CHARACTERISTICS OF COUNTER EXAMPLE LOOPS IN THE COLLATZ CONJECTURE. Thomas W. Allen II

CHARACTERISTICS OF COUNTER EXAMPLE LOOPS IN THE COLLATZ CONJECTURE by Thomas W Allen II An Abstract presented in partial fulfillment of the requirements for the degree of Master of Science in the Department of Mathematics and Computer Science University of Central Missouri May, 01

ABSTRACT by Thomas W Allen II This thesis was undertaken in order to obtain new results about the Collatz Conjecture: that (4,, 1) is the only loop created by the Collatz function We look at some past results and generalize one of them We discover a few general characteristics about loops After odd elements of a loop are discussed, a general formula for them is discovered Using this formula, the Collatz Conjecture is restated in terms of a family of linear Diophantine systems which have unique solutions In particular, we show that (4,, 1) is the only loop created by the Collatz function if and only if every solution set to each of these systems contains an element x such that x R\N or x = k N for some k N

CHARACTERISTICS OF COUNTER EXAMPLE LOOPS IN THE COLLATZ CONJECTURE by Thomas W Allen II A Thesis presented in partial fulfillment of the requirements for the degree of Master of Science in the Department of Mathematics and Computer Science University of Central Missouri May, 01

CHARACTERISTICS OF COUNTER EXAMPLE LOOPS IN THE COLLATZ CONJECTURE by Thomas W Allen II APPROVED: ittee Member ommittee ACCEPTED: epartment of Mathematics and Computer Science Dean, Graduate School UNIVERSITY OF CEhITRAL MISSOURI WARRENSBURG MISSOURI

ACKNOWLEDGMENTS This thesis would not have been possible without the support of many people I would like to thank my supervisor, Dr Dale Bachman, for the tutelage that was provided throughout this process Thanks are due to the members of my thesis committee, Dr Nicholas Baeth and Dr David Ewing, for taking time out of their busy schedules and assisting with edits of this thesis I would also like to thank the Mathematics and Computer Science Department at the University of Central Missouri Without them taking a chance on me, this thesis would have never been completed Last but not least, I would like to express gratitude to my lovely wife Laura She is a strong reason I continue my pursuit of Mathematics further

Contents List of Tables vii 1 Origins 1 11 History 1 111 What is the Collatz Conjecture? 1 1 Previous Results 4 11 About the Collatz Conjecture: Stefan Andrei and Cristian Masalagiu, 1997 [AM97] 4 1 On the minimum cycle lengths of the Collatz Sequences: Matti Sinisalo, 003 [Sin03] 6 Results/Methods 8 1 Plan of Attack 8 Elementary Results 9 3 Advanced Results 15 3 Future work 6 Bibliography 31 vi

List of Tables 11 Cycle Lengths given a Computational Bound R 7 1 Matrix Sizes given a Computational Bound R 5 vii

Chapter 1 Origins 11 History 111 What is the Collatz Conjecture? The Collatz Conjecture (CC) is a classic conjecture in number theory that has plagued 1 mathematicians for over seventy years It is fairly simple to state yet its proof or disproof has so far eluded mathematicians The Collatz function f : N N is defined by 3x + 1 ; if x is odd f(x) = x ; if x is even and a Collatz Sequence (CS) is defined as {x, f(x), f(f(x)),, f n (x) = 1} where x is some starting value, n is a nonnegative integer, and f i (x) f j (x) for all i, j such that 0 i < j n Throughout this entire paper f will refer to the Collatz function and N will denote the set of positive integers A Collatz Sequence is a very specific type of sequence generated by the 1 plagued is used intentionally here According to Joshua Cooper, [the Collatz Problem] is a spectacular way to waste days, weeks, or years of your life [Coo09] 1

CHAPTER 1 ORIGINS Collatz function At times, it will be necessary to discuss a Generalized Collatz Sequence A Generalized Collatz Sequence (GCS) is defined as a sequence of natural numbers {a k } k I where I = {1,,, n} or I = N satisfying a k+1 = f(a k ) for k I, k not a maximum element of I Example 111 A few examples of Collatz Sequences are { 3, 10, 5, 16, 8, 4,, f 7 (3) = 1 }, { 1, 64, 3, 16, 8, 4,, f 7 (1) = 1 }, and { f 0 (1) = 1 } It is evident that if c is an element of a CS then there exists a CS with c as the starting value As an example, the CS { 3, 10, 5, 16, 8, 4,, f 7 (3) = 1 } also gives the CS { 10, 5, 16, 8, 4,, f 6 (10) = 1 } when we consider 10 to be our starting value Example 11 Every Collatz Sequence is a Generalized Collatz Sequence so every sequence seen in Example 111 is a Generalized Collatz Sequence Other examples of GCS s include {3, 10, 5}, {1, 64, 3, 16, 8, 4,, 1, 4}, and {4,, 1, 4,, 1, 4,, 1, } These three GCS s show that GCS s do not have to include 1, elements may repeat, and elements may repeat multiple times In fact, GCS s may be infinite Now that we have discussed what Collatz Sequences and Generalized Collatz Sequences are, we are ready for our first conjecture Conjecture 113 (Collatz Conjecture) If c N, then c is an element of a Collatz Sequence The conjecture originated with mathematician Lothar Collatz in 1937 when he stated it during a lecture given at Syracuse University Although the original problem was stated differently, it has evolved into the more general form stated here Since this problem has been around for so long, it is only natural to question why it hasn t been solved yet It s not for a lack of trying There have been numerous mathematicians who have studied the Collatz Conjecture and numerous papers have been published on the topic Christopher Jones says it best when he states, It is in a sense, a testament to the

