characteristics of counter example loops in the collatz conjecture

CHARACTERISTICS OF COUNTER EXAMPLE LOOPS IN THE COLLATZCONJECTURE

by

Thomas W. Allen II

An Abstractpresented in partial fulfillment

of the requirements for the degree ofMaster of Science

in the Department of Mathematics and Computer ScienceUniversity of Central Missouri

May, 2012

ABSTRACT

by

Thomas W. Allen II

This thesis was undertaken in order to obtain new results about the Collatz Conjecture:

that (4, 2, 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, 2, 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 = 2k ∈ N for some k ∈ N.


by

Thomas W. Allen II

A Thesispresented in partial fulfillment

of the requirements for the degree ofMaster of Science

in the Department of Mathematics and Computer ScienceUniversity of Central Missouri

May, 2012


by

Thomas W. Allen II

ACCEPTED:

Dean, Graduate School

UNIVERSITY OF CEhITRAL MISSOURI

WARRENSBURG. MISSOURI

APPROVED:

ittee Member

ommittee

epartment of Mathematics

and Computer Science

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

1.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 What is the Collatz Conjecture? . . . . . . . . . . . . . . . . . . . . . 1

1.2 Previous Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 About the Collatz Conjecture: Stefan Andrei and Cristian Masalagiu,

1997 [AM97] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 On the minimum cycle lengths of the Collatz Sequences: Matti Sin-

isalo, 2003 [Sin03] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Results/Methods 8

2.1 Plan of Attack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Elementary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Advanced Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Future work 26

Bibliography 31

vi

List of Tables

1.1 Cycle Lengths given a Computational Bound R. . . . . . . . . . . . . . . . . . . 7

2.1 Matrix Sizes given a Computational Bound R . . . . . . . . . . . . . . . . . . . 25

vii

Chapter 1

Origins

1.1 History

1.1.1 What is the Collatz Conjecture?

The Collatz Conjecture (CC) is a classic conjecture in number theory that has plagued1

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

f(x) =

3x + 1 ; if x is odd

x

2; if x is even

and a Collatz Sequence (CS) is defined as {x, f(x), f(f(x)), . . . , fn(x) = 1} where x is some

starting value, n is a nonnegative integer, and f i(x) 6= 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 spectac-ular way to waste days, weeks, or years of your life.”[Coo09]

1

CHAPTER 1. ORIGINS 2

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 {ak}k∈I

where I = {1, 2, . . . , n} or I = N satisfying ak+1 = f(ak) for k ∈ I, k not a maximum

element of I.

Example 1.1.1. A few examples of Collatz Sequences are{

3, 10, 5, 16, 8, 4, 2, f 7(3) = 1}

,{21, 64, 32, 16, 8, 4, 2, f 7(21) = 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, 2, f 7(3) = 1

}also gives the CS

{10, 5, 16, 8, 4, 2, f 6(10) = 1

}when we con-

sider 10 to be our starting value.

Example 1.1.2. Every Collatz Sequence is a Generalized Collatz Sequence so every sequence

seen in Example 1.1.1 is a Generalized Collatz Sequence. Other examples of GCS’s include

{3, 10, 5}, {21, 64, 32, 16, 8, 4, 2, 1, 4}, and {4, 2, 1, 4, 2, 1, 4, 2, 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 1.1.3 (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


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 1.1.4. A natural number c is said to satisfy the Collatz Condition if fn(c) = 1

for some n ∈ N. The smallest positive n such that fn(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 1.1.5. The Collatz Length of 3 is 7 because f 7(3) = 1 and f i(3) 6= 1 for any

natural number i < 7. The Collatz Length of 1 is 3 because f 3(1) = 1 and f i(1) 6= 1 for any

natural number i < 3. The Collatz Length of 8 is 3 because f 3(8) = 1 and f i(8) 6= 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.


1.2 Previous Results

We summarize results from two papers that will be similar to results in our research. Subsec-

tion 1.2.1 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 2.3.9. Subsection

1.2.2 deals with cycles and cycle lengths, the definitions of which are similar to loops as

defined in Section 2.2.

1.2.1 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, 2c ∈ N. Since 2c is even, f(2c) =2c

2= c. Therefore f is

surjective. Also, f(3) = 10 = f(20). Therefore f is not injective.

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 2. If t ∈ N ∪ {0}, then22t+2 − 1

3satisfies the Collatz Condition with Collatz Length

2t + 3. In particular,

f (2t+3)

(22t+2 − 1

3

)= 1.

Example 1.2.1. When t = 0, f (2t+3)

(22t+2 − 1

3

)= f 3(1). Thus we see that 1 satisfies the

Collatz Condition with Collatz Length 3 as we saw in Example 1.1.5.


