples), namely, when a = 7, 3, 9, 3, 37, 43, 6, 67, 73, 79, 97. In each case, there are two independent solutions (that are relevant to our problem of interest). In Table 5.. above, only the solutions of 4a = (c b) + 3b for primes a are shown, with the exceptions when a = 49 = 7 and a = 9 = 3 3; when a = 49, there are four independent solutions, while for a = 9, there are eight independent solutions to the Diophantine equation.

174 73 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function Next, we confine our attention to the Diophantine equation 4a = (c + b) + 3b. In this case, we have the following result. Lemma 5..5 : If (a 0, b 0, c 0 ) is a solution of the Diophantine equation 4a = (c + b) + 3b, (5..3) then a 0 > max{b 0, c 0 }. Proof : We write (5..3) in the equivalent form a = b + bc + c. (5..3a) If (a 0, b 0, c 0 ) be a solution of (5..3), then a 0 = b 0 + b 0 c 0 + c 0. () If b 0 > a 0, then from (), a 0 > a 0 + a 0 c 0 + c 0 0 > a 0 c 0 + c 0, which is impossible. Hence, b 0 < a 0. Again, c 0 > a 0 a 0 > b 0 + a 0 b 0 + a 0 0 > b 0 + a 0 b 0, leading to a contradiction. Hence, c 0 < a 0. The following lemma gives the relationship between the solutions of (5..) and (5..3). Lemma 5..6 : If (a 0, b 0, c 0 ) is a solution of the Diophantine equation 4a = (c b) + 3b, (5..) then (a 0, b 0, c 0 b 0 ) is a solution of the Diophantine equation 4a = (c + b) + 3b. (5..3) Proof : If (a 0, b 0, c 0 ) is a solution of the Diophantine equation (5..), then a 0 = b 0 b 0 c 0 + c 0. (i) Now, rewriting (i) as a 0 = b 0 + b 0 (c 0 b 0 ) + (c 0 b 0 ), we see that (a 0, b 0, c 0 b 0 ) is a solution of a = b + bc + c, which is equivalent to (5..3). The following table gives the solutions of 4a = (c + b) + 3b for a 00. Table 5.. : Solutions of 4a = (c + b) + 3b for a 00 a b c a b c a b c The above Table 5.. shows that, there are solutions of the Diophantine equation 4a = (c + b) + 3b when a = 7, 3, 9, 3, 37, 43, 6, 67, 73, 79, 97 (and multiples of these primes), and in each case, there is only one solution; on the other hand, there are two independent solutions when a = 49, while a = 9 admits of four independent solutions.

175 74 Wandering in the World of Smarandache umbers 5. Families of 60 Degrees Smarandache Function Related and Pseudo Smarandache Function Related Triangles We start with the following result related to the 60 degrees triangles. Lemma 5.. : Let T(a, b, c) be a triangle with A = 60 0, and integer valued sides a, b and c. Then, () a, b and c satisfy the Diophantine equation 4a = (c b) + 3b, (5..) () the Diophantine equation (5..) has an infinite number of solutions. Proof : Part () of the lemma is a restatement of Lemma 5... Now, if (a 0, b 0, c 0 ) is a solution of the Diophantine equation (5..), then (k a 0, k b 0, k c 0 ) is also its solution for any integer k, and hence, (5..) possesses an infinite number of solutions. Now, by inspection, we see that a = 7, b = 3, c = 8, (5..) is a solution of (5..). Note that, a = 7, b = 5, c = 8, (5..3) is a second independent solution of the Diophantine equation (5..), as can readily be verified. Recall that, two solutions are dependent if one is a (non-zero) constant multiple of the other; otherwise, they are independent. Theorem 5.. : There is an infinite number of pairs of dissimilar 60 degrees triangles that are S related. Proof : We consider the following pair of dissimilar triangles T(7p, 3p, 8p) and T(7p, 5p, 8p), where p is a prime. By Lemma 5.., each is a 60 degrees triangle. Now, since for any prime p 7, S(3p) = p = S(5p), it follows that the (dissimilar) triangles T(7p, 3p, 8p) and T(7p, 5p, 8p) are S related. Theorem 5.. : There is an infinite number of pairs of dissimilar 60 degrees triangles that are Z related. Proof : We consider the following pair of dissimilar 60 degrees triangles T(7p, 3p, 8p) and T(7p, 5p, 8p), where p is a prime of the form p = (3 0)n = 30n, n Z +. (5..4) Since 6 (p + ), 0 (p + ), by Lemma 4..4 (in Chapter 4), Z(3p) = p = Z(5p). Thus, the pair of triangles T(7p, 3p, 8p) and T(7p, 5p, 8p) are Z related. Now, since there are an infinite number of primes of the form (5..4) (by Dirichlet s Theorem, see, for example, Hardy and Wright [3], Theorem 5, pp. 3), we have the desired result.

176 75 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function It may be mentioned here that, if p is a prime of the form p = (3 0)n + = 30n +, n Z +, (5..5) then the pair of dissimilar 60 degrees triangles T(7p, 3p, 8p) and T(7p, 5p, 8p) forms a second infinite family of pairs of triangles that are Z related, since for such p, both 6 and 0 divide (p ), and by Lemma 4..4, Z(3p) = p = Z(5p). We now prove the following theorem. Theorem 5..3 : There exists an infinite family of pairs of 60 degrees triangles that are both S related and Z related. Proof : From Theorem 5.. and Theorem 5.., we see that the pair of 60 degrees triangles T(7p, 3p, 8p) and T(7p, 5p, 8p), where p is a prime of the form p = (3 0)n = 30n, n Z +. are both S related and Z related, since for such p, S(3p) = p = S(5p). 5.3 Families of 0 Degrees Smarandache Function Related and Pseudo Smarandache Function Related Triangles We start with the following lemma that characterizes 0 degrees triangles. Lemma 5.3. : Let T(a, b, c) be a triangle with A = 0 0, and integer valued sides a, b and c. Then, () a, b and c satisfy the Diophantine equation 4a = (c + b) + 3b, (5.3.) () the Diophantine equation (5.3.) has an infinite number of solutions. Proof : Part () follows from Lemma 5... Again, since a 0 = 7, b 0 = 3, c 0 = 5 is a solution of the Diophantine equation (5.3.), it has infinite number of solutions of the form (7k, 3k, 5k), k. This proves part () of the lemma. Hence, the proof is complete. Theorem 5.3. : There is an infinite number of pairs of dissimilar 0 degrees triangles that are S related. Proof : We consider the pair of dissimilar triangles T(49p, 6p, 39p) and T(49p, p, 35p), each of which is a 0 degrees triangle, by Lemma 5... Now, for any prime p 7, S(6p) = p = S(39p) = S(p) = S(35p). Thus, for any prime p 7, the (dissimilar) triangles T(49p, 6p, 39p) and T(49p, p, 35p) are S related. This establishes the theorem.

