Let k ≥ 2 be an even integer. Consider the iterates x, t(x), t(t(x)),... of the mapping
\[
t(x)=\left\{\begin{array}{cl}
(k+1)x+1 & \mbox{ if $x$ is odd,}\\
\left\lceil x/k\right\rceil & \mbox{ if $x$ is even,}
\end{array}
\right.
\]
where \(\lceil\quad\rceil\) denotes the ceiling function.
Equivalently
\[
t(x)=\left\{\begin{array}{cl}
(k+1)x+1 & \mbox{ if $x$ is odd,}\\
x/k& \mbox{ if \(x\equiv0\pmod{k}\),}\\
\left\lfloor x/k\right\rfloor+1 & \mbox{ if \(x\) is even and \(x\not\equiv0\pmod{k}\),}
\end{array}
\right.
\]
where \(\lfloor\quad\rfloor\) denotes the floor function.
All trajectories are conjectured by Pruthviraj Hajari to eventually reach 1 if x > 0. (See MathOverflow question.)
It seems certain that if x ≠ 0, then tn(x) will eventually reach one of the cycles 0 → 0, or 1 → k + 2 → 2 → 1, or -(2t + 1) → -2t(k + 1) - k → -(2t + 1), where 0 ≤ t ≤ k/2 - 1.
Last modified 12th December 2020
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