k-tuples of Positive Integers with Restrictions

Transcription

1 International Mathematical Forum, Vol. 13, 2018, no. 8, HIKARI Ltd, k-tuples of Positive Integers with Restrictions Rafael Jakimczuk División Matemática, Universidad Nacional de Luján Buenos Aires, Argentina Copyright c 2018 Rafael Jakimczuk. This article is stributed under the Creative Commons Attribution License, which permits unrestricted use, stribution, and reproduction in any meum, provided the original work is properly cited. Abstract We study the number of k-tuples not exceeng x with very general restrictions. The methods used are elementary. Mathematics Subject Classification: 11A99, 11B99 Keywords: Positive integers, k-tuples, greatest common visor, restrictions 1 Introduction and Preliminary Notes Let k 2 an arbitrary but fixed positive integer. Let us consider a k-tuple of positive integers a 1,..., a k where 1 a i x i = 1,..., k. The number of these k-tuples such that gcda 1,... a k = g will be denoted N k g x. It is well-known that N k g x = 1 g k ζk xk + ox k 1 where ζs denotes, as usual, the Riemann zeta function. Equation 1 was proved by Nymann [4] and another proofs are given by Jakimczuk [3]. Tóth [5] stued the case when the a i are pairwise relatively prime. De Reyna and Heyman [1] stued k-tuples with certain restrictions. In this article we study k-tuples with very general restrictions.

2 376 Rafael Jakimczuk Note that if we have two non empty sets A and B then we can write AUB as an union of at most 3 non empty pairwise sjoint sets that is, a partition, namely A A B, B A B and A B. Besides this partition establish a partition in at most 2 sjoint sets of A, namely A A B and A B, and a partition in at most 2 sjoint sets of B, namely B A B and A B. This is also true for 3 sets A, B and C. If the reader draw three circles then shall see the 7 sets of the partition of A B C and also the three partitions in 4 sets of each set A, B and C. This is true for any number of sets as it is showed in the following lemma. Lemma 1.1 Let us consider the m nonempty sets A 1, A 2,... A m. Then there exists a partition of A 1 A2 Am in at most 2 m 1 nonempty sets such that is partition contains a partition of each set A i i = 1, 2,..., m in at most 2 m 1 sets. If all sets in the partition have at least 2 elements we shall say that the partition is a 2-partition. If the partition of each set A i i = 1, 2,..., m have at least a set with 2 elements we shall say that the partition is a 2 1-partition. Equivalently the partition is a 2 1-partition if there is not a set A i partitioned in unitary sets. Proof. Let us consider the elements of A 1 A2 Am. There exist the following possibilities: There are elements that pertain only a one set, there are elements that pertain only to two sets,..., there are elements that pertain to all sets. This is a partition of A 1 A2 Am in at most m m m + + = 2 m m nonempty sets such that this partition contains a partition of each set A i i = 1, 2,..., m in at most m 1 m 1 m 1 m = 2 m m 1 nonempty sets. The lemma is proved. The following lemmas are well-known and their proofs are simples. Lemma 1.2 Let us consider r finite sets of positive integers A 1, A 2,..., A r such that gcda i = d i i = 1, 2,..., r. Then r gcd A i = gcd gcda 1, gcda 2,..., gcda r. Lemma 1.3 Let us consider a set of positive integers A = {a 1, a 2,..., a n } such that gcda = gcda 1, a 2,..., a n = d and a positive integer a n+1. Then gcda {a n+1 } = gcda 1, a 2,..., a n, a n+1 = gcdd, a n+1

3 k-tuples of positive integers with restrictions 377 We have the following theorem. Theorem 1.4 Let q 1,..., q s be fferent prime numbers and let Bx be the number of positive integers not exceeng x relatively prime to q 1 q s. Then s Bx = 1 1qi x + ox Proof. It is an immeate consequence of the inclusion-exclusion principle, see [2]. Lemma 1.5 Let us consider k 1 sets of positive integer A i = {a i,1, a i,2,..., a i,si } i = 1, 2,..., k such that gcda i = gcda i,1, a i,2,..., a i,si = d i i = 1, 2,..., k. Now, let us consider a positive integer n and the system of equations see Lemma 1.3 gcda i {n} = gcdai,1, a i,2,..., a i,si, n = gcdd i, n = f i i = 1,..., k 2 where the f i i = 1,..., k are positive integers. Then the system has a solution n if and only if gcd, lcmf 1,..., f k = 1 i = 1,..., k 3 f i f i and in this case, the number of solutions n such that n x is 1 1 1qi x + ox = cx + ox 4 lcmf 1,..., f k q i where the product run on the primes q i such that q i vides the product k d i f i. Proof. If the system 2 has a solution n we have gcd f i, n f i = 1 i = 1,..., k Therefore n is multiple of lcmf 1,..., f k, that is, where h is a positive integer. Then gcd n = lcmf 1,..., f k h 5, lcmf 1,..., f k h f i f i = 1 i = 1,..., k 6

