Working with 2s and 3s

TL;DR

This paper presents an elementary approach to establish an equivalent condition for the Collatz conjecture, providing new insights, examples, and open problems related to this longstanding mathematical question.

Contribution

It introduces an equivalent condition for the Collatz conjecture using elementary methods and offers various examples, exercises, and conjectures.

Findings

Derived an equivalent condition for the Collatz conjecture

Provided several examples illustrating the condition

Proposed open problems and conjectures related to the conjecture

Abstract

We establish an equivalent condition to the validity of the Collatz conjecture, using elementary methods. We derive some conclusions and show several examples of our results. We also offer a variety of exercises, problems and conjectures.

Equations

$$T(n)=\left\{\begin{array}[c]{c}\frac{3n+1}{2},\quad n\equiv 1(\operatorname{mod}\ 2)\\ \frac{n}{2},\quad n\equiv 0(\operatorname{mod}\ 2)\end{array}\right.,\quad n\in\mathbb{N},$$

$$T^{\left(k\right)}\left(n\right)=\underset{k\text{ times}}{\underbrace{T\circ T\circ \cdots \circ T}}(n)=1.$$

$$T(3)$$

$$T(7)$$

$$T^{\left(6\right)}(7)$$

$${\sigma }_{\infty }(n)=\inf \left\{k\in \mathbb{N}\cup \left\{\infty \right\}|T^{\left(k\right)}\left(n\right)=1\right\},n\ge 2,$$

$$\mathfrak{C}=\mathbb{N},$$

$$\mathfrak{C}=\left\{n\in \mathbb{N}|{\sigma }_{\infty }(n)<\infty \right\}.$$

$$\begin{tabular}[c]{|c|c|c|c|c|c|c|c|}\hline\cr$n$&$2$&$3$&$4$&$5$&$6$&$7$&$8$\\ \hline\cr$\sigma_{\infty}(n)$&$1$&$5$&$2$&$4$&$6$&$11$&$3$\\ \hline\cr\end{tabular}\ \ .$$

$$S_k=\left\{n\in \mathbb{N}|{\sigma }_{\infty }(n)=k\right\},k\ge 1.$$

$$S_1$$

$$S_6$$

$$\bigcup _{k=0}^{\infty }{S}_k=\mathbb{N}.$$

$$T^{-1}(n)=\left\{\begin{array}[c]{c}\left\{2n\right\},\quad n\equiv 0,1(\operatorname{mod}\ 3)\\ \left\{2n,\frac{1}{3}\left(2n-1\right)\right\},\quad n\equiv 2(\operatorname{mod}\ 3)\end{array}\right..$$

$$x_{n+1}^{\left(m\right)}=T\left(x_n^{(m)}\right),0\le n,$$

$$\left\{x_n^{(2)}\right\}$$

$$\left\{x_n^{(5)}\right\}$$

$$\left\{x_n^{(7)}\right\}$$

$$\bigcap _{m=2}^{\infty }\left\{x_n^{(m)}\right\}=\left\{2,1\right\}.$$

$$x_{n+i}^{\left(m\right)}=T^{\left(i\right)}\left(x_n^{(m)}\right),0\le n,$$

$$T^{\left(i\right)}\left(x\right)=f_{i,j}(x),\text{if }x\equiv j\left(\mathrm{mod}{2}^i\right),0\le j\le 2^i-1.$$

$$f_{1,0}(x)$$

$$f_{2,0}(x)$$

$$f_{3,0}(x)$$

$$f_{3,4}(x)$$

$$\forall n\in \mathbb{N}\exists m\in \mathbb{N}\text{ such that }n\equiv k\left(\mathrm{mod}{2}^m\right)\text{ and }\frac{d}{dx}{f}_{m,k}(x)<1.$$

$$f_{5,10}(x)=\frac{3x+2}{32},f_{5,11}(x)=\frac{27x+23}{32}.$$

$$x_5^{\left(11\right)}=f_{5,11}(11)=10,x_{10}^{\left(11\right)}=f_{5,10}(10)=1.$$