CHAPTER 1 ORIGINS 3 beauty and power of mathematics that, for such a simple set of instructions, an intricate and complex relationship between numbers can be formed [Jon98] There is no clear practical reason for studying the Collatz Conjecture However, that doesn t mean the conjecture shouldn t be studied An article written by Peter Rowlett discusses how the study of abstract conjectures has led to practical applications in fields as varied as video game design, pandemics, and nuclear power [Row11] Before we explore the Collatz Conjecture, a few definitions must be stated Definition 114 A natural number c is said to satisfy the Collatz Condition if f n (c) = 1 for some n N The smallest positive n such that f n (c) = 1 is called the Collatz Length of c If there is no Collatz Sequence containing c, we say the Collatz Length of c is infinite Example 115 The Collatz Length of 3 is 7 because f 7 (3) = 1 and f i (3) 1 for any natural number i < 7 The Collatz Length of 1 is 3 because f 3 (1) = 1 and f i (1) 1 for any natural number i < 3 The Collatz Length of 8 is 3 because f 3 (8) = 1 and f i (8) 1 for any natural number i < 3 Note that two distinct positive integers may have the same Collatz Length We have already seen this with 8 and 1 Both of these integers have Collatz Length 3

CHAPTER 1 ORIGINS 4 1 Previous Results We summarize results from two papers that will be similar to results in our research Subsection 11 has results dealing with natural numbers that satisfy the Collatz Condition as well as their Collatz Lengths One of these results is generalized in Theorem 39 Subsection 1 deals with cycles and cycle lengths, the definitions of which are similar to loops as defined in Section 11 About the Collatz Conjecture: Stefan Andrei and Cristian Masalagiu, 1997 [AM97] This paper gives some basic results on the Collatz Conjecture as well as a few formulas for finding natural numbers that always satisfy the Collatz Condition Each result will be referenced with a prefix of AM for Andrei and Masalagiu AM 1 The function f is surjective, but not injective This is quite easy to verify Proof Let c N Since c N, c N Since c is even, f(c) = c surjective Also, f(3) = 10 = f(0) Therefore f is not injective = c Therefore f is The next result characterizes a subset of the natural numbers that always satisfy the Collatz Condition It also gives the Collatz Length for each element of this subset AM If t N {0}, then t+ 1 satisfies the Collatz Condition with Collatz Length 3 t + 3 In particular, ( ) f (t+3) t+ 1 = 1 3 ( ) Example 11 When t = 0, f (t+3) t+ 1 = f 3 (1) Thus we see that 1 satisfies the 3 Collatz Condition with Collatz Length 3 as we saw in Example 115

CHAPTER 1 ORIGINS 5 ( ) Example 1 When t = 1, f (t+3) t+ 1 = f 5 (5) Therefore 5 satisfies the Collatz 3 Condition with Collatz Length 5 Andrei and Masalagiu have another formula for characterizing a subset of the natural numbers that always satisfy the Collatz Condition This formula also gives the Collatz Length of these numbers AM 3 If m, n N and n is congruent to 1 or 5 modulo 6, then m+1 ( 3mn + 1) 3 m+1 1 satisfies the Collatz Condition with Collatz Length 3 m n + m + In particular, ( ) f (3m n+m+) m+1 ( 3mn + 1) 1 = 1 3 m+1 Example 13 Let m = 1 and n = 7 Then ( ) f (3m n+m+) m+1 ( 3mn + 1) 1 = f 5 (93, 067) 3 m+1 Therefore 93, 067 satisfies the Collatz Condition with Collatz Length 5 Example 14 Let m = 1 and n = 11 Then ( ) f (3m n+m+) m+1 ( 3mn + 1) 1 = f 37 (3, 817, 748, 707) 3 m+1 Therefore 3, 817, 748, 707 satisfies the Collatz Condition with Collatz Length 37 Result AM 3 will be generalized in Chapter with Theorem 39

CHAPTER 1 ORIGINS 6 1 On the minimum cycle lengths of the Collatz Sequences: Matti Sinisalo, 003 [Sin03] Definition 15 Let n N A cycle (c 1, c,, c n ) is a finite sequence of distinct natural numbers such that for 1 i n 1 f(c i ) = c i+1 and f(c n ) = c 1 The cycle (4,, 1) is called the trivial cycle The length of a cycle is the number of distinct elements in the cycle In this paper we use different terminology; see Definition 3 The results in this section will be referenced with a prefix of S for Sinisalo S 1 There are no nontrivial cycles with length less than 75,000 That is, if a cycle other than the trivial cycle exists, then it must be quite large The next result gives a bound for cycle lengths One can manually check any c N by repeatedly applying f If 1 is eventually reached, then c satisfies the Collatz Condition In fact, this has been done for all natural numbers up to R where R = 1508 10 18 and they were all found to satisfy the Collatz Condition We call R the current Computational Bound Using R, a minimum bound on cycle lengths can be obtained By letting n be the number of even elements in a given cycle and k be the number of odd elements in the same cycle we have the following: S Let the Collatz conjecture be verified up to some bound R > 1 Let n be the rational k number with least possible denominator k such that ln 3 ln < n k ln ( ) 3 + 1 R Then the least ln possible cycle length for a nontrivial cycle is n + k Sinisalo uses S and the current computational bound of 1508 10 18 to obtain a minimum bound for the lengths of nontrivial cycles The result includes a small table to illustrate if the Collatz Conjecture is verified to a new computational bound R, then there is a new minimum bound for lengths of nontrivial cycles The particular values of R chosen