177 76 Wandering in the World of Smarandache umbers In proving Theorem 5.3. above, we have made use of the fact that (see Table 5..) () a = 49, b = 6, c = 39, (5.3.) () a = 49, b =, c = 35, (5.3.3) are two independent solutions of the Diophantine equation 4a = (c + b) + 3b. Theorem 5.3. : There is an infinite number of pairs of dissimilar 0 degrees triangles that are Z related. Proof : We start with the pair of dissimilar 0 degrees triangles T(49p, 6p, 39p) and T(49p, p, 35p), where p is a prime of the form p = ( )n = n, n Z +. (5.3.4) Since 6 (p + ), 39 (p + ), (p + ), 35 (p + ), by Lemma 4..4, for any prime p of the form (5.3.4), Z(6p) = Z(39p) = Z(p) = Z(35p) = p. Therefore, for any p of the form (5.3.4), the family of pairs of (dissimilar) 0 degrees triangles T(49p, 6p, 39p) and T(49p, p, 35p) are Z related. All these complete the proof of the theorem. A second infinite family of (dissimilar) 0 degrees triangles that are Z related, can be formed as follows : Let p be a prime of the form p = ( )n + = n +, n Z +. (5.3.5) By Lemma 4..4, for such p, Z(6p) = Z(39p) = Z(p) = Z(35p) = p, since 6 (p ), 39 (p ), (p ), 35 (p ). It then follows that the pair of triangles T(49p, 6p, 39p) and T(49p, p, 35p) are Z related for any prime p of the form (5.3.5). Theorem : There is an infinite number of pairs of dissimilar 0 degrees triangles that are both S related and Z related. Proof : From Theorem 5.3. and Theorem 5.3., we see that the pair of triangles T(49p, 6p, 39p) and T(49p, p, 35p), where p is a prime of the form p = ( )n + = n +, n Z +, are both S related and Z related, since for such p, S(6p) = S(39p) = S(p) = S(35p) = p. All these complete the proof of the theorem.

178 77 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function 5.4 Families of Smarandache Function Related and Pseudo Smarandache Function Related Pythagorean Triangles A triangle is called Pythagorean (or, right angled) if one of its angles is 90 degrees. Recall that a triangle T(a, b, c) (with sides of length a, b and c) is Pythagorean, with 0 C = 90, if and only if a + b = c. Thus, the problem of finding the S related / Z related Pythagorean triangles is related to the solutions of the Diophantine equation a + b = c. In what follows, we shall make use of the following two independent solutions of the Diophantine equation a + b = c : () a 0 = 3, b 0 = 4, c 0 = 5, () a 0 = 5, b 0 =, c 0 = 3. The following result is due to Ashbacher []. Theorem 5.4. : There are an infinite family of pairs of dissimilar Pythagorean triangles that are S related. Proof : For any prime p 7, the two families of dissimilar Pythagorean triangles T(3p, 4p, 5p) and T(5p, p, 3p) are S related, since S(3p) = S(4p) = S(5p) = S(p) = S(3p) = p. Theorem 5.4. : There are an infinite number of pairs of dissimilar Pythagorean triangles that are Z related. Proof : We consider the pair of dissimilar Pythagorean triangles T(3p, 4p, 5p) and T(5p, p, 3p), (5.4.) where p is a prime of the form p = ( )n + = 560n +, n Z +. (5.4.) By Lemma 4..4, Z(3p) = Z(4p) = Z(5p) = Z(p) = Z(3p) = p, so that the triangles T(3p, 4p, 5p) and T(5p, p, 3p) are Z related. Since there is an infinite number of primes of the form (5.4.), we get the desired infinite number of pairs of dissimilar Z related Pythagorean triangles. Theorem : There is an infinite number of pairs of dissimilar Pythagorean triangles that are both S related and Z related. Proof : Follows from Theorem 5.4. and Theorem 5.4., since the pair of triangles T(3p, 4p, 5p) and T(5p, p, 3p), where p is a prime of the form p = ( )n + = 560n +, n Z +, are both S related and Z related.

179 78 Wandering in the World of Smarandache umbers Theorem : There exists a family of infinite number of Pythagorean triangles which are S related, pair wise, to the triangles of a second family of infinite number of 60 degrees triangles. Proof : We consider the Pythagorean triangle T(3p, 4p, 5p) and the 60 degrees triangle T(7p, 3p, 8p). Now, for any prime p, S(3p) = S(4p) = S(5p) = S(7p) = S(8p) = p. Thus, the triangles T(3p, 4p, 5p) and T(7p, 3p, 8p) are S related. Theorem : There exists a family of infinite number of Pythagorean triangles which are Z related, pair wise, to the triangles of a second infinite family of 60 degrees triangles. Proof : The Pythagorean triangle T(3p, 4p, 5p) and the 60 degrees triangle T(7p, 3p, 8p) are Z related for any prime p of the form p = ( )n = 680n, n Z +, (5.4.3) since, for such p, Z(3p) = Z(4p) = Z(5p) = Z(7p) = Z(8p) = p. Note that, if p is a prime of the form p = ( )n + = 680n +, n Z +, (5.4.4) then, for such p, Z(3p) = Z(4p) = Z(5p) = Z(7p) = Z(8p) = p. Thus, the Pythagorean triangle T(3p, 4p, 5p) and the 60 degrees triangle T(7p, 3p, 8p) are Z related for any prime p of the form (5.4.4). Theorem : There exists a family of infinite number of Pythagorean triangles which are both S related and Z related, pair wise, to the triangles of a second family of infinite number of 60 degrees triangles. Proof : We consider the Pythagorean triangle T(3p, 4p, 5p) and the 60 degrees triangle T(7p, 8p, 5p), where p is a prime of the form (5.4.3). Then, by Theorem and Theorem 5.4.5, they are both S related and Z related. Theorem : There exists a family of infinite number of Pythagorean triangles which are S related, pair wise, to the triangles of a second infinite family of 0 degrees triangles. Proof : We consider the Pythagorean triangle T(3p, 4p, 5p) and the 0 degrees triangle T(7p, 3p, 5p), where p is a prime. Since, for such p, S(3p) = S(4p) = S(5p) = S(7p) = p, the triangles T(3p, 4p, 5p) and T(7p, 3p, 5p) are S related. Theorem : There exists a family of infinite number of Pythagorean triangles which are Z related, pair wise, to the triangles of a second infinite family of 0 degrees triangles. Proof : For any prime p of the form p = ( 3 3 5)n = 0n, n + Z, (5.4.5) the Pythagorean triangle T(3p, 4p, 5p) and the 0 degrees triangle T(7p, 3p, 5p) are Z related, since, for such p, Z(3p) = Z(4p) = Z(5p) = Z(7p) = p.

180 79 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function It may be mentioned here that, if p is a prime of the form p = ( 3 3 5)n + = 0n +, n Z +, (5.4.6) then, for such p, Z(3p) = Z(4p) = Z(5p) = Z(7p) = Z(3p) = p, and hence, the Pythagorean triangle T(3p, 4p, 5p) and the 0 degrees triangle T(7p, 3p, 5p) are Z related. Theorem : There exists a family of infinite number of Pythagorean triangles which are both S related and Z related, pair wise, to the triangles of a second family of infinite number of 0 degrees triangles. Proof : By Theorem and Theorem 5.4.8, the Pythagorean triangle T(3p, 4p, 5p) and the 0 degrees triangle T(7p, 3p, 5p) are both S related and Z related, where p is a prime of the form (5.4.5). In passing, we state and prove the following three theorems concerning the 60 degrees and 0 degrees S related / Z related triangles. Theorem : There exists a family of infinite number of 60 degrees triangles which are S related, pair wise, to the triangles of a second infinite family 0 degrees triangles. Proof : The 60 degrees triangle T(7p, 3p, 8p) and the 0 degrees triangle T(7p, 3p, 5p) are S related for any prime p, since, in such a case, S(7p) = S(3p) = S(8p) = S(5p) = p. Theorem 5.4. : There exists a family of infinite number of 60 degrees triangles which are Z related, pair wise, to the triangles of a second infinite family of 0 degrees triangles. Proof : If p is a prime of the form p = ( )n = 680n, n Z +, (5.4.7) then the 60 degrees triangle T(7p, 3p, 8p) and the 0 degrees triangle T(7p, 3p, 5p) are Z related, since, for such p, Z(7p) = Z(3p) = Z(8p) = Z(5p) = p. Note that, if p is a prime of the form p = ( )n + = 680n +, n Z +, (5.4.8) then Z(7p) = Z(3p) = Z(8p) = Z(5p) = p. With such a value of p, we get a second pair of 60 degrees triangle T(7p, 3p, 8p) and the 0 degrees triangle T(7p, 3p, 5p) which are Z related. Theorem 5.4. : There exists a family of infinite number of 60 degrees triangles which are both S related and Z related, pair wise, to the triangles of a second family of infinite number of 0 degrees triangles. Proof : If p is a prime of the form (5.4.7), by Theorems and Theorem 5.4., the 60 degrees triangle T(7p, 3p, 8p) and the 0 degrees triangle T(7p, 3p, 5p) are both S related and Z related.