Example 1.2.2. When t = 1, f (2t+3)

(22t+2 − 1

3

)= f 5(5). Therefore 5 satisfies the Collatz

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, then2m+1(23mn + 1)

3m+1−1 satisfies

the Collatz Condition with Collatz Length 3mn + 2m + 2. In particular,

f (3mn+2m+2)

(2m+1(23mn + 1)

3m+1− 1

)= 1.

Example 1.2.3. Let m = 1 and n = 7. Then

f (3mn+2m+2)

(2m+1(23mn + 1)

3m+1− 1

)= f 25(932, 067).

Therefore 932, 067 satisfies the Collatz Condition with Collatz Length 25.

Example 1.2.4. Let m = 1 and n = 11. Then

f (3mn+2m+2)

(2m+1(23mn + 1)

3m+1− 1

)= f 37(3, 817, 748, 707).

Therefore 3, 817, 748, 707 satisfies the Collatz Condition with Collatz Length 37.

Result AM 3 will be generalized in Chapter 2 with Theorem 2.3.9.


1.2.2 On the minimum cycle lengths of the Collatz Sequences:

Matti Sinisalo, 2003 [Sin03]

Definition 1.2.5. Let n ∈ N. A cycle (c1, c2, . . . , cn) is a finite sequence of distinct natural

numbers such that for 1 ≤ i ≤ n− 1 f(ci) = ci+1 and f(cn) = c1. The cycle (4, 2, 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 2.2.3.

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 275,000.

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 = 1.25208 · 1018 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 2. Let the Collatz conjecture be verified up to some bound R > 1. Letn

kbe the rational

number with least possible denominator k such thatln 3

ln 2<

n

k≤

ln(3 + 1

R

)ln 2

. Then the least

possible cycle length for a nontrivial cycle is n + k.

Sinisalo uses S 2 and the current computational bound of 1.25208 · 1018 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


here are related to specific rational approximations toln 3

ln 2in the following way. Suppose

we have a rational approximationn

k>

ln 3

ln 2. The inequality in S 2 gives us

n

k≤

ln 3 + 1R

ln 2

which can be manipulated to R ≤ 1

2n/k − 3. Thus if it were shown that all numbers less

than1

2n/k − 3satisfy the Collatz Condition, then the minimum cycle length is at least n+k.

The following table gives the values of R =1

2n/k − 3and the corrsponding cycle lengths.

S 3. The length of any nontrivial cycle in a Collatz Sequence is at least 1,027,712,276.

Table 1.1: Cycle Lengths given a Computational Bound R.

Computational Bound Cycle Length3.80765 · 1023 2, 302, 268, 119, 9085.10126 · 1022 355, 504, 839, 9294.35849 · 1021 186, 265, 759, 5952.16891 · 1020 17, 026, 679, 2611.25208 · 1018 1, 027, 712, 276

Chapter 2

Results/Methods

2.1 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 fn(c) = 1. However, in order to disprove the Collatz

Conjecture we must find a natural number c such that fn(c) 6= 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, 2, 1}. This CS is a portion of

the Generalized Collatz Sequence {4, 2, 1, 4, 2, 1, . . . , 4, 2, 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, 2, 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, 2, 1) loop.

8

CHAPTER 2. RESULTS/METHODS 9

2.2 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 2.2.1. If c is odd, then f(c) is even.

Proof. Let c ∈ N be odd. Then c = 2k + 1 for some integer k ≥ 0. Now f(c) = f(2k + 1) =

3(2k + 1) + 1 = 2(3k + 2), and thus f(c) is even.

Lemma 2.2.2. The value of f(c) is odd if and only if c ≡ 2 mod 4.

Proof. Let c = 4k + i for some integer k ≥ 0 and i ∈ {0, 1, 2, 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 2.2.1. If i

is 0, then c is even and thus f(c) = 2k which is even. If i is 2, then c is even and thus

f(c) = 2k + 1 which is odd.

As promised in Section 2.1, we now define a loop.

Definition 2.2.3. Let n ∈ N. A loop (c1, c2, . . . , cn) is a finite sequence of distinct natural

numbers such that for 1 ≤ i ≤ n − 1 f(ci) = ci+1 and f(cn) = c1. The loop (4, 2, 1) is

called the trivial loop. The order of a loop is the number of distinct elements in the loop


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 2.2.4. An arbitrary loop shall be designated with the letter L and a loop other

than (4, 2, 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 2.2.5. 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

2= c, contradicting that c ∈ N. If c is odd, then f(c) = 3c + 1 which

is even, contradicting f(c) = c. Therefore |L| ≥ 2. Now suppose L = (c1, c2). Since L is a

loop, f(c1) = c2 and f(c2) = c1. This leaves three cases to check.

1. Assume c1 and c2 are both odd. This isn’t possible since f(c1) is even and c2 is odd.

2. Assume c1 and c2 are both even. Since f(c1) =c1

2= c2 and f(c2) =

c2

2= c1, we get

c1 = 2c2 and c2 = 2c1. These two equations imply c1 = 0 = c2 which is a contradiction.

