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Does anybody know if the following result is true?

Let $x_{n}$ be a sequence, such that $x_{n+1}= \dfrac{nx_{n}^2+1}{n+1}$ and $x_n>0$ for all $n$.
There is a positive integer $N$ such that $x_n$ is integer for all $n>N$.
Does it follow that $x_n=1$ for all positive integers $n$?

I tried to prove that $x_1 \equiv 1 \text{(mod p)}$ for all prime numbers $p$ but I couldn't make any further progress.

I'm looking for a proof or any reference of this result.
Any help would be appreciated.

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  • $\begingroup$ If it ever equals $1$ it always equals $1$. Once the integers kick in, you get $n+1|x_n^2-1$ but I can't find a contradiction. $\endgroup$ – Ross Millikan Jul 1 '17 at 23:59
  • $\begingroup$ Now posted also on MathOverflow: If $x_{n+1}= \frac{nx_{n}^2+1}{n+1}$ then $x_{n}=1$ $\endgroup$ – Martin Sleziak Aug 12 '17 at 14:34
  • $\begingroup$ By experimenting it seems to me that initial conditions $x_1\in (0, 1)$ produce $\lim_{n\to \infty} x_n=0$, while $x_1\in (1, \infty)$ produces $\lim_{n\to \infty} x_n=\infty$. The blow up rate seems to be exp-exp, that is, (I conjecture that) $$x_n= 2^{2^{An +B}} + o(1), $$ but I do not know the constants $A, B$. (Hope this helps. My plan was to compute $A, B$ and show that they are never integer, except for $x_1=1$). $\endgroup$ – Giuseppe Negro Aug 12 '17 at 19:14
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    $\begingroup$ Here's what I got. If $x_0$ is defined then $x_1=1$ and $x_n=1$ for all positive $n$, so for a contradiction let's take $x_1\ne1$ as the first term. We have $$x_{n+1}-x_n=\frac{nx_n^2-nx_n+1-x_n}{n+1}=\frac{(x_n-1)(nx_n-1)}{n+1}=_{n\ge2}\frac{(n-1)x_{n-1}^2(x_n-1)}{n+1}. \tag{1}$$Putting $n=1$ gives $x_2-x_1=\dfrac{(x_1-1)^2}{2}>0;$ by $(1),$ $x_2<1$ would imply by induction $x_n<1$ for all larger $n$, and since we know $x_n>0$, this contradicts $x_n$ eventually being integer. So $x_2>1$, and by $(1)$ $x_n$ is strictly increasing. ... $\endgroup$ – Vincenzo Oliva Aug 13 '17 at 11:00
  • $\begingroup$ ... As long as $x_n\in\Bbb Z^+$ for $n>N$, these $n$ must actually satisfy $x_{n+1}-x_n\ge1$, hence $x_n\to+\infty$. In fact, we thus get $$\frac{x_{n+1}}{x_n^2}=\frac{n+x_n^{-2}}{n+1}\to1 \iff x_{n+1}\sim x_n^2,$$therefore for $n>N$ one finds $x_n=a^{2^{n+b}} + y_n$, where $a,b$ are fixed integers and $y_n=o(a^{2^n})$ is a sequence of integers. $\endgroup$ – Vincenzo Oliva Aug 13 '17 at 11:01
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It does not. The first term (whether $n=0$ or $n=1$) could be $-1$ and all others would be $1$

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  • $\begingroup$ Thanks, but there was a typo in my question. I've just edited it. $\endgroup$ – jack Jul 1 '17 at 15:27
  • $\begingroup$ @jack Well, according to the sites rules, and I tell you in good faith, you can't do that and expect answerers to correct their answer, they have the right to ignore you. Still, is ok to try to tell them. $\endgroup$ – Santropedro Jul 1 '17 at 15:46
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Taking first term as $a$, every $ n$ th can be written as $(a^2 + n-1)/n$. If you want every term after certain terms to be integer, then only value $a$ can take is $1$ because of the divisibility, given the restriction that first term is greater than $ 0$.

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  • $\begingroup$ I think this is wrong (I made the same mistake). If $x_1=a$ then $x_3=\frac{ \frac{a^2}{2} + a + \frac32}{3}$, which already differs from your formula. I think that the recursion in the OP has no closed form solution. $\endgroup$ – Giuseppe Negro Aug 12 '17 at 18:54
  • $\begingroup$ I rechecked it. its correct. second term will be $a^2+1/2$. so third term will be $a^2+2/3$. unless I have made a huge mistake. can i know what you think would be the second term? $\endgroup$ – jnyan Aug 13 '17 at 4:35
  • $\begingroup$ Of course. If $x_2=\frac{a^2+1}{2}$ then $$x_3=\frac{2 \left( \frac{a^2+1}{2}\right)^2+1}{3}.$$ You are forgetting to take the 2nd power, I think. $\endgroup$ – Giuseppe Negro Aug 13 '17 at 8:42
  • $\begingroup$ Thank you. A huge mistake indeed. $\endgroup$ – jnyan Aug 14 '17 at 9:33

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