181 80 Wandering in the World of Smarandache umbers 5.5 Smarandache Function Related and Pseudo Smarandache Function Related Triangles in Terms of Their Angles In this section, we consider the problem of finding the S related and Z related triangles in terms of their angles, that is, in the sense of Definition 5.3. We consider the case of S related triangles and Z related triangles separately in 5.5. and 5.5. respectively Smarandache Function Related Triangles Using a computer program, Ashbacher [] searched for S related pairs of triangles (in the sense of Definition 5.3). He reports three such pairs. However, in this case, the introduction of S (m) may be helpful. The Smarandache function S(n) is clearly not bijective. However, we can define the inverse S (m) as follows : S (m) = {n Z + : S(n) = m} for any integer m, (5.5.a) with S () =. (5.5.b) Then, for any m Z + {,, 3, 4, }, the set S (m) is non empty and bounded with m! as its largest element. Clearly, n S (m) if and only if the following two conditions are satisfied : () n divides m!, () n does not divide l! for any l with l m. We can look at (5.5.) from a different point of view : On the set R as follows : + Z, we define the relation For any n, n Z +, n R n if and only if S(n ) = S(n ). (5.5.) + It is then straightforward to verify that R is an equivalence relation on Z. It is well known + that an equivalence relation induces a partition (on the set Z ) (see, for example, Gioia [4], Theorem., pp. 3). The sets S (m), m Z +, are, in fact, the equivalence classes induced by + the equivalence relation R on Z, and possess the following two properties : () () S (m) = + m= Z, S (m ) S (m ) =, if m m. Thus, for any n Z +, there is one and only one m Z + such that n S (m). To deal with the condition (5.) (that is, the condition α + β + γ = 80 = α + β + γ ), we consider the restricted sets S (m π ), defined as follows : S (m ) π = { n Z + : S(n) = m, m 80}. (5.5.3)

182 8 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function The following table gives such restricted sets related to our problem. Table 5.5. : Values of m S (m π) = { n Z + : S(n) = m, n 80 } S (m π) m S (m π) {} 59 {59, 8, 77} {} 6 {6, } 3 {3, 6} 67 {67, 34} 4 {4, 8,, 4} 7 {7, 4} 5 {5, 0, 5, 0, 30, 40, 60, 0} 73 {73, 46} 6 {9, 6, 8, 36, 45, 48, 7, 80, 90, 44, 80} 79 {79, 58} 7 {7, 4,, 8, 35, 4, 56, 63, 70, 84, 05,, 6, 40, 68} 83 {83, 66} 8 {3, 64, 96, 8, 60} 89 {89, 78} 9 {7, 54, 8, 08, 35, 6} 97 {97} 0 {5, 50, 75, 00, 50, 75} 0 {0} {,, 33, 44, 55, 66, 77, 88, 99, 0, 3, 54, 65, 76} 03 {03} 3 {3, 6, 39, 5, 65, 78, 9, 04, 7, 30, 43, 56} 07 {07} 4 {49, 98, 47} 09 {09} 5 {5} 3 {3} 7 {7, 34, 5, 68, 85, 0, 9, 36, 53, 70} 7 {7} 9 {9, 38, 57, 76, 95, 4, 33, 5, 7} 3 {3} {} 37 {37} 3 {3, 46, 69, 9, 5, 38, 6} 39 {39} 6 {69} 49 {49} 9 {9, 58, 87, 6, 45, 74} 5 {5} 3 {3, 6, 93, 4, 55} 57 {57} 37 {37, 74,, 48} 63 {63} 4 {4, 8, 3, 64} 67 {67} 43 {43, 86, 9, 7} 73 {73} 47 {47, 94, 4} 79 {79} 53 {53, 06, 59} Clearly, two S related triangles T(α, β, γ) and T(α, β, γ ) must satisfy the following condition : α, α S (m π ) ; β, β S (m π ) ; γ, γ S (m 3 π ) for some m, m, m 3 Z +. Thus, for example, choosing α, α {} = S ( π ) ; β, β {,, 33, 44, 55, 66, 77, 88, 99, 0, 3, 54, 65, 76} = S ( π ) ; γ, γ {7, 4,, 8, 35, 4, 56, 63, 70, 84, 05,, 6, 40, 68} = S (7 π ),

183 8 Wandering in the World of Smarandache umbers we can construct the S related triangles T(,, 68) and T(, 65, 4). Note that, since S ( π ) = {} is a singleton set, if we choose α =, then we must also choose α =. Again, choosing α, α, β, β {5, 50, 75, 00, 50, 75} = S (0 π ), we can form the pair of S related triangles T(5, 50, 5) and T(75, 00, 5) with the characteristic that S(α) = S(α ) = 0 = S(β) = S(β ) (and S(γ) = S(γ ) = 5). Choosing α, α, β, β {5, 0, 5, 0, 30, 40, 60, 0} = S (5 π ), we see that the equilateral triangle T(60, 60, 60) is S related to each of the two triangles T(30, 30, 0) and T(0, 40, 0)! It may be mentioned here that the pair T(60, 60, 60) and T(0, 40, 0) are Z related as well. Ashbacher [] cites the triangles T(3, 7, 70) and T(6,, 53) as an example of a S related pair where all the six angles are different. We found several by trial and error from Table 5.5., some of which are given below : () T(5, 5, 50) and T(30, 50, 00) (S(5) = S(30) = 5, S(5) = S(50) = 0 = S(50) = S(00)), () T(0, 60, 00) and T(0, 0, 50) (S(0) = S(0) = 5 = S(60) = S(0), S(00) = S(50) = 0), (3) T(5, 45, 30) and T(60, 6, 04) (S(5) = S(60) = 5, S(45) = S(6) = 6, S(30) = S(04) = 3), (4) T(5, 85, 90) and T(60, 0, 8) (S(5) = S(60) = 5, S(85) = S(0) = 7, S(90) = S(8) = 6), (5) T(37, 50, 93) and T(74, 75, 3) (S(37) = S(74) = 37, S(50) = S(75) = 0, S(93) = S(3) = 3), (6) T(40, 50, 90) and T(60, 75, 45) (S(40) = S(60) = 5, S(50) = S(75) = 0, S(90) = S(45) = 6). (7) T(0, 70, 90) and T(60, 84, 36) (S(0) = S(60) = 5, S(70) = S(84) = S(90) = S(36) = 6). The last two are examples where all the six angles are acute. An exhaustive computer search by Ashbacher [] for pairs of all dissimilar S related triangles T(α, β, γ) and T(α, β, γ ) (α+ β + γ = 80 = α + β + γ ) with values of α in the range α 78, revealed that α cannot take the following values in any triplet pairs : 83, 97, 07, 3,, 7, 37, 39, 49, 5, (5.5.4) 63, 66, 67, 69, 7, 73, 74, 75, 76, 77, 78. A closer look at the values in (5.5.4) and those in Table 5.5. reveals the following facts : () Generally, the numbers belonging to singleton sets in S (m π ) do not appear in any 80 degrees pairs of triplets, with the exceptions of α =,, 0, 03, 09, 5, 3, 57. For the last six cases, we can form the following examples : (a) T(0, 9, 70) and T(0, 6, 63) (with S(9) = 6 = S(6), S(70) = 7 = S(63)), (b) With α = 03, we have a set of four triangles, T(03, 7, 70), T(03, 4, 63), T(03,, 56) and T(03, 35, 4) (with S(7) = S(4) = S() = S(35) = 7 = S(70) = S(63) = S(56) = S(4)), which are pair wise S related, and a second set of three triangles, T(03,, 66), T(03,, 55) and T(03, 33, 44) (with S() = S() = S(33) =, S(66) = S(55) = S(44) = ), which are pair wise S related, (c) T(09,, 60) and T(09, 66, 5) (with S() = = S(66), S(60) = 5 = S(5)),