$${\Lambda }_m=\left\{n\in \mathbb{N}|n=\frac{2^m}{3^l}-\sum _{k=1}^l\frac{2^{b_k}}{3^k}\right\},m\ge 4,$$

$$0\le l\le m-3\text{and }0\le b_1<b_2<\cdots <b_l\le m-4.$$

$${\Lambda }_4$$

$${\Lambda }_5$$

$${\Lambda }_6$$

$${\Lambda }_4$$

$${\Lambda }_6$$

$${\Lambda }_7$$

Full text

1 Introduction

The Collatz conjecture (also known as the $3n+1$ conjecture, Ulam’s conjecture, the Syracuse problem, Kakutani’s problem, Hasse’s algorithm, etc.) was first proposed by Lothar Collatz in 1937 [2]. In terms of the function $T(n),$ defined by

[TABLE]

the conjecture claims that for all natural numbers $n,$ there exists a natural number $k$ such that

[TABLE]

For example, we have

[TABLE]

We define $T^{\left(\infty\right)}\left(n\right)=1.$

Exercise 1

Prove that if $\forall n\in\mathbb{N}$ $\exists$ $k\in\mathbb{N}$ such that $T^{\left(k\right)}\left(n\right)<n,$ then the Collatz conjecture is true. The number $k$ is called the stopping time of $n.$

As of February 2007, the Collatz conjecture has been verified for numbers up to $13\times 2^{58}=\allowbreak 3,746\,,994,\,889,\,\allowbreak 972,\,252,\,672$ [7]. However, the general case remains open.

Introducing the *total stopping time *function $\sigma_{\infty}(n),$ defined by $\sigma_{\infty}(1)=0$ and

[TABLE]

we can reformulate the Collatz conjecture as

[TABLE]

where

[TABLE]

From (2), we have

[TABLE]

Exercise 2

Find $\sigma_{\infty}(n)$ for $9\leq n\leq 100.$

Hint: The web page http://www.numbertheory.org/php/collatz.html contains an implementation which allows the computation of $\sigma_{\infty}(n)$ for large values of $n.$

One could consider the inverse problem and try to characterize the sets $S_{k},$ defined by $S_{0}=\left\{1\right\}$ and

[TABLE]

The first few $S_{k}$ are

[TABLE]

It is clear from (4) that $2^{k}\in S_{k}$ $\forall k\in\mathbb{N}_{0},$ where $\mathbb{N}_{0}=\mathbb{N}\cup\left\{0\right\}.$ In terms of the sets $S_{k},$ the Collatz conjecture reads

[TABLE]

Exercise 3

Compute $S_{k}$ for $9\leq k\leq 100.$

Hint: Consider the inverse map $T^{-1}:\mathbb{N}\rightarrow\mathbb{P}\left(\mathbb{N}\right),$ given by

[TABLE]

The sequence of natural numbers $\left\{x_{n}^{(m)},\ n\geq 0\right\}$, defined by $x_{0}^{(m)}=m$ and

[TABLE]

is called the trajectory or forward orbit of $m\in\mathbb{N}$. From (2), we have

[TABLE]

Exercise 4

Find $\left\{x_{n}^{(m)}\right\}$ for $9\leq m\leq 100.$

Using the sequences $\left\{x_{n}^{(m)}\right\}$ we can restate Collatz’s conjecture as

[TABLE]

We can also consider higher order recurrences, i.e., instead of (6), use

[TABLE]

where

[TABLE]

For $i=1,2,3,$ we have

[TABLE]

Exercise 5

Prove that if the sequence $\left\{3k+4\ \right\}\subset\mathfrak{C},$ then the Collatz conjecture is true.