3. Assume now that either c1 or c2 is odd. Without loss of generality we may assume

c1 is odd and c2 is even. Therefore f(c1) = 3c1 + 1 = c2 and f(c2) =c2

2= c1. Thus,

3c1 + 1 = 2c1 which implies c1 = −1, contradicting c1 ∈ N.

Therefore |L| 6= 2 and thus |L| ≥ 3.


Since |L| ≥ 3, (4, 2, 1) is one of the smallest loops possible. Later we will show that

(4, 2, 1) is the only loop of order 3 thus making it the smallest loop.

Definition 2.2.6. 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 2.2.7. If L is a loop, then |O(L)| 6= 0 and |E(L)| 6= 0.

Proof. Let L = (c1, c2, . . . , cn) be a loop of order n. By Lemma 2.2.5, n ≥ 3. Assume

|E(L)| = 0 and choose i < n. Since ci is odd, f(ci) is even. This contradicts our assumption,

so |E(L)| 6= 0. Now assume ci is even for all i ≤ n. If each ci is even, thenci

2= ci+1 or

ci = 2ci+1 for all i, 1 < i < n. By induction we have c1 = 2n−1cn, so c1 = 2nc1. This implies

c1 = 0, contradicting c1 ∈ N. Therefore |O(L)| 6= 0.

Lemma 2.2.8. Let L be a loop with |L| = n.

1. If n is even, then |O(L)| ≤ n

2.

2. If n is odd, then |O(L)| ≤ n− 1

2.

Proof. By Lemma 2.2.7, |O(L)| 6= 0 and |E(L)| 6= 0. If c is odd, then f(c) is even. That

is, for every ci ∈ O(L) we have f(ci) ∈ E(L). Therefore 2|O(L)| ≤ n, so |O(L)| ≤ n

2. Now

assume n is odd. In this casen

2/∈ N, and since |O(L)| ∈ N, we have |O(L)| ≤ n− 1

2.

This lemma leads to an obvious corollary whose proof we omit.

Corollary 2.2.9. Let L be a loop with |L| = n.

1. If n is even, then |E(L)| ≥ n

2.

2. If n is odd, then |E(L)| ≥ n + 1

2.


Lemma 2.2.10. Let C be a loop with |C| = n. Then |O(C)| ≥ 2 if and only if the loop is

not (4, 2, 1).

Proof. (⇐) Let C be a loop that is not (4, 2, 1). Suppose |O(C)| = 1. Without loss of

generality we may assume c1 is the odd element of C. We note that c2 = 3c1+1 and ck =ck−1

2

for 2 < k ≤ n since ck−1 is even. If ck−1 =3c1 + 1

2k−3, then ck =

1

2

(3c1 + 1

2k−3

)=

3c1 + 1

2k−2. Thus

by induction ck =3c1 + 1

2k−2for 2 ≤ k ≤ n. Now f(cn) = c1 and cn is even, so

3c1 + 1

2n−1= c1.

Therefore 3c1+1 = c1(2n−1) or c1 =

1

2n−1 − 3. Because c1 ∈ N, c1 = 1. This is a contradiction

and thus |O(C)| ≥ 2.

(⇒) If |O(C)| ≥ 2, then C 6= (4, 2, 1) .

Lemma 2.2.11. If C is a loop other than (4, 2, 1), then |C| ≥ 4.

Proof. Let C be a loop other than (4, 2, 1) with order n. By Lemma 2.2.10, |O(C)| ≥ 2.

Now using the results from Lemma 2.2.8, |C| ≥ 4.

Continuing our examination of the elements of loops, the next two definitions are natural

ones.

Definition 2.2.12. Let L = (c1, c2, . . . , cn) be a loop. The maximum element in L, denoted

by M, is the element such that ci ≤M for all i. The minimum element in L, denoted by m,

is the element such that ci ≥ 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 2.2.13. If m is the minimum element of a loop L, then m is odd.

Proof. If m is even, then f(m) =m

2< m, contradicting the minimumity of m. Therefore m

is odd.


Lemma 2.2.14. 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 = 2k for some odd integer k ≥ 1. Therefore

f(M) =2k

2= k and thus f 2(M) = 3k + 1. Since k ≥ 1, we know 2k < 3k + 1 and thus

M < 3k + 1 which contradicts the maximumity of M . Therefore 4|M .

Corollary 2.2.15. Let L = (c1, c2, . . . , cn) be a loop with |L| = n. Then |E(L)| > n

2. In

particular, |E(L)| > |O(L)|.

Proof. Due to Lemma 2.2.1, for all c ∈ O(L) there exists f(c) ∈ E(L). We know L has a

maximum element M such that f(M) =M

2. However, by Lemma 2.2.14,

M

2∈ E(L). That

is, there is no c ∈ O(L) such that f(c) =M

2. Therefore |E(L)| > n

2.