184 83 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function (d) T(5, 5, 50) and T(5, 30, 5) (with S(5) = 5 = S(30), S(50) = 0 = S(5)), as well as the pair T(5,, 44) and T(5,, 33) (with S() = S() = = S(44) = S(33)), (e) For α = 3, we have three triangles, T(3, 7, 4), T(3, 4, 35) and T(3,, 8) (with S(7) = S(4) = S() = 7 = S(4) = S(35) = S(8)), any two of which are S related, (f) T(57, 9, 4) and T(57, 6, 7) (with S(9) = 6 = S(6), S(4) = 7 = S(7)). () None of 83, 66 S (83 π) appears in any triplet. (3) Both 59, 8 S (59 π ) can appear in one or the other triplet pair, but 77 S (59 π ) cannot appear in any triplet pair. The reason is as follows : If α = 77, then the only option for β and γ are β =, γ =. Noting that both S ( π ) = {} and S ( π ) = {} are singleton sets, we cannot find a second dissimilar triangle S related to T(77,, ). The same is true when α = 73 S (73 π ) (which is a singleton set), α = 75 S (0 π) and α = 76 S ( π ). Again, 43, 86, 9 S (43 π ) can appear in one or the other triplet, but 7 S (59 π ) cannot appear in any triplet. The detail is shown below. α = 7 α = 7 α = 9 α = 86 α = 43 Remark (β,γ) (, 7) Similar Triangle Not Possible Not Possible Not Possible (, 6) Similar Triangle Not Possible Not Possible Not Possible β + γ 8 (3, 5) Similar Triangle Not Possible Not Possible Not Possible (4, 4) Similar Triangle Not Possible Not Possible Not Possible β + γ 48 On the other hand, if α = 7, we can get the S related pair of triangles, T(7, 4, 5) and T(5,, 0) (with S(7) = S(5) = 9, S(4) = S() = 4, S(5) = S(0) = 5), and corresponding to α = 70, we have two pairs, namely, the pair T(70, 3, 7) and T(53, 6, ) (with S(70) = S(53) = 7, S(3) = S(6) = 3, S(7) = S() = 7), and the pair of triangles T(70, 4, 6) and T(53, 4, 3) (with S(4) = S(4) = 4). We looked for 60 degrees and 0 degrees S related pairs of triangles, where a 60 (0) degrees triangle is one whose one angle is 60 (0) degrees, in the sense of Definition 5.3. We got only one pair of the first type, namely, the pair of triangles T(60, 30, 90) and T(60, 40, 80) (with S(30) = S(40) = 5, S(90) = S(80) = 6), and two pairs of the second type, namely, the pair T(0, 4, 36) and T(0,, 48) (with S(4) = S() = 4, S(36) = S(48) = 6), as well as the pair T(0, 0, 40), T(0, 30, 30) (S(0) = S(30) = S(40) = 6). We also found three pairs of Pythagorean triangles that are S related; they are (i) T(90, 40, 50) and T(90, 5, 75) (with S(40) = S(5) = 5, S(50) = S(75) = 0), (ii) T(90, 36, 54) and T(90, 9, 8) (with S(36) = S(9) = 6, S(54) = S(8) = 9), (iii) T(90, 8, 7) and T(90, 45, 45) (with S(8) = S(45) = 6 = S(7)). In (5.5.3), the set S (m) has been defined for any integer m Z +. It might be a problem of interest to study the characteristics of the sets S (m). Of particular interest is the set S (p), where p is an odd prime. Some elementary properties of S (m) are given (without proof) in Vasile Seleacu [5].

185 84 Wandering in the World of Smarandache umbers We first prove the following lemma. Lemma 5.5. : p divides all the elements of S (p). Proof : Let n S (p), so that S(n) = p. Now, n p! p! = na for some integer a. Clearly, p does not divide a, for otherwise, p a a = bp for some integer b p! = nbp nb = (p )! S(n) (p )!, contradicting the definition of S(n). Therefore, p does not divide a, and this, in turn, proves the lemma. Now, we observe the following facts about S (p), where p 3 is a prime : () p, p! S (p), p being the smallest element and p! being the largest element of S (p), () kp S (p) for all k p, (3) (p + k )p S (p) for all k p, (4) if n S (p) then p n, (5) S (p) contains d((p )!) number of elements, where d((p )!) denotes the number of (positive) divisors of (p )! (see Stuparu and Sharpe [6]). Note that, if S(n) = m, then m is prime if n is prime. It then follows that, if m is a positive composite number, then n must also be composite too. Thus, if m is composite, then all the elements of S (m) are also composite numbers. Again, for any fixed prime p 3, S (p) contains exactly one prime (namely, p itself). Thus, the sequence of primes p, p, are contained in the respective sets S (p ), S (p ), (which are pair-wise disjoint). The number of elements of the set S (m) is d(n!) d((n )!) (see Vasile Seleacu [5]). A formula for the least element of S (m), denoted by [5] : If m is a composite number of the form α α α m = p p... p, k k where p, p,, p k are the prime factors of m with p < p < < p k, then where however, S min (m!) α k + (m) = p l, k k S min (m), is given in Vasile Seleacu l (m!) is the exponent of p k k in the prime factors decomposition of m!. Note that, S(S (m)) = m for any m Z + ; min S (S(8)) = S (4) = 4. min min

