Prime 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
- Has this specific prime-factor-reduction and prime-jumping variant been formally studied in number theory?
- 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)
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 $2p-1$.