An Investigation of the Collatz Conjecture

TL;DR

This paper investigates specific initial conditions of Collatz sequences that could prove the conjecture true, combining insights from mathematics and physics experts.

Contribution

It introduces conditions on starting values that imply the Collatz conjecture's validity, offering a novel approach to this longstanding problem.

Findings

Identifies special initial conditions implying the conjecture

Provides theoretical evidence supporting the conjecture's truth

Collaborative interdisciplinary approach enhances understanding

Abstract

This paper explores special conditions on the starting value of a Collatz sequence which imply that the Collatz conjecture is true. This is the result of the collaboration of a retired mathematics professor (Koelzer) and a retired physics professor (Welling).

Equations

$$2^{3^k}$$

$$={\left(3^{k}q-1\right)}^3$$

$$=(3^{k}q{)}^3-3(3^{k}q{)}^2+3(3^{k}q)-1$$

$$=q^3{3}^{3k}-3q^2{3}^{2k}+3(3^{k}q)-1$$

$$=q^3{3}^{2k-1}{3}^{k+1}-q^2{3}^k{3}^{k+1}+3^{k+1}q-1$$

$$=\left(q^3{3}^{2k-1}-q^2{3}^k+q\right)3^{k+1}-1$$

Full text

An Investigation of the Collatz conjecture

John G. Koelzer†, Rockhurst University

Daniel J. Welling

†email: john.koelzer@rockhurst.edu

**Abstract: ** This paper explores special conditions on the starting value of a Collatz sequence which imply that the Collatz conjecture is true. This is the result of the collaboration of a retired mathematics professor (Koelzer) and a retired physics professor (Welling).

The Collatz conjecture was formulated by L. Collatz in 1937. The conjecture concerns the definition of a sequence of positive integers by means of a simple algorithm. The conjecture states that no matter what the starting value is, the sequence eventually equals 1. For more information on the history of the Collatz problem see [1] and [2].

**The Collatz Conjecture: ** Let $n$ be a positive integer. If $n$ is even, divide it by 2 to get $n/2$. If $n$ is odd, multiply it by 3, add 1 and divide the result by 2 to obtain the number $(3n+1)/2$. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always reach 1. Our definition follows that of Terras in [3]. This is equivalent to the original Collatz algorithm but it results in a somewhat shorter sequence. We will assume that our starting value is an odd integer; otherwise we can simply divide by 2 until the number is odd.

**Example: ** The Collatz sequence for $n=19$ is: $19\to 29\to 44\to 22\to 11\to 17\to 26\to 13\to 20\to 10\to 5\to 8\to 4\to 2\to 1$. (14 steps)

In this paper we will present some results concerning the Collatz conjecture. We will assume that the starting integer $N$ in a Collatz sequence is an odd integer, since, if $N$ is even, it can be repeatedly be divided by $2$ until the resulting integer is odd.

The following theorem expresses the relation between an odd integer in a Collatz sequence and the next even integer in the sequence.

**Theorem 1: ** Let $N$ be an odd integer in a Collatz sequence and define $E$ to be $N+1$. The successive odd integers in the sequence starting with $N$ are

$3\left(\dfrac{E}{2}\right)-1,\ 3^{2}\left(\dfrac{E}{2^{2}}\right)-1,\ 3^{3}\left(\dfrac{E}{2^{3}}\right)-1,\ \cdots,\ 3^{k-1}\left(\dfrac{E}{2^{k-1}}\right)-1,$

where $k$ is the highest power of 2 contained as a factor in $E$. The integer $3^{k}\left(\dfrac{E}{2^{k}}\right)-1$ is the first even integer after $N$.

Proof: Let $m$ be an integer such that $1\leq m\leq k-1.$ Then $\dfrac{E}{2^{m}}$ is an even number and hence $3^{m}\left(\dfrac{E}{2^{m}}\right)-1$ will be odd. We will show that these integers are the successive odd entries in the Collatz sequence starting with $N$.