186 85 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function 5.5. Pseudo Smarandache Function Related Triangles Using a computer program, Ashbacher [] searched for Z related pairs of triangles (in the sense of Definition 5.3). He reports three such pairs. However, in this case also, like the S related case in 5.5., the introduction of Z (m) may be helpful. Recall from 4.5 in Chapter 4 that Z (m) is defined as follows : Z (m) = {n Z + : Z(n) = m} for any integer m, (5.5.5) where, n Z (m) if and only if the following two conditions are satisfied : m(m + ) () n divides, l( l + ) () n does not divide for any l with l m. The condition (5.) can be dealt with by considering the restricted sets Z (m ) Z (m ) π : π = {n Z + : Z(n) = m, n 78}. (5.5.6) Table 5.5. gives such restricted sets related to our problem. Clearly, two Z related triangles T(α, β, γ) and T(α, β, γ ) must satisfy the following condition : α, α Z (m π ) ; β, β Z (m π ), γ, γ Z (m 3 π ) for some m, m, m 3 Z +. Thus, for example, choosing α, α {, 6} = Z (3 π ) ; β, β {44, 48, 88, 3, 76} = Z (3 π ), we can construct the Z related triangles T(, 48, 30) and T(6, 44, 30). Again, choosing α, α, β, β {5, 50, 75, 00, 50} = Z (4 π ), we can form the pair of Z related triangles T(5, 50, 5) and T(75, 00, 5) with the characteristic that Z(α) = Z(α ) = 4 = Z(β) = Z(β ) (and Z(γ) = Z(γ ) = 4). Choosing α, α, β, β {8, 0, 4, 30, 40, 60, 0} = Z (5 π ), we see that the equilateral triangle T(60, 60, 60) is Z related to the triangle T(0, 40, 0)! The triangles T(4, 6, 60) and T(4, 6, 04), found by Ashbacher [] by computer search, are an example of a Z related pair where all the six angles are different. We have found three more, given below : () T(8,, 60) and T(40, 36, 04), (with Z(8) = Z(40) = 5, Z() = Z(36) = 8, Z(60) = Z(04) = 64), () T(6, 0, 44) and T(4, 4, 3), (with Z(6) = Z(4) = 3, Z(0) = Z(4) = 5, Z(44) = Z(3) = 63), (3) T(37, 50, 93) and T(74, 75, 3) (with Z(37) = Z(74) = 36, Z(50) = Z(75) = 4, Z(93) = Z(3) = 30).

187 86 Wandering in the World of Smarandache umbers Table 5.5. : Values of Z (m π) = { n Z + : Z(n) = m, n 80 } m Z (m π) m Z (m π) m Z (m π) {} 35 {90} 80 {8, 08, 6, 80} {3} 36 {37, 74, } 8 {83} 3 {, 6} 39 {5, 30, 56} 83 {66} 4 {5, 0} 40 {4, 8, 64} 84 {70} 5 {5} 4 {3} 87 {6, 74} 6 {7, } 4 {43, 9} 88 {89, 78} 7 {4, 4, 8} 43 {86} 95 {5} 8 {9,, 8, 36} 44 {99, 0, 65} 96 {97} 9 {45} 45 {5} 00 {0} 0 {, 55} 46 {47} 0 {03} {, 33, 66} 47 {94, 4} 06 {07} {3, 6, 39, 78} 48 {49, 56, 84, 98, 47, 68} 08 {09} 3 {9} 49 {75} {48} 4 {35, 05} 5 {0} {3} 5 {8, 0, 4, 30, 40, 60, 0} 5 {53, 06} 0 {} 6 {7, 34, 68, 36} 53 {59} 4 {5} 7 {5, 53} 54 {35} 6 {7} 8 {9, 57, 7} 55 {40, 54} 7 {64} 9 {38, 95} 56 {76, 4, 33} 8 {7} 0 {4, 70} 58 {59} 30 {3} {77} 59 {8, 77} 36 {37} {3} 60 {6, } 38 {39} 3 {46, 69, 9, 38} 63 {3, 7, 96,, 44} 48 {49} 4 {5, 50, 75, 00, 50} 64 {80, 04, 60} 50 {5} 5 {65} 65 {43} 56 {57} 6 {7, 7} 66 {67} 6 {63} 7 {54, 63, 6} 67 {34} 66 {67} 8 {9, 58} 69 {6} 68 {69} 9 {87, 45} 70 {7} 7 {73} 30 {3, 93, 55} 7 {4} 78 {79} 3 {6, 6, 4} 7 {73, 46} 55 {8} 3 {44, 48, 88, 3, 76} 78 {79} 34 {85, 9} 79 {58} Ashbacher [], based on an exhaustive computer search for pairs of all dissimilar Z related triangles T(α, β, γ) and T(α, β, γ ) (α + β + γ = 80 = α + β + γ ) with values of α in the range α 78, reports that α cannot take the following values in any pair of triplets :

188 87 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function, 5, 3, 35, 4, 45, 5, 59, 65, 67, 7, 73, 77, 79, 8, 83, 86, 87, 89, 90, 9, 97, 0, 0, 05, 07, 09, 3, 5, 6, 8,, 3, 5, 6, 7, 3, 34, 35, 37, 39, 4, 4, 43, 48, 49, 5, 5, 53, 57, 58, 59, 6, 63, 64, 66, 67, 69, 70, 7, 7, 73, 74, 75, 76, 77, 78. Can the table of the sets Z (m π ), Table 5.5., be utilized in explaining this observation? Very large value of α often forces the two triangles to be similar. For example, if we take α = 74 Z (87 π ), then the three possible pairs of values of (β, γ) are (, 5), (, 4) and (3, 3). Note that the two sets Z ( π ) = {} and Z ( π ) = {3} are each singleton. Thus, if α = 74, then the two triangles T(α, β, γ) and T(α, β, γ ) must be similar. On the other hand, if α = 6 Z (87 π ), then we cannot find β, γ with β + γ = 64. Now, we consider the case when α =64. This case is summarized in the tabular form below : α = 64 α = 64 α = 8 α = 4 Remark (β, γ) (, 5) Similar Triangle Not Possible Not Possible Both belong to singleton sets (, 4) Similar Triangle Not Possible Not Possible β + γ 34 (3, 3) Similar Triangle Not Possible Not Possible 3 belongs to a singleton set (4, ) Similar Triangle Not Possible Not Possible β + γ 64 (5, ) Similar Triangle Not Possible Not Possible β + γ 65 (6, 0) Similar Triangle Not Possible Not Possible β + γ 6 (7, 9) Similar Triangle Not Possible Not Possible β + γ 57 (8, 8) Similar Triangle Not Possible Not Possible However, if α = 65, we can get two dissimilar Z related triangles, namely, T(65, 7, 8) and T(8,, 60). A closer look at the values in (5.5.7) and Table 5.5. reveals the following facts : () The numbers not appearing in any 80 degrees pair of triplets are all belong to singleton sets, with the exception of 3, 47, 64, and 03. For the last three cases, we have the following examples : (a) T(47, 76, 57) and T(47, 4, 9) (with Z(76) = 56 = Z(4), Z(57) = 8 = Z(9)), (b) T(64,, 04) and T(64, 36, 80) (with Z() = 8 = Z(36), Z(04) = 64 = Z(80)), (c) T(03,, 66) and T(03, 55, ) (with Z() = 0 = Z(55), Z(66) = = Z()). () In most of the cases, if a value does not appear in the triplet pair, all other values of the corresponding Z (m π ) also do not appear in other triplet pair, with the exception of 45 Z (9 π ) (87 Z (9 π ) does not appear in any triplet pair), and 46 Z (7 π ) (73 Z (7 π ) does not appear in any triplet pair). In this connection, we may mention the following pair of triplets : (a) T(45, 4, ) and T(45, 8, 7) (with Z(4) = 7 = Z(8), Z() = 6 = Z(7)), (b) T(46, 4, 30) and T(46, 4, 0) (with Z(4) = 7 = Z(4), Z(30) = 5 = Z(0)). (3) The only number in Z (7 ) π that cannot appear in any triplet pair is 6. (5.5.7)