Lemma 2.2.16. Let L = (c1, c2, . . . , cn) and let m be the minimum element of L. Then

4|(3m + 1) if and only if L = (4, 2, 1).

Proof. Assume L = (4, 2, 1). It is clear that m = 1 and 4|(3m+1). Now assume L 6= (4, 2, 1).

Since the minimum element m is odd, f(m) = 3m+ 1 is even by Lemma 2.2.1. Let’s assume

4|(3m + 1). Then f 2(3m + 1) =3m + 1

4. In order to maintain the minimumity of m,

m ≤ 3m + 1

4, or m ≤ 1. This implies that m = 1 which contradicts L 6= (4, 2, 1). Therefore

4 does not divide (3m + 1).

Lemma 2.2.14 and Lemma 2.2.16 lead to an obvious Corollary whose proof we omit.

Corollary 2.2.17. Let L = (c1, c2, . . . , cn) be a loop with maximum element M and minimum

element m. Then M = 3m + 1 if and only if L = (4, 2, 1).

We now discuss the congruence classes of M and m in order to discover more about loop

structure.


Lemma 2.2.18. 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 2.2.19. Let C be a loop with minimum element m. Then m ≡ 3 mod 4 if and only

if C 6= (4, 2, 1).

Proof. (⇐) Let C 6= (4, 2, 1). Then m = 4x + a for some integer x ≥ 0 and a ∈ {0, 1, 2, 3}.

Since m is odd, a /∈ {0, 2}. Suppose a = 1. Then m = 4x+1. We know3m + 1

2is an element

of C because |C| ≥ 4 and 3m + 1 is even. Therefore3m + 1

2=

3(4x + 1) + 1

2which implies

3m + 1

2=

12x + 4

2or

3m + 1

2= 6x + 2. It is clear 6x + 2 is even. However,

3m + 1

2is odd

because 3m + 1 is not divisible by 4 due to Lemma 2.2.16. Therefore a 6= 1 and thus m ≡ 3

mod 4. (⇒) Let C be a loop and suppose m ≡ 3 mod 4. Then clearly C 6= (4, 2, 1).

Definition 2.2.20. If c is an element of a loop L and c has a preimage y /∈ L, then the tail

of c is Tc ={t ∈ N | ∃ n ∈ N with fn(t) = c and fk(t) /∈ L ∀ k, 0 ≤ k < n

}. In this case,

c is called a root of the loop L or the root of the tail Tc.

As an example, {5, 8, 16} ⊂ T4, which is the tail of the root 4 in the loop (4, 2, 1). This

example shows that the (4, 2, 1) loop has a tail, but what about loops other than (4, 2, 1)?

Lemma 2.2.21. If L is a loop, then L has a tail.

Proof. Let L be a loop with maximum element M . We see that 2M is a preimage of M

because f(2M) = M . Since M is maximum, 2M /∈ L. Therefore 2M ∈ TM .

Lemma 2.2.22. 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 ∈ Tc 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 6= b. Therefore by definition of

the function f , a and b are opposite in parity.


2.3 Advanced Results

Theorem 2.3.1. A loop L has a root c if and only if 2|c and c ≡ 1 mod 3.

Proof. (⇒) Since c is a root, c has two preimages opposite in parity. In particular, f(2j+1) =

c for some integer j ≥ 0. Therefore c = f(2j + 1) = 6j + 4 ≡ 1 mod 3. It is also clear that

2|c.

(⇐) Let c ∈ L such that 2|c and c ≡ 1 mod 3. Clearly 2c, which is even, is a preimage of c.

Since 2|c, c = 2k for some k ∈ N. Also, c ≡ 1 mod 3 so 2k = 3x + 1 for some x ∈ N. This

implies x =2k − 1

3which is odd. In particular, x 6= 2c and thus c has a preimage not in L.

Therefore c is a root of L.

Let c and d be roots of disjoint loops. Is it possible for Tc ∩ Td 6= ∅? Is it possible if c

and d are roots of the same loop? These questions will be addressed in Lemma 2.3.2 and

Corollary 2.3.3.

Lemma 2.3.2. 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 fn(b) ∈ X and fm(b) ∈ Y for some n, m ∈ N. Without loss of generality

assume m ≥ n. Since fn(b) ∈ X implies fn+i(b) ∈ X for all i ∈ N, we see fm(b) ∈ X ∩ Y .

This contradicts X and Y being disjoint and thus their tails do not intersect.

Corollary 2.3.3. 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 Tx ∩ Ty 6= ∅

and let c ∈ Tx ∩ Ty. Since c ∈ Tx, fn(c) = x for some n ∈ N and fk(c) /∈ L for all k < n.

Likewise, since c ∈ Ty, fm(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 = fn(c) = y, and the tails are


not distinct. Therefore n < m and since fm(c) = y and f j(c) /∈ L for all j < m we have

x = fn(c) /∈ L, which contradicts x being a root. Thus Tx ∩ Ty = ∅.