4 378 Rafael Jakimczuk and consequently gcd, lcmf 1,..., f k = 1 f i f i i = 1,..., k 7 That is, contion 3. On the other hand, if contion 7 holds then contion 6 holds if and only if gcd, h = 1 i = 1,..., k 8 f i Equations 5, 8 and Theorem 1.4 give equation 4. The lemma is proved. Let us consider a tuple A with, for example, 5 positive integers a 1, a 2, a 3, a 4, a 5, we shall call subtuple of A a tuple whose elements pertain to A, for example, a 1, a 2, a 3, a 4, a 5, a 2, a 4, a 5, a 1, a 2, a 3 and a 3, a 5 are subtuples of A. Let k be an arbitrary but fixed positive integer. In this article we study the number of k-tuples a 1, a 2,..., a k not exceeng x, that is 1 a i x i = 1,..., k, such that the elements of certain subtuples have a greatest common visor fixed depenng of the subtuple. 2 Main Results Theorem 2.1 Let k 2 be an arbitrary but fixed positive integer. Let us consider the k-tuple A = a 1, a 2,..., a k and t subtuples T j = a i1, a i2,..., a isj j = 1,..., t 2 s j k. The sets of elements in these subtuples T j are the sets R j = {a i1, a i2,..., a isj } j = 1,..., t where we establish the restrictions gcdr j = d j j = 1,..., t. We also put the following restriction: the partition see Lemma 1.1 that correspond to the R j sets j = 1,..., t is a 2-partition. Then if Ax is the number of k-tuples not exceeng x that satisfy these restrictions we have Ax = ρx k + o x k 9 where the positive constant ρ 1 is defined bellow in the proof. Remark 2.2 In the former theorem we assume that there exists a k-tuple that satisfies the restrictions. In contrary case the problem has not solution. Proof. Suppose that the R j are pairwise sjoint that is, they already are a partition then we have see equation 1 t x s j t Ax = d j s j ζsj + o k xs j j=1 x s j t 1 = x k + o x k ζs j d j s j j=1 j=1

5 k-tuples of positive integers with restrictions 379 and the theorem is proved. If the R j are not pairwise sjoint then we use Lemma 1.1. Therefore there exists a partition of the set t j=1 R j in n 2 t 1 sets S such that each set R j is partitioned in M j sets S 1,j, S 2,j,..., S Mj,j. The sets S have e 2 elements, since by hypothesis the partition is a 2-partition. For each set S we consider the associated subtuple U formed with the elements of S. Suppose that the problem has solution. Then there exist n d positive integers such that gcds = d and such that the restrictions gcdr j = gcd d 1,j, d 2,j,..., d Mj,j = dj are fulfilled see Lemma 1.2. This set of n d positive integers we shall call a compatible set with the restrictions and it contribute to Ax with see equation 1 x e k U ζe d + o e e xe U x = 1 x k + ox k 10 U ζe d e Let V be the set such that its elements are the sets of all possible compatible sets with the restrictions. Since the total number of k-tuples not exceeng x is x k = x k + ox k we have see equation 10 that the following sum finite or infinite of positive terms has a sum 0 < ρ 1. ρ = 1 11 V U ζe d e If the sum is finite the theorem is proved. Suppose that the sum is infinite that is, 11 is a series of positive terms. Let g be the maximum d in each compatible set with the restrictions. Clearly there are a finite number of sets compatible with the restrictions with the same g. Let ɛ > 0. There exists G such that 1 < ɛ 12 V, g>g U ζe d e g>g 1 g 2 < ɛ 13 Now, we have see 9, 10 and 11 Ax = 1 x k + o x k + F x V, g G U ζe d e = ρx k 1 x k + o x k + F x 14 V, g>g U ζe d e

6 380 Rafael Jakimczuk Let u s x be the number of subtuples U not exceeng x with the same greatest common visor s. Since the total number of these subtuples is x e we have On the other hand, we have g u s x x e 15 s=1 u g x xe g e xe g 2 16 Therefore the contribution Bx to F x of the compatible sets with the restrictions such that g is obtained for i, j = a, b will be bounded by see 15 and 16 0 Bx g u k x u g a,b xxk e u a,b k=1 Hence, we have see 17 and 13 0 F x G<g g x n xk g 2 nxk g>g x k g g 2 nɛxk 18 Since g can be in all n possible i, j and g x is the maximum greatest common visor of the subtuples U not exceeng x. Equations 14, 12 and 18 give Ax x k ρ ɛ + ɛ + nɛ = n + 2ɛ x x ɛ That is, equation 9. Since ɛ can be arbitrarily small. The theorem is proved. Theorem 2.3 Let k 2 be an arbitrary but fixed positive integer. Let us consider the k-tuple A = a 1, a 2,..., a k and t subtuples T j = a i1, a i2,..., a isj j = 1,..., t 2 s j k. The sets of elements in these subtuples T j are the sets R j = {a i1, a i2,..., a isj } j = 1,..., t where we establish the restrictions gcdr j = d j j = 1,..., t. We also put the following restriction: the partition see Lemma 1.1 that correspond to the R j sets j = 1,..., t is a 2 1- partition. Then if Ax is the number of k-tuples not exceeng x that satisfy these restrictions we have Ax = ρx k + o x k 19 where the positive constant ρ 1 is defined bellow in the proof.