189 88 Wandering in the World of Smarandache umbers We searched for 60 degrees, 0 degrees and Pythagorean Z related pairs of triangles, in the sense of Definition 5.3 (where a 60 (0) degrees triangle is one whose one angle is 60 (0) degrees). We have found the following two pairs of Z related 60 degrees triangles : () T(8, 60, ) and T(4, 60, 96) (with Z(8) = Z(4) = 5, Z() = Z(96) = 63), () T(3, 60, 88) and T(7, 60, 48) (with Z(3) = Z(7) = 63, Z(88) = Z(48) = 3). Note that, in the second example, all the six angles are acute. As for the 0 degrees Z related triangles, we have found only one pair, namely, the triangles T(0, 40, 0) and T(30, 30, 0) (with (Z(0) = Z(30) = Z(40) = Z(0) = 5), while our search for a pair of Z related Pythagorean triangles went in vain. We also looked for dissimilar pairs of Z related triangles where all the six angles are acute. In addition to the second pair listed above, we have the following pairs : (i) T(7, 0, 88), T(7, 60, 48) (with Z(0) = 5 = Z(60), Z(88) = 3 = Z(48)), (ii) T(88, 0, 7), T(88, 60, 3) (with Z(0) = 5 = Z(60), Z(7) = 63 = Z(3)), (iii) T(54, 4, 84), T(54, 70, 56) (with Z(4) = 0 = Z(70), Z(84) = 48 = Z(56)), (iv) T(70,, 88), T(70, 66, 44) (with Z() = = Z(66), Z(88) = 3 = Z(44)), (v) T(80, 5, 75), T(80, 50, 50) (with Z(5) = 4 = Z(50) = Z(75)), (vi) T(8,, 88), T(8, 55, 44) (with (Z() = 0 = Z(55), Z(88) = 3 = Z(44)). It may be mentioned here that the last four pairs are S related as well, since S(4) = 7 = S(70) = S(84) = S(56), S() = = S(66) = S(88) = S(44), S(5) = 0 = S(50) = S(75), S() = = S(55) = S(88) = S(44). Using computer search, we looked for Z related pairs of triangles under the additional condition that α = α. Our search resulted in 59 such pairs, of which 5 pairs are S related as well. Our search shows that α = α cannot take the following values in any pair :,, 3, 6, 7, 8, 3, 4, 5, 6, 7, 8,,, 3, 4, 9, 3, 34, 35, 38, 39, 40, 4, 43, 45, 46, 49, 50, 5, 5, 53, 56, 57, 58, 59, 6, 6, 65, 66, 67, 69, 73, 74, 76, 77, 78, 79, 8, 83, 84, 85, 86, 87, 89, 90, 9, 9, 93, 95, 96, 97, 98, 99, 0, 0, 04, 05, 06, 07, 09, 0,, 3, 4, 5, 6, 7, 8, 9,,, 3, 4, 5, 6, 7, 33, 34, 35, 4, 4, 43, 44, 47, 48, 49, 50, 5, 5, 53, 55, 56, 57, 58, 59, 60, 6, 6, 63, 64, 65, 66, 67, 68, 69, 70, 7, 7, 73, 74, 75, 76, 77, 78. Similar search for S related pairs of triangles (under the same condition that α = α ) found 07 pairs, with the maximum number of 46 pairs when α = α = 37. The search shows further that α = α cannot take the following values in any pair : 58, 7, 73, 77, 83, 86, 9, 97, 07,, 3, 8,, 3, 7, 3, 37, 39, 4, 4, 44, 47, 48, 49, 5, 53, 54, 56, 58, 59, 6, 6, 63, 65, 66, 67, 68, 69, 70, 7, 7, 73, 74, 75, 76, 77, 78.

190 89 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function 5.6 Some Observations and Remarks Theorem 5.. and Theorem 5.. in 5. prove respectively the existence of pairs of infinite families of 60 degrees triangles that are S related and Z related, the corresponding results in the case of the 0 degrees triangles are given in Theorem 5.3. and Theorem 5.3. respectively in 5.3, while Theorem 5.4. and Theorem 5.4. respectively in 5.4 settles the cases with the Pythagorean triangles. However, note that, the values of a, b and c, as well as the values a, b, and c in the relevant pairs of triangles T(a, b, c) and T(a, b, c ) are quite large, particularly for the case of Z related triangles. This is unavoidable if we want to prove the infinitude of the families of pairs of dissimilar 60 degrees, 0 degrees and Pythagorean triangles that are S related or Z related. It may be mentioned that, it is possible to find pairs of dissimilar 60 degrees (and 0 degrees) triangles T(a, b, c) and T(a, b, c ) that are S related or Z related, with smaller values of a, b, c, and a, b, c. To find such pairs of triangles manually (without using any computer program), we have to look at the solutions of the Diophantine equations (5..) and (5.3.) more closely. First, we consider the Diophantine equation 4a = (c b) + 3b. (5..) By Lemma 5.., the problem of finding the pair of dissimilar 60 degrees triangles is equivalent to the problem of finding (non trivial) independent solutions of the Diophantine equation (5..). Lemma 5..4 guarantees that the solutions of the equation (5..) occur in independent pairs. Once we can find a pair of independent solutions of (5..), we can manipulate with them to find a pair of triangles that are S related or Z related. We illustrate it with the help of an example below. Example 5.6. : Two independent solutions of the Diophantine equation (5..) are () a = 7, b = 3, c = 8, (5..) () a = 7, b = 5, c = 8. (5..3) Using these two solutions, we can form the family of (dissimilar) pairs of 60 degrees triangles, T(7k, 3k, 8k) and T(7k, 5k, 8k), where k is an integer. Now, if we choose k such that Z(3k) = Z(5k), () then the corresponding pair of triangles becomes Z related. Choosing k = 8 (so that Z(3k) = Z(4) = 5 = Z(40) = Z(5k)), we get the pair of 60 degrees triangles T(56, 4, 64) and T(56, 40, 64), which are Z related. The choice k = 4 gives a second pair of triangles T(98, 4, ) and T(98, 70, ) (since Z(4) = 0 = Z(70)). And k = 33 gives the third pair (since Z(99) = 44 = Z(65)). We can still get more, besides those corresponding to k = 9, 3 (from (5..4) and (5..5) with n = ), for example, those corresponding to k = 36, 5. Thus, for this particular pair of independent solutions of the Diophantine equation (5..), the problem of finding pair(s) of (dissimilar) 60 degrees Z related triangles reduces to the problem of finding the solutions of (). So, the question arises : How to find the solutions of ()?

