# All terms of this sequence are equal to 1: $x_{n+1}= \frac{nx_{n}^2+1}{n+1}$

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.

• 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. – Ross Millikan Jul 1 '17 at 23:59
• Now posted also on MathOverflow: If $x_{n+1}= \frac{nx_{n}^2+1}{n+1}$ then $x_{n}=1$ – Martin Sleziak Aug 12 '17 at 14:34
• 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 $x_1=1$). – Giuseppe Negro Aug 12 '17 at 19:14
• 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. ... – Vincenzo Oliva Aug 13 '17 at 11:00
• ... 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. – Vincenzo Oliva Aug 13 '17 at 11:01

It does not. The first term (whether $n=0$ or $n=1$) could be $-1$ and all others would be $1$
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 $0$.
• 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. – Giuseppe Negro Aug 12 '17 at 18:54
• 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? – jnyan Aug 13 '17 at 4:35
• 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. – Giuseppe Negro Aug 13 '17 at 8:42