**Case 1: ** $m=1$. The next term in the Collatz sequence after $N$ is

$\dfrac{3N+1}{2}=\dfrac{3}{2}N+\dfrac{1}{2}=\dfrac{3}{2}\left(E-1\right)+\dfrac{1}{2}=\dfrac{3}{2}E-1$,

and so the theorem is proven for $m=1$.

Case 2: $2\leq m\leq k-1.$ Since $3^{m-1}\left(\dfrac{E}{2^{m-1}}\right)-1$ is an odd integer, the next term in the Collatz sequence is $\dfrac{3\left(3^{m-1}\left(\dfrac{E}{2^{m-1}}\right)-1\right)+1}{2}$, which simplifies algebraically to $3^{m}\left(\dfrac{E}{2^{m}}\right)-1$ and the theorem is proved in this case.Finally, since $k$ is the highest power of 2 contained as a factor in $E$, $\dfrac{E}{2^{k}}$ will be an odd integer. This means that $3^{k}\left(\dfrac{E}{2^{k}}\right)-1$ is even and it is the first even integer in the Collatz sequence after $N$.

Given an odd integer $N$, let $u$ be the number of steps to the next even integer, say $M$, and let $d$ be the number of steps from $M$ to the next odd integer $N^{\prime}$. By Theorem 1 $u$ is the largest power of 2 in $N+1$ and, since $M$ is even, $d$ is the largest power of 2 in $M$. The relation between $N$ and $N^{\prime}$ is given by the formula

$N^{\prime}=\left[(\frac{3}{2})^{u}(N-1\right](\frac{1}{2})^{d}$

Here is a simple corollary to illustrate the theorem:

Corollary 1: Suppose $k$ is a positive integer. Let $N=10^{k}-1$ be an odd integer in a Collatz sequence. Then the next even integer in the sequence will be $15^{k}-1$.

Proof: Using the notation from Theorem 1, $E=10^{k}$ and the largest power of 2 in $E$ is $2^{k}$. So by the Theorem, the first even integer following $10^{k}-1$ is $3^{k}(\tfrac{E}{2^{k}})-1=15^{k}-1$.

Another result following from Theorem 1 is shown below:

Theorem 2: Let $k$, $l$ and $m$ be positive integers with $m$ odd and let $N=\left(\frac{2}{3}\right)^{k}\left(2^{l}m+1\right)-1$. If $N$ is an odd integer then $N$ reaches $m$ in $l+k$ steps.

Proof: Let $E$ be the even integer $N+1$. Note that $E$ can be written as $\dfrac{2^{k}\left(2^{l}m+1\right)}{3^{k}}$. We claim that $k$ is the highest power of 2 contained as a factor in $E$. Suppose $p$ is the highest power; Since $2^{k}$ is a factor of $E$, we must have $k\leq p$.

Furthermore, we can write $\dfrac{E}{2^{p}}$ as $\dfrac{2^{k}\left(\dfrac{2^{l}m+1}{3^{k}}\right)}{2^{p}}$. Since $2^{l}m+1$ and $3^{k}$ are odd, their quotient is odd and so $2^{p}$ must divide $2^{k}$. This implies $p\leq k$ and we conclude $k=p$.

By Theorem 1, $N$ is followed by $k-1$ odd integers in the Collatz sequence and the even number $3^{k}\left(\dfrac{E}{2^{k}}\right)-1$. This even number can be rewritten as $\left(\dfrac{3}{2}\right)^{k}\left[\dfrac{2^{k}(2^{l}m+1)}{3^{k}}\right]-1$, which algebraically reduces to $2^{l}m$. This means that the next $l$ terms of the Collatz sequence are all products of $m$ multiplied by powers of 2 terminating at $m$ and the total number of steps from $N$ to $m$ is $l+k$.

**Example: ** $k=3,l=2,d=47\ \to N=(8/27)(4\cdot 47+1)-1=55$. The Collatz sequence from 55 to 47 is: $55\to 83\to 125\to 188\to 94\to 47$. (5 steps)

**Remarks: ** To apply Theorem 1, it is required to find integers $k$, $l$ and $m$ that make $N$ odd. Also, it would be nice to show that there is an infinite number of values of $k$, $l$ and $m$ satisfying the hypotheses of the theorem, but that is not essential to the proof.