In terms of (7), the Collatz conjecture reads

[TABLE]

For example, we have $11\equiv 11\left(\operatorname{mod}2^{5}\right)$ and

[TABLE]

Thus,

[TABLE]

The literature on the Collatz conjecture is vast and growing rapidly. Rather than attempting to cover it, we refer the reader to the excellent survey papers [5] and [6].

2 Representation of natural numbers

Let the sets $\Lambda_{m}$ be defined by $\Lambda_{m}=\left\{2^{m}\right\},\quad 0\leq m\leq 3$ and

[TABLE]

for some $m,l,b_{1},\ldots,b_{l}\in\mathbb{N}_{0},$ with

[TABLE]

The first few $\Lambda_{m}$ are

[TABLE]

Using the $\left(l+2\right)-$tuple $\left(l,b_{1},\ldots,b_{l},m\right)$ to represent the number $\frac{2^{m}}{3^{l}}-{\displaystyle\sum\limits_{k=1}^{l}}\frac{2^{b_{k}}}{3^{k}},$ we can write

[TABLE]

Exercise 6

Compute $\Lambda_{m}$ for $9\leq m\leq 100.$

Hint: (a) If $\left(v_{1},v_{2},\ldots,v_{l+2}\right)\in\Lambda_{m},$ then $\left(v_{1},v_{2}+1,\ldots,v_{l+2}+1\right)\in\Lambda_{m+1}.$

(b) $\left(1,0,2m\right)\in\Lambda_{2m}$ for all $m\geq 2.$

Comparing (5) with (8), it seems that $S_{m}=\Lambda_{m}.$ The next results will show this to be true.

Lemma 7

For all $m\in\mathbb{N}_{0},$ we have

[TABLE]

Proof. Let $n\in\Lambda_{m+1}.$ Then,

[TABLE]

and

[TABLE]

if $b_{1}>0$ ($n$  even) or

[TABLE]

if $b_{1}=0$ ($n$  odd). In either case, $T(n)\in\Lambda_{m}.$

Lemma 8

For all $m\in\mathbb{N}_{0},$ we have

[TABLE]

Proof. We use induction on $m.$ The case of $m=0$ is clearly true, since $\Lambda_{0}=\left\{1\right\}=S_{0}.$

Assuming (10) to be true for $m$, let $n\in\Lambda_{m+1}.$ From (9) we have $T(n)\in\Lambda_{m}$ and therefore $\sigma_{\infty}\left[T(n)\right]=m.$ Thus, $\sigma_{\infty}(n)=m+1$ and the result follows.

Exercise 9

Show that

[TABLE]

The other inclusion is also true.

Theorem 10

For all $m\in\mathbb{N}_{0},$

[TABLE]

Proof. Clearly, $S_{m}=\left\{2^{m}\right\}=\Lambda_{m},\quad 0\leq m\leq 3.$

Let $m\geq 4$ and $s\in S_{m}.$ Using (11) we can write the recurrence (6) as

[TABLE]

where

[TABLE]

Assuming $\left\{\theta_{n}\right\}$ to be a known sequence, the solution of (12) is [1]

[TABLE]

or using (13)

[TABLE]

with

[TABLE]

Setting $n=m$ and solving for $s$ in (14), we obtain $\left(x_{m}^{(s)}=1\right)$

[TABLE]

Let $l=\Theta\left(m-1\right).$ From (13) and (15), we see that $\Theta\left(x\right)$ is a step function with unit jumps at $x=b_{1},b_{2},\ldots,b_{l},m,$ where $0\leq b_{1}<b_{2}<\cdots<b_{l}<m.$ Therefore, we can rewrite (16) as

[TABLE]

Finally, since $x_{m-3}^{(s)}=8,$ $x_{m-2}^{(s)}=4,$ $x_{m-1}^{(s)}=2$ and $x_{m}^{(s)}=1,$ the penultimate jump must occur before or at $x=m-4.$ Thus, $b_{l}\leq m-4$ and $l=\Theta\left(m-1\right)\leq m-3.$