191 90 Wandering in the World of Smarandache umbers Note that, since S(4) = 7 = S(70), the pair of triangles T(98, 4, ) and T(98, 70, ) is S related as well. That is, T(98, 4, ) and T(98, 70, ) are both S related and Z related. With the solutions (5..) and (5..3) of (5..), we can think of other possibilities as well : Case () : Are there integers k and k, with k k (k, k ), such that S(7k) = S(7k ), S(3k) = S(3k ), S(8k) = S(8k )? Case () : Are there integers k and k, with k k (k, k ), such that S(7k) = S(7k ), S(5k) = S(5k ), S(8k) = S(8k )? Case (3) : Are there integers k and k (k, k ), such that S(7k) = S(7k ), S(3k) = S(5k ), S(8k) = S(8k )? In Case (), choosing k = and k = 3 = 6, we get the pair of triangles T(7, 5, 8) and T(, 5, 4), which are S related (with S(7) = 7 = S(), S(5) = 5 = S(5), S(8) = 4 = S(4)). In Case (3), choosing k = 5 = 0 and k =, we get the pair of triangles T(70, 30, 80) and T(4, 0, 6), which are S related (with S(70) = 7 = S(4), S(30) = 5 = S(0), S(80) = 6 = S(6)). In the case of the Z related triangles, the three possibilities are Case (i) : Are there integers k and k, with k k (k, k ), such that Z(7k) = Z(7k ), Z(3k) = Z(3k ), Z(8k) = Z(8k )? Case (ii) : Are there integers k and k, with k k (k, k ), such that Z(7k) = Z(7k ), Z(5k) = Z(5k ), Z(8k) = Z(8k )? Case (iii) : Are there integers k and k (k, k ), such that Z(7k) = Z(7k ), Z(3k) = Z(5k ), Z(8k) = Z(8k )? These cases are more difficult to deal with. Case (iii) with k = 4, k = 4, gives the pair of Z related triangles T(98, 4, ) and T(94, 0, 336), since Z(98) = 48 = Z(94), Z(4) = 0 = Z(0), Z() = 63 = Z(336). Example 5.6. : Since () a 3 = 3, b 3 = 7, c 3 = 5, () a 4 = 3, b 4 = 8, c 4 = 5. are two independent solutions of the Diophantine equation (5..) (as can easily be verified), the two triangles T(3k, 7k, 5k) and T(3k, 8k, 5k) are 60 degrees Z related if k is such that Z(7k) = Z(8k). Since, for k = 7, Z(49) = 48 = Z(56), we get the desired pair of triangles T(9, 49, 05) and T(9, 56, 05). Choosing k =, we get the pair of triangles T(43, 77, 65) and T(43, 88, 65), which are S related. Example : We now consider the following pair : () a = 7, b = 3, c = 8, () a 4 = 3, b 4 = 7, c 4 = 5. By Example 5.6. and Example 5.6., each is a solution of the Diophantine equation (5..). Now, using these two solutions, we can form pairs of 60 degrees triangles that are S related. One such pair is T(9, 5, 36) and T(, 9, 55), since S(9) = S(7 7) = 7 = S(3 7) = S(), S(5) = S(3 7) = 7 = S(7 7) = S(9), S(36) = S(8 7) = 7 = S(5 7) = S(55).

192 9 Chapter Chapter 5 : Smarandache 4 : The Pseudo umber Smarandache Related Triangles Function Next, we confine our attention to the Diophantine equation 4a = (c + b) + 3b. (5..) A look at Table 5.. shows that, it is more difficult to form pairs of 0 degrees triangles that are S related / Z related, even with the two independent solutions of the Diophantine equation (5..) corresponding to a = 49. From Theorem 5.3. with p = 7, we get the pair of 0 degrees triangles T(833, 7, 663) and T(833, 357, 595) that are S related. The following example illustrates how to construct pairs of 0 degrees S related triangles, using the solutions of (5..). Example : Two independent solutions of the Diophantine equation (5..) are () a = 7, b = 3, c = 5, () a = 3, b = 7, c = 8. Using these two solutions, we can form a pair of (dissimilar) 0 degrees triangles, T(7k, 3k, 5k) and T(3k, 7k, 8k ), where k and k are integers. Now, if we choose k and k such that S(7k) = S(3k ), S(3k) = S(7k ), S(5k) = S(8k ), then the corresponding pair of triangles becomes S related. Choosing k = 7, k = 7 (so that S(7k) = S(9) = 7 = S() = S(3k ), S(3k) = S(5) = 7 = S(9) = S(7k ), S(5k) = S(85) = 7, 7 = S(36) = S(8k )), we get the pair of 0 degrees triangles T(9, 5, 85) and T(, 9, 36), which are S related. We can now form more; in fact, each member of the family of similar triangles T(9m, 5m, 85m), m 6, is S related to each of the family of similar triangles T(n, 9n, 36n), n 6. Remark 5.6. : To construct a pair of (dissimilar) 0 degrees S related triangles, using the (independent) solutions (3, 7, 8) and (49, 6, 39) of the Diophantine equation (5..), we consider the triangles T(3k, 7k, 8k) and T(49k, 6k, 39k ), where k and k are integers such that S(3k) = S(49k ), S(7k) = S(6k ), S(8k) = S(39k ). Choosing k = k = 7, we get the desired pair T(, 9, 36) and T(833, 7, 663). However, if we agree that the order of the sides (of the two triangles) can be disregarded (that is, the triangles T(a, b, c) and T(a, b, c ) are S related if S(a) equals any one of S(a ), S(b ) and S(c ), S(b) equals any one of the remaining two, and S(c) equals the third one), then we can get another pair of S related triangles from T(3k, 7k, 8k) and T(49k, 6k, 39k ) as follows : We choose k = 7 and k =. The resulting pair, also found by Ashbacher [] by computer search, is T(364, 96, 4) and T(98, 3, 78), where S(a) = S(364) = 3 = S(78) = S(c ), S(b) = S(96) = 4 = S(98) = S(a ), S(c) = S(4) = 8 = S(3) = S(b ). The construction of a pair of 0 degrees Z related triangles, using the solutions of the Diophantine equation (5..), is much more difficult, particularly because only very little is known about the values of Z(n). We illustrate below how this pair, also found by Ashbacher [], can be obtained from the solutions of the Diophantine equation (5..). Example : The pair of 0 degrees triangles T(3k, 7k, 8k) and T(49k, 6k, 39k ) (considered in Remark 5.6.) are Z related if we choose k = 4 and k =. The resulting pair is T(3, 68, 9) and T(588, 9, 468), with Z(3) = 43 = Z(468), Z(68) = 48 = Z(588). Note that, in this case, the order of the sides of the two triangles is disregarded.

193 9 Wandering in the World of Smarandache umbers We close this last chapter of the book with the following open problems and conjectures. Question 5.6. : Is it possible to find a general explicit formula for all the solutions of the Diophantine equation 4a = (c b) + 3b? Table 5.. shows that, for a 00, the solutions of the equation 4a = (c b) + 3b exists when a is a prime of the form 3n +. Note that, if a solution of 4a = (c b) + 3b is known, then a second independent solution of the equation is also known, by Lemma 5..4; moreover, this enables us to find a solution of the Diophantine equation 4a = (c + b) + 3b as well, by virtue of Lemma Question 5.6. : Is it possible to find a good upper bound to the number of pairs of dissimilar triangles that are S related / Z related, in the sense of Definition 5.? Question : Is it possible to find a good upper bound to the number of pairs of dissimilar 60 degrees triangles that are S related / Z related, in the sense of Definition 5.? Question : Is it possible to find better upper bounds to the number of elements of the set S (m π )? Conjecture 5.6. : There is no pair of dissimilar 90 degrees triangles that are Z related, in the sense of Definition 5.3. Question : Given a triangle T(a, b, c), what is the condition that a similar triangle T(a, b, c ) exists such that T(a, b, c) and T(a, b, c ) are S related / Z related? Given the right-angled triangle T(3, 4, 5), we can find a similar triangle, namely, T(6, 8, 0), such that T(3, 4, 5) and T(6, 8, 0) are S related; but, there is no triangle which is similar and Z related to T(3, 4, 5). However, the right angled triangles T(8, 80, 8) and T(36, 60, 64) are Z related. Note that, the triangles T(8, 80, 8) and T(36, 60, 64) are not S related. References [] Sastry, K. R. S., Smarandache Number Related Triangles, Smarandache otions Journal,, 07 09, 000. [] Ashbacher, Charles, Solutions to Some Sastry Problems on Smarandache Number Related Triangles, Smarandache otions Journal,, 0, 000. [3] Hardy, G.H. and Wright, E.M., An Introduction to the Theory of Numbers, 5 th Edition, Oxford Science Publications, Clarendon Press, Oxford, UK. 00. [4] Gioia, A. A., The Theory of Numbers, An Introduction, Dover Publications, Inc., 00. [5] Vasile Seleacu, On the Numerical Function S min, Smarandache otions Journal,, 3 4, 000. [6] Stuparu, A. and Sharpe, D.W., Problem of Number Theory, Smarandache Function (Book Series), Vol. 4 5, 4, 994. [7] Majumdar, A. A. K., On Some Pseudo Smarandache Function Related Triangles, Scientia Magna, 4(3), 95 05, 008.