7 k-tuples of positive integers with restrictions 381 Remark 2.4 In the former theorem we assume that there exists a k-tuple that satisfies the restrictions. In contrary case the problem has not solution. Proof. Suppose that the R j are pairwise sjoint that is, they already are a partition, then we proceed as in the former theorem. If the R j are not pairwise sjoint then we use Lemma 1.1. Therefore there exists a partition of the set t j=1 R j in n 2 t 1 sets such that n 1 > 0 are unitary, namely {a h1 },..., {a hn1 } and n 2 > 0 have at least 2 elements. Besides each set R j is partitioned. There are t 1 0 sets R j such that in their partition there are not unitary sets and t 2 > 0 sets R j such that in their partition there are unitary sets t 1 + t 2 = t. In each R j there are M j 1 not unitary sets, namely S 1,j, S 2,j,..., S Mj,j. The sets S have e 2 elements. For each set S we consider the associated subtuple U formed with the elements of S. In each t 2 > 0 sets R j there are N j 1 unitary sets, namely u 1,j, u 2,j,..., u Nj,j and also there is the set S 1,j, this set will be called pivot set. Each unitary set {a hi } i = 1,..., n 1 will be repeated in m i 1 sets R j, namely R j1, R j2,..., R jmi. We put gcds = d and, by Lemma 1.3, gcds 1,j u = gcdd 1,j, u = f d 1,j. Therefore the restriction contions for the t 1 0 sets R j will be Lemma 1.2 M j gcdr j = gcd = gcd d 1,j, d 2,j,..., d Mj,j = dj S and the restriction contions for the t 2 > 0 sets R j will be Lemma 1.2 and Lemma 1.3 M j gcdr j = gcd S N j u = gcd d 1,j, d 2,j,..., d Mj,j, f 1,j,..., f Nj,j = dj We shall say that the numbers f and the n 2 > 0 numbers d are a set compatible with the restrictions and it contribute to Ax with see 1 and Lemma x e U ζe d + o n1 e xe c hi x + ox x k e U n 1 = 1 1 n1 c U ζe d e hi x k + o x k 20 Let V be the set such that its elements are the sets of all possible compatible sets with the restrictions. Since the total number of k-tuples not exceeng x is x k = x k + ox k we have see equation 20 that the following sum finite or infinite of positive terms has a sum 0 < ρ 1. ρ = 1 1 n1 c V U ζe d e hi 21

8 382 Rafael Jakimczuk If the sum is finite the theorem is proved. Suppose that the sum is infinite that is, 21 is a series of positive terms. Let g be the maximum d in each compatible set with the restrictions. Clearly there are a finite number of sets compatible with the restrictions with the same g. Let ɛ > 0. There exists G such that 1 1 n1 c V, g>g U ζe d e hi < ɛ 22 g>g Now, we have see 20, 21 and 22 1 g 2 < ɛ 23 Ax = 1 1 n1 c V, g G U ζe d e hi x k + o x k + F x = ρx k 1 1 n1 c V, g>g U ζe d e hi x k + o x k + F x 24 Let u s x be the number of subtuples U not exceeng x with the same greatest common visor s. Since the total number of these subtuples is x e we have Besides we have g u s x x e 25 s=1 u g x xe g e xe g 2 26 Each unitary set see above {a hi } i = 1,..., n 1 is repeated in m i 1 sets R j, namely R j1, R j2,..., R jmi. Therefore to each unitary set correspond a finite sequence seq hi of m i numbers f and the number of n x that correspond to this sequence seq hi will be denoted seq hi x. Note that the number of all possible sequences seq hi is bounded by g h i. Suppose that the n 2 numbers d are fixed, then we have seqhi x x n 1, 27 where the product is on the n 1 unitary sets and the sum is on all possible sequences seq hi.

9 k-tuples of positive integers with restrictions 383 Therefore the contribution Bx to F x of the compatible sets with the restrictions such that g is obtained for i, j = a, b will be bounded by see 25, 26 and 27 0 Bx g u k x u g a,b xxn 1 x k e u n 1 x k 28 a,b k=1 g 2 Hence, we have see 23 0 F x G<g g x n 2 xk g 2 n 2x k g>g 1 g 2 n 2ɛx k 29 Since g can be in all n 2 possible i, j and g x is the maximum greatest common visor of the subtuples U not exceeng x. The rest of the proof is as the former theorem. The Theorem is proved. Acknowledgements. The author is very grateful to Universidad Nacional de Luján. References [1] J. A. de Reyna and R. Heyman, Counting tuples restricted by pairwise coprimality contions, Journal of Integer Sequences, , Article 15. [2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, [3] R. Jakimczuk, The greatest common visor of k positive integers, International Mathematical Forum, , no. 5, [4] J. E. Nymann, On the Probability that k positive integers are relatively prime, Journal of Number Theory, , no. 5, [5] L. Tóth, The Probability that k positive integers are pairwise relatively prime, The Fibonacci Quarterly, , no. 1, Received: July 7, 2018; Published: July 31, 2018