CHAPTER 1 ORIGINS 7 here are related to specific rational approximations to ln 3 ln we have a rational approximation n k > ln 3 which can be manipulated to R than 1 n/k 3 in the following way Suppose ln The inequality in S gives us n k ln 3 + 1 R ln 1 Thus if it were shown that all numbers less n/k 3 satisfy the Collatz Condition, then the minimum cycle length is at least n + k The following table gives the values of R = 1 n/k 3 and the corrsponding cycle lengths S 3 The length of any nontrivial cycle in a Collatz Sequence is at least 1,07,71,76 Table 11: Cycle Lengths given a Computational Bound R Computational Bound Cycle Length 380765 10 3, 30, 68, 119, 908 51016 10 355, 504, 839, 99 435849 10 1 186, 65, 759, 595 16891 10 0 17, 06, 679, 61 1508 10 18 1, 07, 71, 76

Chapter Results/Methods 1 Plan of Attack In order to prove the Collatz Conjecture we must show that for every natural number c there exists a natural number n such that f n (c) = 1 However, in order to disprove the Collatz Conjecture we must find a natural number c such that f n (c) 1 for all n N Note that this requires the existence of a natural number with infinite Collatz Length thus creating an infinite subset of the natural numbers each of which does not satisfy the Collatz Condition Let us take a closer look at the Collatz Sequence {4,, 1} This CS is a portion of the Generalized Collatz Sequence {4,, 1, 4,, 1,, 4,, 1, } This GCS will motivate our definition of a loop By assuming that the Collatz function produces another loop (a sequence that repeats) that is distinct from the (4,, 1) loop, general characteristics about loops can be found These characteristics shall lead to restrictions in order for another loop to coexist with the (4,, 1) loop 8

CHAPTER RESULTS/METHODS 9 Elementary Results In order to consider more advanced topics associated with the Collatz Conjecture, a strong understanding of basic results is required Every result and proof in this section is fairly straightforward, but are necessary for use in later sections A few of these results are results that can be seen in almost every research paper on the Collatz Conjecture, though they are often left without proof However, each of the results in this section pertaining to loops is original We first consider the range of the function f In particular, if c is an odd or even number, what can we say about f(c)? Our first pair of results categorizes f(c) based on the parity of c Lemma 1 If c is odd, then f(c) is even Proof Let c N be odd Then c = k + 1 for some integer k 0 Now f(c) = f(k + 1) = 3(k + 1) + 1 = (3k + ), and thus f(c) is even Lemma The value of f(c) is odd if and only if c mod 4 Proof Let c = 4k + i for some integer k 0 and i {0, 1,, 3} There are three cases to check If i is 1 or 3, then c is odd and thus f(c) = 3c + 1 is even by Lemma 1 If i is 0, then c is even and thus f(c) = k which is even If i is, then c is even and thus f(c) = k + 1 which is odd As promised in Section 1, we now define a loop Definition 3 Let n N A loop (c 1, c,, c n ) is a finite sequence of distinct natural numbers such that for 1 i n 1 f(c i ) = c i+1 and f(c n ) = c 1 The loop (4,, 1) is called the trivial loop The order of a loop is the number of distinct elements in the loop

CHAPTER RESULTS/METHODS 10 and is denoted by L = n Note that loops A and B are equal if and only if A is a cyclic permutation of B These definitions are identical to the definitions of cycles and cycle lengths for Sinisalo A survey of literature reveals that both the words cycle and loop are used for this concept We have chosen to use loop throughout this paper Since we have chosen to use loop instead of cycle, we feel the word order is a reasonable substitute for cycle length Notation 4 An arbitrary loop shall be designated with the letter L and a loop other than (4,, 1) with the letter C Since a loop is a finite sequence of distinct numbers it is only natural to discuss the order of a loop Lemma 5 If L is a loop, then L 3 Proof Suppose L = (c) By definition of a loop, f(c) = c We now consider two cases If c is even, then f(c) = c = c, contradicting that c N If c is odd, then f(c) = 3c + 1 which is even, contradicting f(c) = c Therefore L Now suppose L = (c 1, c ) Since L is a loop, f(c 1 ) = c and f(c ) = c 1 This leaves three cases to check 1 Assume c 1 and c are both odd This isn t possible since f(c 1 ) is even and c is odd Assume c 1 and c are both even Since f(c 1 ) = c 1 = c and f(c ) = c = c 1, we get c 1 = c and c = c 1 These two equations imply c 1 = 0 = c which is a contradiction 3 Assume now that either c 1 or c is odd Without loss of generality we may assume c 1 is odd and c is even Therefore f(c 1 ) = 3c 1 + 1 = c and f(c ) = c = c 1 Thus, 3c 1 + 1 = c 1 which implies c 1 = 1, contradicting c 1 N Therefore L and thus L 3