This is an interesting find. Every element x of Tc has a nonempty preimage f−1(x) ⊆ Tc.

Thus the nonempty set f−j(x) ={a ∈ N|f j(a) = x

}⊆ Tc for all j ≥ 1. Note that if i 6= j,

then f−i(x) ∩ f−j(x) = ∅, so Tc has an infinite number of elements. That is, if a loop other

than (4, 2, 1) exists, then there exist at least two infinite disjoint subsets of N defined by the

function f .

Definition 2.3.4. The set of roots in a loop L is denoted by R(L).

Theorem 2.3.5. 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 soM

2∈ L. However,

M

2is

even andM

2≡ 2 mod 3. Thus,

M

2is not a root. Therefore there exist even elements in L

that are not roots and so |R(L)| < |E(L)|.

Theorem 2.3.5 leads to three immediate corollaries. The proofs of Corollary 2.3.6 and

Corollary 2.3.7 are omitted.

Corollary 2.3.6. Let L be a loop. Then |R(L)| < |E(L)|.

Corollary 2.3.7. Let L = (4, 2, 1). Then |R(L)| = 1.


Corollary 2.3.8. A loop C has multiple roots if and only if C 6= (4, 2, 1).

Proof. Let C be a loop with multiple roots. Then C 6= (4, 2, 1) because (4, 2, 1) has only

one root. Now assume C 6= (4, 2, 1). By Lemma 2.2.10, |O(C)| ≥ 2. From Theorem 2.3.5,

|R(C)| ≥ 2.

Assuming a loop other than (4, 2, 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 2.3.9. If P − 1 satisfies the Collatz Condition, P − 1 /∈ {1, 2, 4}, and 3Z |P for

some Z ∈ N, then2ZP

3Z− 1 satisfies the Collatz Condition. Also, if the Collatz Length for

P − 1 is k, then the Collatz Length for2ZP

3Z− 1 is k + 2Z.

Proof. Let P − 1 satisfy the Collatz Condition, P − 1 /∈ {1, 2, 4}, and 3Z |P for some Z ∈ N.

Since 3Z |P , we see2ZP

3Z− 1 is a natural number and thus is in the domain of f . Since

2ZP

3Z− 1 is odd,

f

(2ZP

3Z− 1

)= 3

(2ZP

3Z− 1

)+ 1

=2ZP

3Z−1− 3 + 1

=2ZP

3Z−1− 2.

This number is even, so applying f yields f

(2ZP

3Z−1− 2

)=

2Z−1P

3Z−1−1. The number obtained

after applying f twice is another number that appears in the same form as the original


number. Since 3Z |P , we see 3Z−1|P . In general,

f 2j

(2ZP

3Z− 1

)=

2Z−jP

3Z−j− 1

for 0 ≤ j ≤ Z.

In particular,

f 2Z

(2ZP

3Z− 1

)=

2Z−ZP

3Z−Z− 1

=20P

30− 1

= P − 1.

Since P − 1 satisfies the Collatz Condition,2ZP

3Z− 1 satisfies the Collatz Condition.

We also need to verify that fn

(2ZP

3Z− 1

)6= 1 for 0 ≤ n < 2Z. Assume fn

(2ZP

3Z− 1

)=

1 for some n, 0 ≤ n < 2Z. Since fn

(2ZP

3Z− 1

)= 1, fn+i

(2ZP

3Z− 1

)∈ {1, 2, 4} for all

i ∈ N. We have just seen f 2Z

(2ZP

3Z− 1

)= P − 1 which contradicts P − 1 /∈ {1, 2, 4}. Now

since k is the Collatz Length for P − 1,2ZP

3Z− 1 has Collatz Length k + 2Z.

The loop (4, 2, 1) can be rewritten as (1, 4, 2). 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 2.2.3, a loop L with order n has the characteristic fn(c) = c. There-

fore the loop (1, 4, 2) written as a Generalized Collatz Sequence is {1, 4, 2, 1, 4, 2, . . .} or

{odd, even, even, odd, even, even, . . .}. Does any other loop follow a pattern like this?

Definition 2.3.10. A loop L with |L| = n is said to be parity periodic if the elements of L fol-

low the pattern of(d1, e1, . . . , ek, d2, ek+1, . . . , e2k, . . . , d n

k+1, e kn

k+1−(k−1), e kn

k+1−(k−2), . . . , e kn

k+1

)where di ∈ O(L), ei ∈ E(L), for all i ∈ {1, . . . , k}.


Theorem 2.3.11. A loop L is parity periodic if and only if L = (4, 2, 1).

Proof. (⇐) Let L = (4, 2, 1). Clearly (4, 2, 1) = (1, 4, 2) is parity periodic.

