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The following is a problem we were unable to prove and left stated in the paper "Arithmetical properties of a sequence arising from an arctangent sum", J. Numb. Theory 128 (2008) 1807–1846.

Define the sequence $\{x_n\}$ by $x_1=1$ and for $n\geq2$ by $$x_n=\frac{x_{n-1}+n}{1-nx_{n-1}}.$$ We believe and claim that $x_n$ is not an integer, for $n\geq5$. Any ideas?

Note: $x_1=1, x_2=-3, x_3=0, x_4=4$. Among other things, we proved $x_n\neq0$ for $n\geq4$.

Special case: Since the problem is a bit hard and in view of no success being reported, I propose the following special case to attend to: prove (or disprove, which I doubt) that $$x_n\neq 1 \qquad \text{for any $n\geq2$}.$$

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  • $\begingroup$ Some trivialities: So if one proves that for integers larger than four $x_n$ is non-integer, then that yields the statement in "... then $x_n\not= 0$" statement. $\endgroup$
    – BigM
    Sep 18, 2016 at 3:40
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    $\begingroup$ Oh, yes. It becomes a corollary. $\endgroup$ Sep 18, 2016 at 3:43
  • $\begingroup$ For the special case,suppose there exists $n_0 \ge 2$ s.t. $x_{n_0}=1$ then $x_{n_0-1}=(1-n_0)/(1+n_0)$; now you need to find an equation for $x_{n_0-k}$ by iterating the recursive relation. Then you need to find for which values $x_{n_0-k} = x_1=1$, and then search for a contradiction. BTW, the denominator of your recursive relation should be well defined, i.e. it's not defined for: $x_n = 1/(n+1)$. $\endgroup$
    – Alan
    Sep 24, 2016 at 7:16
  • $\begingroup$ In the above-mentioned paper, it is proved that $1-nx_{n-1}\neq0$ for $n>1$. So, $x_n$ is well-defined, no worries. $\endgroup$ Sep 24, 2016 at 14:15
  • $\begingroup$ (Title: maybe you mean Riccati?) $\endgroup$ Apr 12, 2017 at 0:38

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