CHAPTER RESULTS/METHODS 11 Since L 3, (4,, 1) is one of the smallest loops possible Later we will show that (4,, 1) is the only loop of order 3 thus making it the smallest loop Definition 6 The set of odd elements in a loop L is denoted by O(L) whereas the set of even elements in a loop L is denoted by E(L) Note that E(L) + O(L) = L Lemma 7 If L is a loop, then O(L) 0 and E(L) 0 Proof Let L = (c 1, c,, c n ) be a loop of order n By Lemma 5, n 3 Assume E(L) = 0 and choose i < n Since c i is odd, f(c i ) is even This contradicts our assumption, so E(L) 0 Now assume c i is even for all i n If each c i is even, then c i = c i+1 or c i = c i+1 for all i, 1 < i < n By induction we have c 1 = n 1 c n, so c 1 = n c 1 This implies c 1 = 0, contradicting c 1 N Therefore O(L) 0 Lemma 8 Let L be a loop with L = n 1 If n is even, then O(L) n If n is odd, then O(L) n 1 Proof By Lemma 7, O(L) = 0 and E(L) = 0 If c is odd, then f(c) is even That is, for every c i O(L) we have f(c i ) E(L) Therefore O(L) n, so O(L) n Now assume n is odd In this case n n 1 / N, and since O(L) N, we have O(L) This lemma leads to an obvious corollary whose proof we omit Corollary 9 Let L be a loop with L = n 1 If n is even, then E(L) n If n is odd, then E(L) n + 1

CHAPTER RESULTS/METHODS 1 Lemma 10 Let C be a loop with C = n Then O(C) if and only if the loop is not (4,, 1) Proof ( ) Let C be a loop that is not (4,, 1) Suppose O(C) = 1 Without loss of generality we may assume c 1 is the odd element of C We note that c = 3c 1 +1 and c k = c k 1 for < k n since c k 1 is even If c k 1 = 3c 1 + 1, then c k 3 k = 1 ( ) 3c1 + 1 = 3c 1 + 1 Thus k 3 k by induction c k = 3c 1 + 1 for k n Now f(c k n ) = c 1 and c n is even, so 3c 1 + 1 = c n 1 1 Therefore 3c 1 +1 = c 1 ( n 1 1 ) or c 1 = n 1 3 Because c 1 N, c 1 = 1 This is a contradiction and thus O(C) ( ) If O(C), then C (4,, 1) Lemma 11 If C is a loop other than (4,, 1), then C 4 Proof Let C be a loop other than (4,, 1) with order n By Lemma 10, O(C) Now using the results from Lemma 8, C 4 ones Continuing our examination of the elements of loops, the next two definitions are natural Definition 1 Let L = (c 1, c,, c n ) be a loop The maximum element in L, denoted by M, is the element such that c i M for all i The minimum element in L, denoted by m, is the element such that c i m for all i The first few results for maximum and minimum elements are trivial results They are listed below because they will be used extensively throughout the rest of the paper Lemma 13 If m is the minimum element of a loop L, then m is odd Proof If m is even, then f(m) = m < m, contradicting the minimumity of m Therefore m is odd

CHAPTER RESULTS/METHODS 13 Lemma 14 If M is the maximum element of a loop L, then 4 M Proof If M is odd, then f(m) = 3M + 1 Since M < 3M + 1, we get a contradiction that M is the maximum element of L Therefore M is even Suppose M is not divisible by 4 Since M is even and not divisible by 4, M = k for some odd integer k 1 Therefore f(m) = k = k and thus f (M) = 3k + 1 Since k 1, we know k < 3k + 1 and thus M < 3k + 1 which contradicts the maximumity of M Therefore 4 M Corollary 15 Let L = (c 1, c,, c n ) be a loop with L = n Then E(L) > n In particular, E(L) > O(L) Proof Due to Lemma 1, for all c O(L) there exists f(c) E(L) We know L has a maximum element M such that f(m) = M However, by Lemma 14, M E(L) That is, there is no c O(L) such that f(c) = M Therefore E(L) > n Lemma 16 Let L = (c 1, c,, c n ) and let m be the minimum element of L Then 4 (3m + 1) if and only if L = (4,, 1) Proof Assume L = (4,, 1) It is clear that m = 1 and 4 (3m+1) Now assume L (4,, 1) Since the minimum element m is odd, f(m) = 3m + 1 is even by Lemma 1 Let s assume 4 (3m + 1) Then f (3m + 1) = 3m + 1 In order to maintain the minimumity of m, 4 m 3m + 1, or m 1 This implies that m = 1 which contradicts L (4,, 1) Therefore 4 4 does not divide (3m + 1) Lemma 14 and Lemma 16 lead to an obvious Corollary whose proof we omit Corollary 17 Let L = (c 1, c,, c n ) be a loop with maximum element M and minimum element m Then M = 3m + 1 if and only if L = (4,, 1) We now discuss the congruence classes of M and m in order to discover more about loop structure

CHAPTER RESULTS/METHODS 14 Lemma 18 If M is the maximum element of a loop L, then M 1 mod 3 Proof Since M is even and maximum, M = f(x) for some odd natural number x That is, f(x) = 3x + 1 and thus M 1 mod 3 Lemma 19 Let C be a loop with minimum element m Then m 3 mod 4 if and only if C (4,, 1) Proof ( ) Let C (4,, 1) Then m = 4x + a for some integer x 0 and a {0, 1,, 3} Since m is odd, a / {0, } Suppose a = 1 Then m = 4x+1 We know 3m + 1 is an element of C because C 4 and 3m + 1 is even Therefore 3m + 1 3(4x + 1) + 1 = which implies 3m + 1 = 1x + 4 or 3m + 1 = 6x + It is clear 6x + is even However, 3m + 1 is odd because 3m + 1 is not divisible by 4 due to Lemma 16 Therefore a 1 and thus m 3 mod 4 ( ) Let C be a loop and suppose m 3 mod 4 Then clearly C (4,, 1) Definition 0 If c is an element of a loop L and c has a preimage y / L, then the tail of c is T c = { t N n N with f n (t) = c and f k (t) / L k, 0 k < n } In this case, c is called a root of the loop L or the root of the tail T c As an example, {5, 8, 16} T 4, which is the tail of the root 4 in the loop (4,, 1) This example shows that the (4,, 1) loop has a tail, but what about loops other than (4,, 1)? Lemma 1 If L is a loop, then L has a tail Proof Let L be a loop with maximum element M We see that M is a preimage of M because f(m) = M Since M is maximum, M / L Therefore M T M Lemma If c is a root of the loop L, then c has two preimages opposite in parity Proof Since c is a root, there exists a T c such that f(a) = b Likewise, there exists b L such that f(b) = c We can see that f(a) = f(b) = c but a b Therefore by definition of the function f, a and b are opposite in parity