194 93 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) APPE DIX A. : Values of Z(n) for n = ()5000 Table A.. : Values of Z(n) for n = ()35 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

195 94 Wandering in the World of Smarandache umbers Table A.. : Values of Z(n) for n = 353()70 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

196 95 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A..3 : Values of Z(n) for n = 7()088 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

197 96 Wandering in the World of Smarandache umbers Table A..4 : Values of Z(n) for n = 089()456 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

198 97 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A..5 : Values of Z(n) for n = 457()84 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

199 98 Wandering in the World of Smarandache umbers Table A..6 : Values of Z(n) for n = 85()9 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

200 99 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A..7 : Values of Z(n) for n = 93()558 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

201 00 Wandering in the World of Smarandache umbers Table A..8 : Values of Z(n) for n = 56()98 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

202 0 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A..9 : Values of Z(n) for n = 99()396 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

203 0 Wandering in the World of Smarandache umbers Table A..0 : Values of Z(n) for n = 397()3664 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

204 03 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A.. : Values of Z(n) for n = 3665()403 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

205 04 Wandering in the World of Smarandache umbers Table A.. : Values of Z(n) for n = 4033()4400 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

206 05 Chapter 4 : The Pseudo Appendix Smarandache A. : Values Function of Z(n) Table A..3 : Values of Z(n) for n = 440()4768 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

207 06 Wandering in the World of Smarandache umbers Table A..4 : Values of Z(n) for n = 4769()5000 n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n) n Z(n)

208 07 Chapter 4 : The Appendix Pseudo Smarandache A. : Values Function of Z(np) APPE DIX A. : Values of Z(np); n = 6, 7, 8, 0,,, 3, 6, 4, 3, 48, 96 Table. : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for p 8 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

209 08 Wandering in the World of Smarandache umbers Table. : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for 9 p 449 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

210 09 Chapter 4 : The Appendix Pseudo Smarandache A. : Values Function of Z(np) Table.3 : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for 457 p 743 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

211 0 Wandering in the World of Smarandache umbers Table.4 : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for 75 p 05 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

212 Chapter 4 : The Appendix Pseudo Smarandache A. : Values Function of Z(np) Table.5 : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for 06 p 399 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

213 Wandering in the World of Smarandache umbers Table.6 : Values of Z(6p), Z(7p), Z(8p), Z(0p), Z(p), Z(p), Z(3p), Z(6p), Z(4p), Z(3p), Z(48p) and Z(96p) for 409 p 5 p Z(6p) Z(7p) Z(8p) Z(0p) Z(p) Z(p) Z(3p) Z(6p) Z(4p) Z(3p) Z(48p) Z(96p)

214 3 Chapter 4 : The Appendix Pseudo Smarandache A.3 : Values of Function Z(p. k ) APPE DIX A.3 : Values of Z(p. k ) for p =, 3, 7, 9, 3, 9, 3 Table A.3. : Values of Z(. k ), Z(3. k ), Z(7. k ), Z(9. k ), Z(3. k ), Z(9. k ), Z(3. k ) k Z(. k ) Z(3. k ) Z(7. k ) Z(9. k ) Z(3. k ) Z(9. k ) Z(3. k )

215 4 Wandering in the World of Smarandache umbers Acknowledgements Part of the work was done during the Academic Development Leave (ADL) for six months from September 00 to March 0 from the Ritsumeikan Asis-Pacific University, Japan. The author acknowledges with thanks the Ritsumeikan APU for granting the ADL. Thanks are also due to the Department of Mathematics, Jahangirnagar University, Bangladesh, for allowing their research resources during the author s ADL period. Last, but not the least, my colleague at APU, Professor Dr. Hary Gunarto, deserves my thanks for his computer works in the preparation of the tables in the Appendices at the end of the book. Khan Md. Anwarus Salam, a graduate student at the University of Electro-Communications, Japan, volunteered for the design of the cover page with the theme in accordance with the title of the book.

216 5 Chapter 4 : The Pseudo Smarandache Function SUBJECT I DEX Arithmetical function d(n) divisor function 8,7,77,80,50,5,53 φ(n) Euler phi function 8,73,74,53 σ(n) 8,55 Pseudo Smarandache function 89 additive analogue of 40 definition of 89 dual of 38 explicit expressions of 93 properties of 90 palindromes 67 related triangles 69 S(n) = Z(n) 56 S(Z(n)) = S(n) 58 S(Z(n)) = Z(n) 58 series involving 47 Z(n) = n 4 Z(kn) = n 4 Z(n) 43,6 Z(n) Z(n + ) 44 Z(n) Z(n) = d(n) 50 Z(n) + φ(n) = d(n) 53 Z(n) + d(n) = n 53 Z(n) = φ(n) 53,60 Z(n) + φ(n) = n 53 Z(n) = σ(n) 55 [Z(n)] + Z(n) = n 56 Z(S(n)) = S(n) 58 Z(n) = n 6 Z(n) = n 63 Smarandache determinant sequences 39 bisymmetric arithmetic determinant sequence 46,50 bisymmetric determinant natural sequence 47,48 cyclic arithmetic determinant sequence 4 cyclic determinant natural sequence 45 Smarandache function 53 additive analogue of 64 completely S perfect numbers 80 definition of 53 dual of 6,85 explicit expressions of 58,59,60,6 number of primes 8 properties of 54,55,56,57 related triangles 69 S(n) = p 66 S(n) = n 67 S(kn) = n 67 S(n) S(n + ) = n 68,84 S(n) + d(n) = n 7 S(n) = φ(n) 73 [S(n)] + S(n) = kn 75 S(n + ) = S(n) 87 Smarandache perfect numbers 76,77 series involving 69 Smarandache sequences 9 circular sequence 9,6 determinant sequences 39 even sequence 9,4 higher power product sequences 9, odd sequence 9, permutation sequence 9,3 pierced chain sequence 9,3 prime product sequence 9,7 reverse sequence 9,8 series involving 36 square product sequences 9,0 symmetric sequence 9,30 S related triangles 69,74,75,77,80 60 degrees triangles 74 0 degrees triangles 75 S (m) 80 Z related triangles 69,74,75,77,85 Z (m) 64,85

217 6 Wandering in the World of Smarandache umbers This book covers only a part of the wide and diverse field of the Smarandache Notions, and contains some of the materials that I gathered as I wandered in the world of Smarandache. Most of the materials are already published in different journals, but some materials are new and appear for the first time in this book. All the results are provided with proofs. Chapter gives eleven recursive type Smarandache sequences, namely, the Smarandache Odd, Even, Prime Product, Square Product (of two types), Higher Power Product (of two types), Permutation, Circular, Reverse, Symmetric and Pierced Chain sequences Chapter deals with the Smarandache Cyclic Arithmetic Determinant and Bisymmetric Arithmetic Determinant sequences, and series involving the terms of the Smarandache bisymmetric determinant natural and bisymmetric arithmetic determinant sequences Chapter 3 treats the Smarandache function S(n) Chapter 4 considers, in rather more detail, the pseudo Smarandache function Z(n) And the Smarandache S-related and Z-related triangles are the subject matter of Chapter 5. To make the book self-contained, some well-known results of the classical Number Theory are given in Chapter 0. In order to make the book up-to-date, the major results of other researchers are also included in the book. At the end of each chapter, several open problems are given.