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Define $f(n) = \lfloor {ne}\rfloor$ if $n$ is odd and $f(n) = \lfloor {n/e}\rfloor$ if $n$ is even. Is the set $\{n, f(n), f(f(n)),\dots\}$ bounded for every $n$?

Computer sampling suggests that each such set is finite - indeed, that the iterates reach 0 - but that the set of cardinalities of the sets is infinite.

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  • $\begingroup$ just curious: how far ($n$-wise) did you test? How much precision for $e$ did you use? $\endgroup$ Commented May 11, 2015 at 20:15
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    $\begingroup$ It should be noted that Clark asked a question about these sequences when $e$ is replaced by an arbitrary real numbers: mathoverflow.net/questions/204764/… Not that that makes the current question uninteresting, but I thought it would be useful to provide a link. $\endgroup$ Commented May 11, 2015 at 20:26
  • $\begingroup$ Yaakov - Mathematica checked for $n$ up $50000$ using "MaxExtraPrecision = Infinity". Here's the set for $n=5$: ${5,13,35,95,258,94,34,12,4,1,2,0}$ $\endgroup$ Commented May 11, 2015 at 20:29

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Write the iteration as $x(j+1) = f(x(j))$ with $x(0) = n$. Heuristically, while $x(j)$ is large, $\log x(j)$ undergoes a "random walk" with a slight bias to the left; with probability $1$ you eventually get to a small enough value of $x(j)$, and then you're trapped in a fixed point or cycle. This heuristic should not be taken too seriously, but it does suggest that the set should always be finite, although sometimes very large.

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