CHAPTER RESULTS/METHODS 15 3 Advanced Results Theorem 31 A loop L has a root c if and only if c and c 1 mod 3 Proof ( ) Since c is a root, c has two preimages opposite in parity In particular, f(j+1) = c for some integer j 0 Therefore c = f(j + 1) = 6j + 4 1 mod 3 It is also clear that c ( ) Let c L such that c and c 1 mod 3 Clearly c, which is even, is a preimage of c Since c, c = k for some k N Also, c 1 mod 3 so k = 3x + 1 for some x N This implies x = k 1 which is odd In particular, x c and thus c has a preimage not in L 3 Therefore c is a root of L Let c and d be roots of disjoint loops Is it possible for T c T d? Is it possible if c and d are roots of the same loop? These questions will be addressed in Lemma 3 and Corollary 33 Lemma 3 If X and Y are disjoint loops, then their tails do not intersect Proof Since X and Y are disjoint loops, X Y = Assume X and Y have tails that share an element b Then f n (b) X and f m (b) Y for some n, m N Without loss of generality assume m n Since f n (b) X implies f n+i (b) X for all i N, we see f m (b) X Y This contradicts X and Y being disjoint and thus their tails do not intersect Corollary 33 If a loop L has more than one tail, then the tails do not intersect Proof Let L be a loop with two distinct tails having roots x and y Assume T x T y and let c T x T y Since c T x, f n (c) = x for some n N and f k (c) / L for all k < n Likewise, since c T y, f m (c) = y for some m N and f j (c) / L for all j < m Without loss of generality we may assume n m If n = m, then x = f n (c) = y, and the tails are

CHAPTER RESULTS/METHODS 16 not distinct Therefore n < m and since f m (c) = y and f j (c) / L for all j < m we have x = f n (c) / L, which contradicts x being a root Thus T x T y = This is an interesting find Every element x of T c has a nonempty preimage f 1 (x) T c Thus the nonempty set f j (x) = { a N f j (a) = x } T c for all j 1 Note that if i j, then f i (x) f j (x) =, so T c has an infinite number of elements That is, if a loop other than (4,, 1) exists, then there exist at least two infinite disjoint subsets of N defined by the function f Definition 34 The set of roots in a loop L is denoted by R(L) Theorem 35 Let L be a loop with L = n and O(L) = k Then O(L) R(L) < E(L) Proof Since O(L) = k, E(L) = n k Because a root is even, R(L) E(L) Let x O(L) Then f(x) = 3x + 1 is even and f(x) 1 mod 3 Therefore f(x) is a root, and thus O(L) R(L) Therefore O(L) R(L) E(L) Since L has a maximum element M E(L) and M 1 mod 3, M is a root Also, 4 M so M L However, M is even and M mod 3 Thus, M is not a root Therefore there exist even elements in L that are not roots and so R(L) < E(L) Theorem 35 leads to three immediate corollaries The proofs of Corollary 36 and Corollary 37 are omitted Corollary 36 Let L be a loop Then R(L) < E(L) Corollary 37 Let L = (4,, 1) Then R(L) = 1

CHAPTER RESULTS/METHODS 17 Corollary 38 A loop C has multiple roots if and only if C (4,, 1) Proof Let C be a loop with multiple roots Then C (4,, 1) because (4,, 1) has only one root Now assume C (4,, 1) By Lemma 10, O(C) From Theorem 35, R(C) Assuming a loop other than (4,, 1) exists implies that there are at least three tails of infinite length that all lie within the natural numbers yet never intersect at all Now let us take a second to recall the paper by Andrei and Masalagiu They discovered different classes of natural numbers that always satisfy the Collatz Condition We derive a generalization of AM 3 here Theorem 39 If P 1 satisfies the Collatz Condition, P 1 / {1,, 4}, and 3 Z P for some Z N, then Z P 1 satisfies the Collatz Condition Also, if the Collatz Length for 3Z P 1 is k, then the Collatz Length for Z P 1 is k + Z 3Z Proof Let P 1 satisfy the Collatz Condition, P 1 / {1,, 4}, and 3 Z P for some Z N Since 3 Z P, we see Z P 1 is a natural number and thus is in the domain of f Since 3Z Z P 1 is odd, 3Z f ( Z P ) 3 Z 1 ( ) Z P = 3 3 1 + 1 Z = Z P 3 Z 1 3 + 1 = Z P 3Z 1 ( ) Z P This number is even, so applying f yields f 3 = Z 1 P 1 The number obtained Z 1 3Z 1 after applying f twice is another number that appears in the same form as the original