Corollary 11

The Collatz conjecture is true if and only if every natural number $n$ can be represented in the form

[TABLE]

for some $m,l,b_{1},\ldots,b_{l}\in\mathbb{N}_{0},$ with

[TABLE]

Corollary 11 is not a proof of the Collatz conjecture, but it provides a lot of information on the set $\mathfrak{C}$ and the function $\sigma_{\infty}(n).$ When $l=0,$ we recover the known fact that $2^{m}\in S_{m},$ $\forall m\in\mathbb{N}_{0}.$ For $l=1,$ we have the following result.

Lemma 12

For all $m\in\mathbb{N},$ we have

[TABLE]

Proof. Let $n\in\Lambda_{m},$ with $l=1.$ We have

[TABLE]

Thus, $2^{m-b_{1}}\equiv 1\left(\operatorname{mod}2\right)$ and therefore $m-b_{1}\equiv 0\left(\operatorname{mod}2\right).$ Considering the cases $m$ even and $m$ odd, the result follows.

When $l=2,$ the situation is slightly more complicated. To simplify matters, we restrict ourselves to the case of $n$ being odd.

Proposition 13

For all $m\geq 5,$ with $m\neq 6,8,$ we have

[TABLE]

Proof. Let $n\in\Lambda_{m},$ odd, with $l=2.$ Then,

[TABLE]

and therefore

[TABLE]

Considering all possible cases, we have

1. $2^{b_{2}}\equiv 1\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 3\left(\operatorname{mod}9\right)$, which implies

[TABLE]

2. $2^{b_{2}}\equiv 2\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 6\left(\operatorname{mod}9\right)$, which implies

[TABLE]

3. $2^{b_{2}}\equiv 4\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 3\left(\operatorname{mod}9\right)$, which implies

[TABLE]

4. $2^{b_{2}}\equiv 5\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 6\left(\operatorname{mod}9\right)$, which implies

[TABLE]

5. $2^{b_{2}}\equiv 7\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 3\left(\operatorname{mod}9\right)$, which implies

[TABLE]

6. $2^{b_{2}}\equiv 8\left(\operatorname{mod}9\right)$ and $2^{m-b_{2}}-1\equiv 6\left(\operatorname{mod}9\right)$, which implies

[TABLE]

Thus, for $m$ even we shall have $m-b_{2}\equiv 2\left(\operatorname{mod}6\right)$ or $b_{2}\equiv m-2\left(\operatorname{mod}6\right)$ and for $m$ odd we need $m-b_{2}\equiv 4\left(\operatorname{mod}6\right)$ or $b_{2}\equiv m-4\left(\operatorname{mod}6\right),$ with $1\leq b_{2}\leq m-4.$ Writing $b_{2}$ in terms of $m,$ the result follows.

From Corollary 11, we can also get an idea of how the total stopping time $\sigma_{\infty}(n)$ behaves if the Collatz conjecture is true. Solving for $m$ in (C5) we have

[TABLE]

In other words, $\sigma_{\infty}(n)$ lies on the family of parametric curves

[TABLE]

For example, we have

[TABLE]

Exercise 14

Prove that

[TABLE]

2.0.1 Binary sequences

Another approach is to study the sequence $\left\{\theta_{k}^{(s)},\ k\geq 0\right\},$ which contains a wealth of information.