Here is a special case of Theorem 2:

**Corollary 2: **Let $k$ and $l$ be positive integers and let $N=\left(\frac{2}{3}\right)^{k}\left(2^{l}+1\right)-1$. If $N$ is an odd integer then $N$ reaches $1$ in $l+k$ steps.

**Proof: ** Let $m=1$ in Theorem 1.

**Examples:

$k=1,l=5\to N=\frac{2}{3}\left(2^{5}+1\right)-1=21.$** The Collatz sequence for 21 is: $21\to 32\to 16\to 8\to 4\to 2\to 1.$ (6 steps)

$k=2,l=9\to N=\left(\frac{2}{3}\right)^{2}\left(2^{9}+1\right)-1=227.$ The Collatz sequence for 227 is: $227\to 341\to 512\to 256\to 128\to 64\to 32\to 16\to 8\to 4\to 2\to 1$. (11 steps)

$k=3,l=9\to N=\left(\frac{2}{3}\right)^{3}\left(2^{9}+1\right)-1=151.$ The Collatz sequence for 151 is: $151\to 227\to 341\to 512\to 256\to 128\to 64\to\cdots\to 4\to 2\to 1.$ (12 steps)

We prove a result that will be useful later in the paper.

**Lemma 1: ** Let $n$ be a positive integer. Then $2^{(3^{n-1})}\equiv-1\mod{3^{n}}$.

**Proof: ** We will use mathematical induction  [4].

Case 1: If $n=1$, $2^{(3^{0})}=2\equiv-1\mod{3^{1}}$

Case 2: Assume the lemma is true for $n=k$; i.e., $2^{(3^{k-1})}\equiv-1\mod{3^{k}}$. We may write $2^{(3^{k-1})}=3^{k}q-1$ for some integer $q$. To show the equation is true for $n=k+1$, we cube both sides of this equation using the binomial theorem:

[TABLE]

So $2^{3^{k}}-(-1)$ is a multiple of $3^{k+1}$ and the formula is verified for $n=k+1$.

**Example: **For $n=4$, $2^{(3^{n-1})}=2^{27}=134,217,728\equiv 80\equiv-1\mod{81}$.

Theorem 3: Suppose $k$ is a positive integer and $l=3^{k-1}\cdot r$, where $r$ is an odd integer. If $N=\left(\frac{2}{3}\right)^{k}\left(2^{l}+1\right)-1$ then the Collatz sequence beginning with $N$ terminates at 1 after $l+k$ steps.

**Proof: ** By invoking Corollary 1, it suffices to show that $N$ is an odd integer under the given hypotheses. By Lemma 2, $2^{(3^{k-1})}\equiv-1\mod{3^{k}}$.

$2^{\left(3^{k-1}\right)}\equiv-1\mod{3^{k}}$

$\Rightarrow 2^{3^{\left(k-1\right)}m}\equiv\left(-1\right)^{m}\mod{3^{k}}$

$\Rightarrow 2^{l}+1\equiv\left(-1\right)^{m}+1\equiv 0\mod{3^{k}}$, since $m$ is an odd integer.

Because $2^{l}+1$ is a multiple of $3^{k}$, it follows that $N$ is an odd integer.

Example: $k=2,\ l=3^{1}\cdot 5=15\to N=\left(\frac{2}{3}\right)^{2}\left(2^{15}+1\right)-1=14563.$ The Collatz sequence for 14563 is: $14563\to 21845\to 32768\to 16384\to 8192\to 4096\to\cdots\to 8\to 4\to 2\to 1$ (17 steps)

Corollaries 3 and 4 provide special cases of starting values for Collatz sequences that satisfy the Collatz conjecture and for which we can provide the number of steps required to reach 1.

**Corollary 3: ** If $q$ is an integer and $N=\dfrac{4^{q}-1}{3}$, then the Collatz sequence starting with $N$ reaches 1 in $2q$ steps.

Proof: Let $k=1$ in Theorem 2 and simplify algebraically.

