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When analyzing the collatz conjecture in reverse. I came across this argument that attempts to show that an integer $4x+1$ resolves in $n$ steps if $x$ is odd and resolves in $n$ steps and likewise if $x$ is odd an resolves in $n$ steps, then $4x+1$ likewise resolves in $n$ steps.

Is my reasoning correct?

Let:

  • $\nu_2(x)$ be the 2-adic valuation of $x$
  • $x_1, x_2, \dots, x_n$ be $n$ distinct odd integers such that:
    • $x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$
  • $x_1$ resolves in $n$ steps if $\dfrac{3x_n+1}{2^{\nu_2(3x_n + 1)}} = 1$

Observation 1: if an odd integer $x$ resolves to $1$ in $n$ steps, then there exist $n$ positive integers $a_1, a_2, \dots, a_n$ where $x = f(a_1, a_2, \dots, a_n)$

(1) If $n=1$, then $\dfrac{3x+1}{2^{a_1}} = 1$ and $x = f(a_1) = \dfrac{2^{a_1}-1}{3}$

(2) Assume that for any $x$ that resolves in $n-1$ steps, there exist $n-1$ integers $a_1, \dots, a_{n-1}$ with $f_{n-1}(a_1, a_2, \dots, a_{n-1})= x$

(3) If $x$ resolves in $n$ steps, there exists a positive integer $a_n$ such that $\dfrac{3x+1}{2^{a_n}}$ is an integer that resolves in $n-1$ steps and $x = f(a_1, a_2, \dots, a_n) = \dfrac{3f_{n-1}(a_1, a_2, \dots, a_{n-1}) + 1}{2^{a_n}}$

Observation 2: if an odd integer $x$ resolves in $n$ steps, then it follows that $4x+1$ also resolves in $n$ steps

(1) From Observation 1, there exist $a_1, a_2, \dots, a_n$ such that $\dfrac{3x+1}{2^{a_n}} = f_{n-1}(a_1, \dots, a_{n-1})$

(2) $\dfrac{3(4x+1)+1}{2^{a_n+2}} = \dfrac{12x+4}{2^{a_n+2}} = \dfrac{4(3x+1)}{2^{a_n+2}} = f_{n-1}(a_1, \dots, a_{n-1})$

Corollary: If $x$ is a positive odd integer and $4x+1$ is a solution in $n$ steps, then $x$ is also a solution in $n$ steps:

By assumption, there exist $a_1, a_2, \dots, a_n$ such that:

$$\frac{3(4x+1)+1}{2^{a_n}} = \frac{4(3x+1)}{2^{a_n}} = \frac{3x+1}{2^{a_n-2}}= f_{n-1}(a_1, a_2, \dots, a_{n-1})$$

Update 1: corrected an ambiguity -- my definitions only apply to odd numbers.

Update 2: I wrote a simple java app to validate. I have been able to validate this result up to $90\times10^6$.

My code iterates through odd numbers $x$ and validates that the count of steps for $x$ matches the count of steps for $4x+1$.

Larry Freeman
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    Without weighing in on the details of the argument (I'm so out of touch on details) I would wonder if this conjecture has been attempted to be verified empirically for at least some range of $x$, and if so, how large that range would be. – leslie townes Feb 21 '21 at 18:19
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    @leslietownes I confirmed it up to $90\times10^6$. Thanks very much for the suggestion. – Larry Freeman Feb 21 '21 at 19:17
  • Cool! I just love anything empirical before I worry about formulas. Even if being empirical is dumber than formulas and goes nowhere, I feel like there's no substitute for the "real thing." – leslie townes Feb 21 '21 at 19:22

1 Answers1

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If $x$ is odd then after one (×3+1) step we have $3x+1$. We then complete the op's step by applying $k$ (/2) steps. Where $k$ is the largest integer such that $2^k|3x+1$. $4x+1$ is also odd so after one (×3+1) step we have $12x+4$ then if we apply two (/2) steps we get $3x+1$ then we divide by $2^k$ and get to the same number. So after one of the op's steps $x$ and $4x+1$ arrive at the same number $\frac{3x+1}{2^k}$.

quantus14
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