# Give me such a $“3n±Q”$ problem that we do not know a counter-example [closed]

Maybe, This question is stupid.But, I want to ask.Because, really I dont Know answer.This problem may be similar to others. My Question is:

$$f(n) = \begin{cases} Pn±Q & \text {if n is odd} \\ \frac{n}{2} & \text {if n is even} \end{cases} ,$$ and we can find such $k$ $$f^{k}(n)=1$$ Here $P,Q\in \mathbb{N}$

For Example: We know counter examples, for $"3n-5","3n-1",3n+5"$ problems. So that $f^{k}(n)≠1$

Give me such a problem that we do not have a counter-example,(In shortly $f^{k}(n)=1$)

If the example you give is $"3n ± Q"$, and "$Q≠1"$ it was very good.

(Please, edit or improve question for me,because Unfortunately, I'm not as knowledgeable as you.) Thanks so much!

## closed as unclear what you're asking by Did, kingW3, Sahiba Arora, Xam, user251257Jul 17 '17 at 0:09

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• I think it's not so interesting. Collatz took already his place with his $3n+1$ conjecture. – Michael Rozenberg Jul 16 '17 at 7:59
• It is not my question. I want to know , is there a problem like a collatz conjecture ("$3n+K"$) ,which that we dont know counter-examples.$"K≠1"$ – Elvin Jul 16 '17 at 8:05
• You wrote that “We all know that,There is no counter-examples for Collatz Conjecture”. Actually, if we all knew that, then it wouldn't be a conjecture anymore, right? – José Carlos Santos Jul 16 '17 at 8:09
• @Idontknow There are several things that I do not understand, even after your latest edition. – José Carlos Santos Jul 16 '17 at 9:01
• @Idontknow I am sorry, but I do not have time for this right now. – José Carlos Santos Jul 16 '17 at 9:07

Surely the article "On the '3x+1'-Problem" of R.E.Crandall of 1978 fits your question well (it is online you can find it). Here is a screenshot of a part of that article dealing with the general $qx+r$-problem-variant:
As an interesting sidenote: I've found a second cycle with the $q=181$ - problem. And also the $q=3511$ (having a similar property as $q=1093$ being a wieferich prime) should have been looked at.
• You say that, there are not $P$ and $Q$?? Always $f^{k}(n)=1$?? ($P≠3$ and $Q≠1$) – Elvin Jul 16 '17 at 15:10