(⇒) Let L be a loop other than (4, 2, 1) that is parity periodic with |L| = n. It is impossible

for L = (d1, d2, . . . , dn) by Lemma 2.2.1. It is impossible for L =(d1, e1, d2, e2, . . . , dn

2, en

2

)by Lemma 2.2.14. Now assume

L =(d1, e1, . . . , ek, d2, ek+1, . . . , e2k, . . . , d n

k+1, e kn

k+1−(k−1), e kn

k+1−(k−2) . . . , e kn

k+1

)with k ≥ 2.

Without loss of generality, suppose d1 is the minimum element of L. Therefore f 2(d1) =

f(e1) = e2, which is even since k ≥ 2. Thus3d1 + 1

2= e2 = 2l for some integer l, so

4|(3d1 + 1) which contradicts Lemma 2.2.16. Thus a loop L is parity periodic if and only if

L = (4, 2, 1).

Definition 2.3.12. Let O(L) = {d0, d1, . . . , dk−1} and write

L =(d0, e0,1, . . . , e0,j0 , d1, e1,1, . . . , e1,j1 , . . . , dk−1, ek−1,1, . . . , ek−1,jk−1

)where el,i ∈ E(L) for

all l ∈ {0, . . . , k − 1} and all i ∈ {1, . . . , jl}. Then for each l ∈ {0, . . . , k − 1}, define a

subsequence of L to be (dl, el,1, . . . , el,jl). The order of a subsequence is the number of

elements in the subsequence. If (dl, el,1, . . . , el,jl) is a subsequence of L, we denote the pair

[dl, jl] as the index of the subsequence. Note that for l ∈ {0, . . . , k − 1} f jl(3dl + 1) =

d(l+1 mod k) or3dl + 1

2jl= d(l+1 mod k). We say L is written as a juxtaposition of indices if

L = [d0, j0] [d1, j1] . . . [dk−1, jk−1] where d0 is the minimum element of L.

By Theorem 2.3.11, (4, 2, 1) is the only parity periodic loop. That is, if C 6= (4, 2, 1),

then there exist two indices of C [di, ji] and [dk, jk] such that ji 6= jk. Also, it should be

noted that j0 = 1 whenever L 6= (4, 2, 1) due to Lemma 2.2.16.


Theorem 2.3.13. Let L be a loop. Then if da ∈ O(L) with 0 ≤ a ≤ k − 1, then

da =

∑k−1i=0

(3k−1−i · 2

∑ i−1l=0j(l+a mod k)

)2

∑ k−1i=0 ji − 3k

.

Proof. Let L be a loop and write L = [d0, j0] [d1, j1] . . . [dk−1, jk−1] with dk = d0. Since

di+1 =3di + 1

2jifor each i ∈ {0, . . . , k − 1}, we see

d1 =3d0 + 1

2j0

and

d2 =3d1 + 1

2j1=

32d0 + 3

2j0+j1+

1

2j1.

If we assume

di =3id0 + 3i−1

2j0+···+ji−1+

3i−2

2j1+···+ji−1+ · · ·+ 1

2ji−1

for some i, then

di+1 =3di + 1

2ji

=3i+1d0 + 3i

2j0+···+ji+

3i−1

2j1+···+ji+ · · ·+ 1

2ji.


Thus

d0 = dk =3kd0 + 3k−1

2j0+···+jk−1+

3k−2

2j1+···+jk−1+ · · ·+ 1

2jk−1.

Finding a common denominator and simplifying,

d0 · 2∑ k−1

i=0 ji = d0 · 3k +∑

k−1i=0

(3k−1−i · 2

∑ i−1l=0jl

)

or

d0 =

∑k−1i=0

(3k−1−i · 2

∑ i−1l=0jl

)2

∑ k−1i=0 ji − 3k

.

Since d0 was arbitrary, any di could have been chosen to play the role of d0. That is, for

any integer a ∈ {0, . . . , k − 1}

da =

∑k−1i=0

(3k−1−i · 2


)2

∑ k−1i=0 ji − 3k

.

We can now set up a system of linear Diophantine equations. For notational ease, let

B = 2∑ k−1

i=0 ji − 3k be the denominator of the expression in Theorem 2.3.13. If

∑k−1i=0

(3k−1−i · 2

∑ i−1l=0j(l+p mod k)

)= B

for some p ∈ {0, . . . , k − 1}, then dp = 1. Therefore L = (4, 2, 1). If

∑k−1i=0

(3k−1−i · 2

∑ i−1l=0j(l+p mod k)

)< B

for some p ∈ {0, . . . , k − 1}, then dp < 1. Therefore

∑k−1i=0

(3k−1−i · 2


)


is a positive multiple of B for every p ∈ {0, . . . , k − 1}.

Alternatively we can use di+1 =3di + 1

2jifor i = 0, . . . , k−1 to create the following matrix

equation.

−3 2j0 0 · · · · · · 0

0 −3 2j1 0 · · · ...

