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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!

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closed as unclear what you're asking by Did, kingW3, Sahiba Arora, Xam, user251257 Jul 17 '17 at 0:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I think it's not so interesting. Collatz took already his place with his $3n+1$ conjecture. $\endgroup$ – Michael Rozenberg Jul 16 '17 at 7:59
  • $\begingroup$ 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.$"K≠1"$ $\endgroup$ – Elvin Jul 16 '17 at 8:05
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    $\begingroup$ 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? $\endgroup$ – José Carlos Santos Jul 16 '17 at 8:09
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    $\begingroup$ @Idontknow There are several things that I do not understand, even after your latest edition. $\endgroup$ – José Carlos Santos Jul 16 '17 at 9:01
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    $\begingroup$ @Idontknow I am sorry, but I do not have time for this right now. $\endgroup$ – José Carlos Santos Jul 16 '17 at 9:07
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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:


picture

source: MATHEMATICS OF COMPUTATION, VOLUME 32, NUMBER 144, Oct 1978
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.
But I think there was no substantial progress over this material of 1978... (it should then also be mentioned in the wikipedia, btw.)

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  • $\begingroup$ You say that, there are not $P$ and $Q$?? Always $f^{k}(n)=1$?? ($P≠3$ and $Q≠1$) $\endgroup$ – Elvin Jul 16 '17 at 15:10

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