**Example: ** $q=4\to N=\dfrac{4^{4}-1}{3}=85$.

The Collatz sequence for 85 is: $85\to 128\to 64\to 32\to 16\to 8\to 4\to 2\to 1$. (8 steps).

**Corollary 4: ** If $r$ is an odd integer and $N=\dfrac{2^{3r+2}-5}{9}$, then the Collatz sequence starting with $N$ reaches 1 in $3r+2$ steps.

Proof: Let $k=2$ in Theorem 2 and simplify algebraically.

**Example: ** $r=3\to N=\dfrac{2^{11}-5}{9}=227$.

The Collatz sequence for 227 is: $227\to 341\to 512\to 256\to 128\to 64\to 32\to 16\to 8\to 4\to 2\to 1$. (11 steps)

We now derive some results which identifies some properties of general Collatz sequences. We begin by categorizing positive odd integers.

**Definitions: ** Every odd positive integer $N$ falls into one of three categories:

(i)   $N=6r+1,r=0,1,2,...,$ (call this type A) Examples: 1, 7, 13,…

(ii)   $N=6r+3,r=0,1,2,...,$ (call this type B) Examples: 3, 9, 15,…

(iii) $N=6r+5,r=0,1,2,...,$ (call this type C) Examples: 5, 11, 17,…

**Theorem 4: ** Every odd integer in a Collatz sequence except possibly the first one is of type A or C.

Proof: Let $N$ be an odd integer of type B in a Collatz sequence. Then there is an integer $r$ such that $N=6r+3$. Suppose the number in the Collatz sequence preceding $N$ in the sequence is the odd number $N^{\prime}$. Then by the Collatz algorithm, $\dfrac{3N^{\prime}+1}{2}=N=6r+3$. We then have $3N^{\prime}+1=12r+6$. Since 3 divides the right-hand side of this equation it must divide the left-hand side, which is impossible. We conclude that $N$ cannot be immediately preceded by an odd integer.

Therefore we can assume that either $N$ is at the start of the Collatz sequence or there is an integer $k$ such that the even numbers $2^{k}N,\ 2^{k-1}N,\ \cdots,\ 2N$ comprise the sequence before $N$. This implies that there is an odd number $N^{\prime}$ preceding $2^{k}N$. This means that $\dfrac{3N+1}{2}=2^{k}N=2^{k}(6r+3)=2^{k}3(2r+1)$. So $3N^{\prime}+1=3(2^{k+1})(2r+1)$. Since 3 divides the right hand side of this equation, it must divide the left side, which is a contradiction and the theorem is proved.

Remark: This means that no Collatz sequence contains an odd multiple of three; i.e., Type B except possibly at the start of the sequence.

**Example: ** The Collatz sequence for $N=9$: $9$(type B)$\to 14\to 7$(type A) $\to 11$(type C)$\to 17$(type C)$\to 26\to 13$(type A)$\to 20\to 10\to 5$(type C) $\to 8\to 4\to 2\to 1$(type A).

The following is a conjecture that expresses a relationship between an odd integer in a Collatz sequence and the next odd integer.

**Conjecture: ** Let $N$ be an odd integer in a Collatz sequence and let $N^{\prime}=\tfrac{3N+1}{2}$ be the next term in the sequence. If $N^{\prime}$ factors into $2^{l}m$ then $m$, the next odd integer in the sequence, is of type A if $l$ is odd and is of type C if $l$ is even.

**Examples:

$N=15\to N^{\prime}=23=2^{0}\cdot 23$**; $l=0$ (even) and $m=23$ (type C).

$N=117\to N^{\prime}=176=2^{4}\cdot 11$; $l=4$ (even) and $m=11$ (type C).

$N=49\to N^{\prime}=74=2^{1}\cdot 37$; $l=1$ (odd) and $m=37$ (type A).

$N=133\to N^{\prime}=200=2^{3}\cdot 25$; $l=3$ (odd) and $m=25$ (type A).

$N=341\to N^{\prime}=512=2^{9}\cdot 1$; $l=9$ (odd) and $m=1$ (type A).