... 0 −3. . . . . .

...

...... 0

. . . . . . 0

0 · · · ... · · · −3 2jk−2

2jk−1 0 · · · · · · 0 −3

d0

d1

...

...

dk−2

dk−1

=

1

1

...

...

1

1

.

For notational ease, let

A =

−3 2j0 0 · · · · · · 0

0 −3 2j1 0 · · · ...

... 0 −3. . . . . .

...

...... 0

. . . . . . 0

0 · · · ... · · · −3 2jk−2

2jk−1 0 · · · · · · 0 −3

, ~d =

d0

d1

...

...

dk−2

dk−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(2∑k−1

i=0 ji)

= (−1)k−1(2∑k−1

i=0 ji − 3k)

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?


Theorem 2.3.14. [GP90] To solve the system of Linear Diophantine equations AX = B,

unimodular row reduce [At|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 RtK = B has integer solutions for

K, and all the solutions of AX = B are of the form X = T tK.

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.

2. Interchange two rows of the matrix.

3. Multiple one row of the matrix by −1.

Solving A~d = ~1 would show whether every di, with i ∈ {0, . . . , k − 1}, is an odd natural

number or not. If there exists some di /∈ N or di 6= 2j + 1 for some integer j ≥ 0, then

(4, 2, 1) is the only loop possible by Theorem 2.3.13.

With the aid of Mathematica, a few elementary examples were explored.

Example 2.3.15. Let L be a loop with |O(L)| = 1.

This leads to the 1× 1 matrix

A = [2j0 − 3].

It is evident that j0 = 2 and thus L = (4, 2, 1). This is not surprising since we already know

(4, 2, 1) is the only loop such that |O(L)| = 1.




A =

−3 2

2j1 −3

.

As you can see, j0 = 1 because L 6= (4, 2, 1). Now solving for ~d,

d0

d1

=

5

−9 + 2j1+1

3 + 2j1

−9 + 2j1+1

.

Since each entry must be an odd positive integer,5

−9 + 2j1+1is an odd positive integer which

is impossible. Therefore |O(L)| 6= 2. In particular, there are no loops with exactly two odd

elements.



A =

−3 2 0

0 −3 2j1

2j2 0 −3

.

As before, j0 = 1. Now solving for ~d,

d0

d1

d2

=

15 + 2j1+1

−27 + 2j1+j2+1

9 + 32j1 + 2j1+j2

−27 + 2j1+j2+1

9 + 52j2

−27 + 2j1+j2+1

.


It is evident that |E(L)| = j0 + j1 + j2, but is |E(L)| ∈ N? Using the result S 2, |O(L)| = 3,

and the current computational bound of R = 1.25208 ·1018 we can see that |E(L)| /∈ N. That

is: the inequalityln 3

ln 2<

j0 + j1 + j2

3≤

ln(3 + 1

R

)ln 2

does not hold if |E(L)| ∈ N. Therefore

|O(L)| 6= 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)

ln 2(cf S 2) we obtain the inequality

|O(L)| ≥ |L| ln 2

ln (6R+2R

).

We know from S 3 that |L| ≥ 1, 027, 712, 276 and R = 1.25208 · 1018. Therefore |O(L)| ≥

397, 573, 379. That is, the smallest potential matrix A has size 397, 573, 379× 397, 573, 379.

Table 1.1 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 2.1: Matrix Sizes given a Computational Bound R

Computational Bound Loop Order Matrix Size3.80765 · 1023 2, 302, 268, 119, 908 890, 638, 885, 193× 890, 638, 885, 1935.10126 · 1022 355, 504, 839, 929 137, 528, 045, 312× 137, 528, 045, 3124.35849 · 1021 186, 265, 759, 595 72, 057, 431, 991× 72, 057, 431, 9912.16891 · 1020 17, 026, 679, 261 6, 586, 818, 670× 6, 586, 818, 6701.25208 · 1018 1, 027, 712, 276 397, 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 di is crucial in order to see if the Collatz function produces

more than one loop or not.

2. Discuss how tails can coexist in N without intersecting.

How can multiple tails exist in N? A loop C 6= (4, 2, 1) has |R(C)| ≥ 2 by Corollary

2.3.8. Therefore, if there exists a loop C 6= (4, 2, 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.

26

CHAPTER 3. FUTURE WORK 27

The General Collatz function fn(x) : N→ N with integer n ≥ 0 is defined by

fn(x) =

(2n − 1)x + 1 ; if x is odd

x

2; if x is even

.

The first three General Collatz functions are:

f0(x) =

1 ; if x is odd

x

2; if x is even

,

f1(x) =

x + 1 ; if x is odd

x

2; if x is even

,

and

f2(x) =

3x + 1 ; if x is odd

x

2; if x is even

.

