preprints.org > computer science and mathematics > algebra and number theory > doi: 10.20944/preprints202308.0106.v1

Preprint Communication Version 1 Preserved in Portico This version is not peer-reviewed

Version 1 : Received: 25 July 2023 / Approved: 1 August 2023 / Online: 2 August 2023 (08:01:29 CEST)

Starting with a positive odd integer $n$, apply a function $f(n)$ repetitively such that $f(n) = 3n + 1$ if $n \not\equiv 0 (\text{mod}\:2)$ or $f(n) = n/2$ if $n \equiv 0 (\text{mod}\:2)$. Let the sequence of all the odd integers obtained form an orbit $n, f^1(n) , f^2(n), \ldots, f^k(n), f^{k + 1}(n)$. The $k^{th}$ odd integer is written as $f^k(n) = (3^k + \mathbf{B(k)})/2^{\mathbf{z(k)}}$ where $\mathbf{B(k)}$ is the collection of terms without the coefficient $n$. We show that when $1 - \mathbf{B(k)}/(2^{\mathbf{z(k)}} n) \approx 1$, the condition for unbounded orbit is given by OEIS A022921. Further, if there exists at least one orbit for which $f^{k + 1}(n) = n$ when $(k + 1) < 3$, we show that there exists no orbit with $k + 1 > 3$ for which $f^{k + 1}(n) = n$. The final result is that the odd integers that can violate the Collatz conjecture are smaller than 5.

Collatz conjecture; 3n+1 problem; inequality relations

Computer Science and Mathematics, Algebra and Number Theory

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