CHAPTER RESULTS/METHODS 18 number Since 3 Z P, we see 3 Z 1 P In general, ( ) f j Z P 3 1 = Z j P 1 Z 3 Z j for 0 j Z In particular, ( ) f Z Z P 3 1 Z = Z Z P 3 Z Z 1 = 0 P 3 0 1 = P 1 Since P 1 satisfies the Collatz Condition, Z P 1 satisfies the Collatz Condition 3Z ( ) ( ) We also need to verify that f n Z P 3 1 1 for 0 n < Z Assume f n Z P Z 3 1 = ( ) ( ) Z 1 for some n, 0 n < Z Since f n Z P 3 1 = 1, f n+i Z P Z 3 1 {1,, 4} for all ( ) Z i N We have just seen f Z Z P 3 1 = P 1 which contradicts P 1 / {1,, 4} Now Z since k is the Collatz Length for P 1, Z P 1 has Collatz Length k + Z 3Z The loop (4,, 1) can be rewritten as (1, 4, ) Disregarding the exact numbers in the loop and considering only the parity of each element, the loop can be written as (odd, even, even) By Definition 3, a loop L with order n has the characteristic f n (c) = c Therefore the loop (1, 4, ) written as a Generalized Collatz Sequence is {1, 4,, 1, 4,, } or {odd, even, even, odd, even, even, } Does any other loop follow a pattern like this? Definition 310 A loop L with L = n is said to be parity periodic if the elements of L follow the pattern of d 1, e 1,, e k, d, e k+1,, e k,, d n, e ( ) k+1 kn (k 1), e kn (k ),, e kn k+1 k+1 k+1 where d i O(L), e i E(L), for all i {1,, k}

CHAPTER RESULTS/METHODS 19 Theorem 311 A loop L is parity periodic if and only if L = (4,, 1) Proof ( ) Let L = (4,, 1) Clearly (4,, 1) = (1, 4, ) is parity periodic ( ) Let L be a loop other than (4,, 1) that is parity periodic with L = n It is impossible for L = (d 1, d,, d n ) by Lemma 1 It is impossible for L = ( d 1, e 1, d, e,, d n, e ) n by Lemma 14 Now assume ( ) L = d 1, e 1,, e k, d, e k+1,, e k,, d n, e k+1 kn (k 1), e kn (k ), e kn with k k+1 k+1 k+1 Without loss of generality, suppose d 1 is the minimum element of L Therefore f (d 1 ) = f(e 1 ) = e, which is even since k Thus 3d 1 + 1 = e = l for some integer l, so 4 (3d 1 + 1) which contradicts Lemma 16 Thus a loop L is parity periodic if and only if L = (4,, 1) Definition 31 Let O(L) = {d 0, d 1,, d k 1 } and write L = ( d 0, e 0,1,, e 0,j0, d 1, e 1,1,, e 1,j1,, d k 1, e k 1,1,, e k 1,jk 1 ) where el,i E(L) for all l {0,, k 1} and all i {1,, j l } Then for each l {0,, k 1}, define a subsequence of L to be (d l, e l,1,, e l,jl ) The order of a subsequence is the number of elements in the subsequence If (d l, e l,1,, e l,jl ) is a subsequence of L, we denote the pair [d l, j l ] as the index of the subsequence Note that for l {0,, k 1} f j l (3d l + 1) = d (l+1 mod k) or 3d l + 1 = d (l+1 mod k) We say L is written as a juxtaposition of indices if j l L = [d 0, j 0 ] [d 1, j 1 ] [d k 1, j k 1 ] where d 0 is the minimum element of L By Theorem 311, (4,, 1) is the only parity periodic loop That is, if C (4,, 1), then there exist two indices of C [d i, j i ] and [d k, j k ] such that j i j k Also, it should be noted that j 0 = 1 whenever L (4,, 1) due to Lemma 16

CHAPTER RESULTS/METHODS 0 Theorem 313 Let L be a loop Then if d a O(L) with 0 a k 1, then d a = ( k 1 i=0 3 k 1 i ) i 1 l=0 j (l+a mod k) k 1 i=0 j i 3 k Proof Let L be a loop and write L = [d 0, j 0 ] [d 1, j 1 ] [d k 1, j k 1 ] with d k d i+1 = 3d i + 1 for each i {0,, k 1}, we see j i = d 0 Since d 1 = 3d 0 + 1 j 0 and d = 3d 1 + 1 j 1 = 3 d 0 + 3 j 0+j 1 + 1 j 1 If we assume d i = 3i d 0 + 3 i 1 3 i j 0+ +j i 1 + j 1+ +j i 1 + + 1 j i 1 for some i, then d i+1 = 3d i + 1 j i = 3i+1 d 0 + 3 i j 0+ +j i + 3i 1 j 1+ +j i + + 1 j i

CHAPTER RESULTS/METHODS 1 Thus d 0 = d k = 3k d 0 + 3 k 1 3 k + + + 1 j 0+ +j k 1 j 1+ +j k 1 j k 1 Finding a common denominator and simplifying, d 0 k 1 i=0 j i = d 0 3 k + k 1 i=0 (3 k 1 i ) i 1 l=0 j l or d 0 = ( k 1 i=0 3 k 1 i ) i 1 l=0 j l k 1 i=0 j i 3 k Since d 0 was arbitrary, any d i could have been chosen to play the role of d 0 That is, for any integer a {0,, k 1} d a = ( k 1 i=0 3 k 1 i ) i 1 l=0 j (l+a mod k) k 1 i=0 j i 3 k We can now set up a system of linear Diophantine equations For notational ease, let B = k 1 i=0 j i 3 k be the denominator of the expression in Theorem 313 If k 1 i=0 (3 k 1 i ) i 1 l=0 j (l+p mod k) = B for some p {0,, k 1}, then d p = 1 Therefore L = (4,, 1) If k 1 i=0 (3 k 1 i ) i 1 l=0 j (l+p mod k) < B for some p {0,, k 1}, then d p < 1 Therefore k 1 i=0 (3 k 1 i ) i 1 l=0 j (l+a mod k)