The f2 function is the Collatz function f that has been discussed throughout this

paper. It is clear that the f0 function produces the loop (1) since f0(1) = 1. The proof

that this is the only loop is quite simple and will be given later. It is also clear that

the f1 function produces the loop (2, 1) because f1(2) = 1 and f1(1) = 2. The proof

that this is the only loop is also quite simple and shall be given later.


Each of the functions in the family {fn : n ∈ N ∪ {0}} satisfies several of the properties

we have discussed earlier. These properties will be stated here without proof.

Theorem 3.0.1. Let fk ∈ {fn : n ∈ N ∪ {0}}. Suppose L is a loop generated by the

function fk with maximum element M and minimum element m. Then

a) If c ∈ O(L), then fk(c) ∈ E(L).

b) If c ∈ E(L), then either fk(c) ∈ E(L) or fk(c) ∈ O(L).

c) M ∈ E(L).

d) m ∈ O(L).

e) M ≡ 1 mod 2k − 1.

A few less elementary results on this family of functions follow.

Theorem 3.0.2. If g is a function in the {fn : n ∈ N ∪ {0}} family, then g has a loop.

Proof. Let g be a function in the {fn : n ∈ N ∪ {0}} family. Then g = fk for some

k ≤ n.

g(1) = (2k − 1)(1) + 1

= 2k − 1 + 1

= 2k.

Since gk(2k) = 1, g has a loop.

As previously stated, the function f0 and the function f1 only have one loop. The

proofs of these follow but first we need a definition.


Definition 3.0.3. We say c ∈ N satifies the General Collatz Condition for the function

fk, if fnk (c) = 1 for some integer n ≥ 0.

Theorem 3.0.4. If c ∈ N, then c satisfies the General Collatz Condition for the

function f0. In particular, (1) is the only loop in f0.

Proof. Let c ∈ N. By the fundamental theorem of arithmetic c = 2k1pk22 pk3

3 · · · pknn

where ki ≥ 0 for all i ∈ {1, . . . , n} and each pi is an odd prime. We can see that

fk10 (c) = pk2

2 pk33 · · · pkn

n which is odd. Therefore fk1+10 (c) = 1 and thus c satisfies the

General Collatz Condition for the function f0.

Theorem 3.0.5. If c ∈ N, then c satisfies the General Collatz Condition for the

function f1. In particular, (2, 1) is the only loop in f1.

Proof. This proof is by strong induction. We see that f1(1) = 2, f1(2) = 1, f1(3) = 4,

and f1(4) = 2. Therefore 1,2,3 and 4 satisfy the General Collatz Condition for the

function f1. Assume every natural number up to k for some k > 4 satisfies the General

Collatz Condition for the function f1. Assume k + 1 is even. It is clear that k + 1 < 2k

as long as k > 1. Thereforek + 1

2< k and thus f1(k+1) =

k + 1

2< k. By the induction

hypothesis, for all c ∈ N, c < k, there exists n ∈ N such that fn1 (c) = 1. Hence,

k + 1

2

satisfies the General Collatz Condition for the function f1. Therefore k + 1 satisfies

the General Collatz Condition for the function f1. Now assume k + 1 is odd. It is clear

that k + 2 < 2k as long as k > 2 and thusk + 2

2< k. Since f1(k + 1) = k + 2 is even,

f1(k + 2) =k + 2

2< k. By the induction hypothesis, for all c ∈ N, c < k, there exists

n ∈ N such that fn1 (c) = 1. Hence,

k + 2

2satisfies the General Collatz Condition for

the function f1. Therefore k+2 satisfies the General Collatz Condition for the function

f1 and thus k + 1 satisfies the General Collatz Condition for the function f1.


Why is it so easy to verify that f0 and f1 have a unique loop yet it is so difficult in the

f2 case?

Looking at the family of functions {fn : n ∈ N ∪ {0}} leads to a new more General

Collatz Conjecture.

Conjecture 3.0.6 (General Collatz Conjecture). If c ∈ N, then c satisfies the General

Collatz Condition for {fn : n ∈ N ∪ {0}}. In particular, (2n, 2n−1, . . . , 2, 1) is the only

loop in {fn : 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, 2, 1) is the only loop or not. We also conjectured about {fn : 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 2.1,

these characteristics shall lead to restrictions in order for another loop to coexist with the

(4, 2, 1) loop. We hope that these restrictions will lead to a contradiction, thus showing that

(4, 2, 1) is the only loop.

Bibliography

[AM97] Stefan Andrei and Cristian Masalagiu. About the Collatz conjecture. Acta Infor-

matica, 35:167–179, 1997.

[Coo09] Josh Cooper, Nov. 2009. http://www.math.sc.edu/ cooper/combprob.html.

[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, 2011.

[Sin03] Matti K. Sinisalo. On the Minimal Cycle Lengths of the Collatz Sequence. Master’s

thesis, University of Oulu, Finland, 2003.

31

characteristics of counter example loops in the collatz conjecture

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