Collatz-like problem involving prime factors Unfortunately I am not well-versed in LaTeX so I will try my best to keep this looking presentable.
As an overview, I was investigating a variation of the Collatz conjecture:
Define $f(1) = 1$
Then, if $n$ is even, $f(n) = \frac{n}{2}$
Otherwise, let $n$'s smallest prime factor be $p$, then $f(n) = pn+1$
So then I was trying it out for small values:
$f(1) = 1$
$f(2) = 1, f(1) = 1$
$f(3) = 10, f(10) = 5, f(5) = 26, f(26) = 13, ...$ and so on. Eventually you get to $f(213) = 640$ which reduces down to $10$. Evidently, loops/cycles can form.
$f(4) = 2, f(2) = 1$
$f(5) = 26$ which is already a part of $3$'s cycle
$f(6) = 3$ which is, again, part of $3$'s cycle
By now, I was thinking that this function had a nice trend of either reducing to $1$ or joining onto another number's cycle. However, $f(7)$ causes an issue. I have iterated for $f(7)$ around $100$ times using python, however it is still unclear whether $f(7)$ converges or not (for want of a better word).
I also noticed that convergence of $f(9)$ depended on convergence of $f(7)$.
There is no reason why it should converge, I saw on another page that the average smallest prime factor of integers up to $n$ is asymptotic to $\frac{n}{2\log(n)}$. So then $f(n)$ has average order $\frac{n^2}{2\log(n)}$.
My question then is ultimately: for any integer n, will $f(n)$ eventually reach 1 or form a cycle? And, if not, is there an interesting reason/proof why not? This question could be (and probably is) fairly difficult, so as a weaker question, does $f(7)$ converge?
If this question is similar to, or a corollary of, another question asked here, I apologise. Please let me know and I will remove the question.
Edit: I wrote some basic python code to do the computations for me, however, it is not very efficient. With that being said, I also have an interest in the computational complexity of this problem e.g. finding smallest prime factor, checking for repeated values (i.e. checking if a cycle has been formed) etc.
 A: Summary of new information that has been commented so far:

*

*The scale of which the iterations can grow really bogs down total computation time, the problem isn't quite prime factorisation but is a weaker form. In most cases, the numbers grow so large that computation beyond them is extremely slow (Thanks to Peter/Gottfried, it is confirmed that $f(7)$ goes as large as ~$10^{100}$)

*Thanks to Gottfried, we have a list of non-powers of two that also converges to 1. Namely $\small \begin{array} {rr} n & :&factorization \\ \hline    14941629968793 & : & 3.11.17.53.79.313.20323 \\   11206222476595 & : & 5.2241244495319 \\   109435766373 & : & 3^2.12159529597 \\   20519206195 & : & 5.2129.1927591 \\   200382873 & : & 3.19.3515489 \\   150287155 & : & 5.37.812363 \\   5733 & : & 3^2.7^2.13 \\   1075 & : & 5^2.43 \\   21 & : & 3.7 \\   1 & : & 1 \\   \end{array}$
Any of these numbers multiplied by a power of two will also converge. However, it appears that the smallest prime factor of each of the numbers in the table alternates between 3 and 5 (unknown whether its coincidental or not)


*Further exploration of cycles forming, including 1-cycles i.e powers of 2 and Gottfried's list, leads the conclusion to be that convergence onto a cycle is rare. Making divergence the average outcome, rather than an anomalous one.

Perhaps a better question to ask is: other than the cycle generated by 3, do other cycles even exist (not including the 1-cycle)?