CHAPTER RESULTS/METHODS is a positive multiple of B for every p {0,, k 1} Alternatively we can use d i+1 = 3d i + 1 j i for i = 0,, k 1 to create the following matrix equation 3 j 0 0 0 0 3 j 1 0 0 3 0 0 0 3 j k j k 1 0 0 3 d 0 d 1 d k d k 1 = 1 1 1 1 For notational ease, let A = 3 j 0 0 0 0 3 j 1 0 0 3 0 0 0 3 j k j k 1 0 0 3, d = d 0 d 1 d k d k 1, and 1 = 1 1 1 1 By using cofactor expansion on the first column of A, det(a) = 3(( 3) k 1 ) + ( 1) k 1 ( k 1 i=0 j i ) = ( 1) k 1 ( k 1 i=0 j i 3 k ) which is not 0 Therefore A 1 exists and the system has the unique solution d = A 1 1 It is also evident that det(a) = ±B How do we solve a linear Diophantine system?

CHAPTER RESULTS/METHODS 3 Theorem 314 [GP90] To solve the system of Linear Diophantine equations AX = B, unimodular row reduce [A t I] to [R T ], where R is in row-echelon form Then the system AX = B has integer solutions if and only if the system R t K = B has integer solutions for K, and all the solutions of AX = B are of the form X = T t K Note: Row-echelon form allows the pivots of the matrix to be any integer, not necessarily 1 Also, an elementary unimodular row operation on a matrix consists one of the following three types of operations 1 Add an integer multiple of one row of the matrix to another row Interchange two rows of the matrix 3 Multiple one row of the matrix by 1 Solving A d = 1 would show whether every d i, with i {0,, k 1}, is an odd natural number or not If there exists some d i / N or d i j + 1 for some integer j 0, then (4,, 1) is the only loop possible by Theorem 313 With the aid of Mathematica, a few elementary examples were explored Example 315 Let L be a loop with O(L) = 1 This leads to the 1 1 matrix A = [ j 0 3] It is evident that j 0 = and thus L = (4,, 1) This is not surprising since we already know (4,, 1) is the only loop such that O(L) = 1

CHAPTER RESULTS/METHODS 4 Example 316 Let L be a loop with O(L) = This leads to the matrix A = 3 j 1 3 As you can see, j 0 = 1 because L (4,, 1) Now solving for d, d 0 d 1 = 5 9 + j 1+1 3 + j 1 9 + j 1+1 5 Since each entry must be an odd positive integer, is an odd positive integer which 9 + j 1+1 is impossible Therefore O(L) In particular, there are no loops with exactly two odd elements Example 317 Let L be a loop with O(L) = 3 This leads to the 3 3 matrix A = 3 0 0 3 j 1 j 0 3 As before, j 0 = 1 Now solving for d, d 0 d 1 d = 15 + j 1+1 7 + j 1+j +1 9 + 3 j 1 + j 1+j 7 + j 1+j +1 9 + 5 j 7 + j 1+j +1

CHAPTER RESULTS/METHODS 5 It is evident that E(L) = j 0 + j 1 + j, but is E(L) N? Using the result S, O(L) = 3, and the current computational bound of R = 1508 10 18 we can see that E(L) / N That is: the inequality ln 3 ln < j 0 + j 1 + j ln ( ) 3 + 1 R does not hold if E(L) N Therefore 3 ln O(L) 3 In particular, there are no loops with exactly three odd elements A similar analysis can be done for larger sized matrices Since E(L) + O(L) = L and E(L) O(L) ln (3 + 1 ) R (cf S ) we obtain the inequality ln O(L) L ln ln ( 6R+ R ) We know from S 3 that L 1, 07, 71, 76 and R = 1508 10 18 Therefore O(L) 397, 573, 379 That is, the smallest potential matrix A has size 397, 573, 379 397, 573, 379 Table 11 can be used to generate the size of a potential matrix once a new computational bound R is obtained A new table is listed which gives the sizes of such matrices Table 1: Matrix Sizes given a Computational Bound R Computational Bound Loop Order Matrix Size 380765 10 3, 30, 68, 119, 908 890, 638, 885, 193 890, 638, 885, 193 51016 10 355, 504, 839, 99 137, 58, 045, 31 137, 58, 045, 31 435849 10 1 186, 65, 759, 595 7, 057, 431, 991 7, 057, 431, 991 16891 10 0 17, 06, 679, 61 6, 586, 818, 670 6, 586, 818, 670 1508 10 18 1, 07, 71, 76 397, 573, 379 397, 573, 379

Chapter 3 Future work Some potential future explorations of the Collatz Conjecture include: 1 Find the unique solution of A d = 1 for every matrix A Knowing the value for every d i is crucial in order to see if the Collatz function produces more than one loop or not Discuss how tails can coexist in N without intersecting How can multiple tails exist in N? A loop C (4,, 1) has R(C) by Corollary 38 Therefore, if there exists a loop C (4,, 1), then there are at least three tails in N In-depth research on tails could lead to a result about tails not being able to coexist in N 3 Extend the Collatz function to a more general function In order to prove the Collatz Conjecture it may be easier to look at a more general function 6

