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Prime collatz-conjecture

数论 Math StackExchange 8 票 1 回答 325 浏览 提问者: 黃曦永 2026-07-06 06:21
sequences-and-series number-theory collatz-conjecture

问题内容

I would like to propose a prime-based variant of the Collatz conjecture that I came up with. I am interested to know if this specific variation has been studied before, or if there are any known counterexamples to the behavior I observed.

Definition of the Function

Let $n > 1$ be a positive integer. We define $p_{min}(n)$ as the smallest prime factor of $n$. If $n$ itself is a prime number, let $p_{next}(n)$ denote the next immediate prime number greater than $n$.

We define the transition function $f(n)$ as follows: $$f(n) = \begin{cases} \frac{n}{p_{min}(n)} & \text{if } n \text{ is composite} \\ p_{next}(n) + 1 & \text{if } n \text{ is prime} \end{cases}$$

Examples of Sequences

  • Starting with $n = 7$ (Prime):

    • $7$ is prime $\implies f(7) = 11 + 1 = 12$
    • $12$ is composite $\implies f(12) = 12 / 2 = 6$
    • $6$ is composite $\implies f(6) = 6 / 2 = 3$
    • $3$ is prime $\implies f(3) = 5 + 1 = 6$
    • This leads to the repeating cycle: $6 \to 3 \to 6 \to 3 \dots$
  • Starting with $n = 2$ (Prime):

    • $2$ is prime $\implies f(2) = 3 + 1 = 4$
    • $4$ is composite $\implies f(4) = 4 / 2 = 2$
    • This leads to a small cycle: $2 \to 4 \to 2 \to 4 \dots$ (which eventually connects to $4 \to 2$ forever).

My Conjecture

Conjecture: For any starting positive integer $n > 1$, the sequence defined by repeated applications of $f(n)$ will always eventually enter a finite periodic cycle. Specifically, based on my initial testing, it seems to always fall into the $\{3, 6\}$ cycle or the $\{2, 4\}$ cycle.

My Questions

  1. Has this specific prime-factor-reduction and prime-jumping variant been formally studied in number theory?
  2. Are there any known starting values of $n$ that escape to infinity, or enter a much larger/different cycle than the ones mentioned above?

Any insights, computer search data, or theoretical proofs regarding the bounds of this function would be highly appreciated!

回答 (1)

dialegou 13 票 2026-07-06 06:35 原文

There is the following fact: There is always a prime number strictly between $n$ and $2n$ for every integer greater than $1$. This means for your map that $f(f(n))\leq n$. Thus, every every starting value will end in a finite periodic cycle. And every cycle has length $2$ and the smaller values of such a cylcle are exactly those primes $p$ for which the next prime number is $2p-1$.