Definition 15

Let $\tau:\mathfrak{C}\rightarrow\mathbb{N}$ be defined by

[TABLE]

For example, we have

[TABLE]

Clearly, $\tau\left(2^{n}\right)=2^{n}$, $\forall n\in\mathbb{N}_{0}.$

Exercise 16

Find $\tau(n)$ for $9\leq n\leq 100.$

Let’s study the image of $\Lambda_{m}$ by $\tau.$ We have

[TABLE]

Exercise 17

Let

[TABLE]

where $N_{m}=\#\Lambda_{m}$ denotes the number of elements in the set $\Lambda_{m}.$ Prove that $\forall m\in\mathbb{N}_{0}$ there exist a sequence

[TABLE]

such that

[TABLE]

From (18), we have

[TABLE]

It follows from (19) that

[TABLE]

Using (16), we can define an inverse function for $\tau(n).$

Definition 18

Let $\phi:\mathbb{N}\rightarrow\mathbb{Q}$ be defined by

[TABLE]

where $\beta_{m-1}^{(n)}\ldots\beta_{1}^{(n)}\beta_{0}^{(n)}$ is the binary representation of $n,$ i.e.,

[TABLE]

For example, we have

[TABLE]

Exercise 19

Find $\phi(n)$ for $9\leq n\leq 100.$

It follows from Theorem 10 that

[TABLE]

while (19) implies that

[TABLE]

In terms of $\phi,$ the Collatz conjecture reads

[TABLE]

With (C6), we finally reach a statement equivalent to the Collatz conjecture, which is independent of the original formulation in terms of $T(x).$ Although we have not succeeded in proving (C6), we hope that studying the function $\phi(n)$ will shed new light on the Collatz problem.

3 Further problems

In the spirit of the Monthly, we offer a series of problems to the curious reader. Those labeled ”Exercise” are relatively easy to prove, ”Problems” denote results strongly supported by numerical evidence and ”Conjectures” are those that we would really wish to prove, but that may turn out to be false.

Conjecture 20

Prove that

[TABLE]

where $\delta(n)$ is a slowly varying function, which might be eventually constant.

Definition 21

The Abby-Normal numbers ($\mathcal{AN}$ numbers). Let the scaled total stopping time $\gamma\left(n\right)$ be defined by

[TABLE]

We say that $a_{k}$ is the $k-$th $\mathcal{AN}$ number if

[TABLE]

with $a_{0}=2.$ In other words, $\left\{\gamma\left(a_{k}\right)\right\}$ is an increasing sequence of sharp lower bounds for the function $\delta(n)$ defined in (20).

Exercise 22

Show that

[TABLE]

From the results obtained by Eric Roosendaal [8], it follows that

[TABLE]

are possible $\mathcal{AN}$ numbers. We have

[TABLE]

Exercise 23

Find all $\mathcal{AN}$ numbers in the interval $[1,000,000,\quad 100,759,293,214,567].$

Conjecture 24

Prove that there exist infinitely many $\mathcal{AN}$ numbers.

Problem 25

Let $V_{m}$ be the $m-$vector

[TABLE]

and $L:\mathbb{R}^{N}\rightarrow\mathbb{R}^{N}$ the linear operator defined by

[TABLE]

Let $\theta\left(m\right)$ be the angle between $V_{m}$ and $L\left(V_{m}\right).$ Prove that

(i)

[TABLE]

(ii)

[TABLE]

(iii)

[TABLE]

Hint: See [4].

Problem 26

Prove that:

(i)

$\forall k\geq 6$* $\exists$ $m_{k}\in S_{k}$ such that $m_{k}+1\in S_{k}.$*

(ii)

$\forall k\geq 7$* $\exists$ $m_{k}\in S_{k}$ such that $m_{k}+2\in S_{k}.$*

(iii)

$\forall l\in\mathbb{N}$* $\exists$ $K\in\mathbb{N}$ such that $\forall k\geq K$ $\exists$ $m_{k}\in S_{k}$ such that $m_{k}+l\in S_{k}$ .*

Hint: See [3].

Exercise 27

Prove that

[TABLE]

Exercise 28

Let

[TABLE]

Show that

[TABLE]

Conjecture 29

Prove that

[TABLE]

Problem 30

Prove that

[TABLE]

and

[TABLE]

where $\left\lfloor\cdot\right\rfloor$ denotes the greatest integer function.

Problem 31

Let

[TABLE]

with

[TABLE]

Prove that

[TABLE]