CHAPTER 3 FUTURE WORK 7 The General Collatz function f n (x) : N N with integer n 0 is defined by f n (x) = ( n 1)x + 1 ; if x is odd x ; if x is even The first three General Collatz functions are: and f 0 (x) = 1 ; if x is odd x ; if x is even x + 1 ; if x is odd f 1 (x) = x ; if x is even,, f (x) = 3x + 1 ; if x is odd x ; if x is even The f function is the Collatz function f that has been discussed throughout this paper It is clear that the f 0 function produces the loop (1) since f 0 (1) = 1 The proof that this is the only loop is quite simple and will be given later It is also clear that the f 1 function produces the loop (, 1) because f 1 () = 1 and f 1 (1) = The proof that this is the only loop is also quite simple and shall be given later

CHAPTER 3 FUTURE WORK 8 Each of the functions in the family {f n : n N {0}} satisfies several of the properties we have discussed earlier These properties will be stated here without proof Theorem 301 Let f k {f n : n N {0}} Suppose L is a loop generated by the function f k with maximum element M and minimum element m Then a) If c O(L), then f k (c) E(L) b) If c E(L), then either f k (c) E(L) or f k (c) O(L) c) M E(L) d) m O(L) e) M 1 mod k 1 A few less elementary results on this family of functions follow Theorem 30 If g is a function in the {f n : n N {0}} family, then g has a loop Proof Let g be a function in the {f n : n N {0}} family Then g = f k for some k n g(1) = ( k 1)(1) + 1 = k 1 + 1 = k Since g k ( k ) = 1, g has a loop As previously stated, the function f 0 and the function f 1 only have one loop The proofs of these follow but first we need a definition

CHAPTER 3 FUTURE WORK 9 Definition 303 We say c N satifies the General Collatz Condition for the function f k, if f n k (c) = 1 for some integer n 0 Theorem 304 If c N, then c satisfies the General Collatz Condition for the function f 0 In particular, (1) is the only loop in f 0 Proof Let c N By the fundamental theorem of arithmetic c = k 1 p k p k 3 3 p kn n where k i 0 for all i {1,, n} and each p i is an odd prime We can see that f k 1 0 (c) = p k p k 3 3 p kn n which is odd Therefore f k 1+1 0 (c) = 1 and thus c satisfies the General Collatz Condition for the function f 0 Theorem 305 If c N, then c satisfies the General Collatz Condition for the function f 1 In particular, (, 1) is the only loop in f 1 Proof This proof is by strong induction We see that f 1 (1) =, f 1 () = 1, f 1 (3) = 4, and f 1 (4) = Therefore 1,,3 and 4 satisfy the General Collatz Condition for the function f 1 Assume every natural number up to k for some k > 4 satisfies the General Collatz Condition for the function f 1 Assume k + 1 is even It is clear that k + 1 < k as long as k > 1 Therefore k + 1 < k and thus f 1 (k+1) = k + 1 < k By the induction hypothesis, for all c N, c < k, there exists n N such that f1 n (c) = 1 Hence, k + 1 satisfies the General Collatz Condition for the function f 1 Therefore k + 1 satisfies the General Collatz Condition for the function f 1 Now assume k + 1 is odd It is clear that k + < k as long as k > and thus k + < k Since f 1 (k + 1) = k + is even, f 1 (k + ) = k + < k By the induction hypothesis, for all c N, c < k, there exists n N such that f1 n (c) = 1 Hence, k + satisfies the General Collatz Condition for the function f 1 Therefore k + satisfies the General Collatz Condition for the function f 1 and thus k + 1 satisfies the General Collatz Condition for the function f 1

CHAPTER 3 FUTURE WORK 30 Why is it so easy to verify that f 0 and f 1 have a unique loop yet it is so difficult in the f case? Looking at the family of functions {f n : n N {0}} leads to a new more General Collatz Conjecture Conjecture 306 (General Collatz Conjecture) If c N, then c satisfies the General Collatz Condition for {f n : n N {0}} In particular, ( n, n 1,,, 1) is the only loop in {f n : n N {0}} In this paper we reviewed some past results on the Collatz Conjecture Andrei, Masalagiu, and Sinisalo gave us some nice results about classes of numbers that satisfy the Collatz Condition as well as potential orders of future loops We discovered a linear Diophantine system which, if solved, would definitively show whether (4,, 1) is the only loop or not We also conjectured about {f n : n N {0}}, a family of functions to which f belongs The purpose of this thesis was to explore the Collatz Conjecture in a new way To the best of my knowledge, there are no papers in which the author assumes the existence of another loop and then discovers characteristics about said loop These characteristics include maximum/minimum elements, tails, and roots As previously stated in Section 1, these characteristics shall lead to restrictions in order for another loop to coexist with the (4,, 1) loop We hope that these restrictions will lead to a contradiction, thus showing that (4,, 1) is the only loop

Bibliography [AM97] Stefan Andrei and Cristian Masalagiu About the Collatz conjecture Acta Informatica, 35:167 179, 1997 [Coo09] Josh Cooper, Nov 009 http://wwwmathscedu/ cooper/combprobhtml [GP90] William J Gilbert and Anu Pathria Linear Diophantine Equations Master s thesis, University of California at Berkeley, 1990 [Jon98] Christopher Jones The Collatz Problem: An Interactive Algorithm International Baccalaureate Extended Essay, Mathematics, pages 1 11, 1998 [Row11] Peter Rowlett The unplanned impact of mathematics Nature, 475:166 169, 011 [Sin03] Matti K Sinisalo On the Minimal Cycle Lengths of the Collatz Sequence Master s thesis, University of Oulu, Finland, 003 31