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Does there exist a function $f:\Bbb N\to\Bbb N$ such that \begin{align}a_{n+1}&=f(a_n)\\a_{f(n)+1}&=a_n\end{align} implies $\{a_n\}_{n\ge0}$ is a non-constant, positive integer sequence? Evidently $f$ cannot be monotonic.

If there is such a function, can all such $f$ be classified?

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  • $\begingroup$ I'm assuming you also want such a sequence $a_n$ to exist - otherwise the implication is vacuously true. $\endgroup$
    – Wojowu
    Commented May 10, 2023 at 22:27
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    $\begingroup$ $f$ defined by $f(0)=1,f(2k-1)=2k,f(2k)=2k-1$ for all $k>0$ seems to work. I have no idea about classification. $\endgroup$
    – Wojowu
    Commented May 10, 2023 at 23:00
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    $\begingroup$ If there is a fixpoint $f(m)=m$ then $f(a_m) = a_{m+1} = a_{f(m)+1} = a_m$ and the sequence has a non-constant prefix followed by a constant tail. $\endgroup$
    – Peter Taylor
    Commented May 10, 2023 at 23:49

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