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I found a conjecture of some mathematics teacher. Is this known or solved problem? Is there a non-constant function $f$ such that for any $n$, iterating $f(n)$ end to the sequence $1,1,1,1,1,\ldots$. By iteration I mean the similar way as in Collatz problem we end to the sequence $2,1,2,1,\ldots$. For example, does the function $f(n)=\begin{cases}n/3\text{ if }n\equiv 0\pmod 3 \\(4n-1)/3\text{ if }n\equiv 1\pmod 3\\(5n-7)/3\text{ if} n\equiv 2\pmod 3\end{cases}$ end to the sequence $\ldots, 1,1,1,\ldots$ This is from the Finnish mathematical journal from college students called Solmu, http://matematiikkalehtisolmu.fi/2017/1/otaksuma.pdf

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    $\begingroup$ You mean, like: $$f(n)=\begin{cases}n/2&n\text{ even}\\(n+1)/2&n\text{ odd}\end{cases}$$ $\endgroup$ – Thomas Andrews Mar 1 '17 at 17:44
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    $\begingroup$ You'll probably want to state some more conditions on $f$ than you have here -- as currently written the constant function that always returns $1$ will satisfy your criteria. $\endgroup$ – Henning Makholm Mar 1 '17 at 17:45
  • $\begingroup$ Can you understand the article you linked ? $\endgroup$ – A---B Mar 1 '17 at 17:49
  • $\begingroup$ @HenningMakholm Hmm. It looks like the author and me forgot to check the easiest possible solution. $\endgroup$ – Jaakko Seppälä Mar 1 '17 at 18:52
  • $\begingroup$ @user2219896: How about Thomas's example, then? Or $ f(n) = \max(1,n-1)$? $\endgroup$ – Henning Makholm Mar 1 '17 at 18:58

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