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Let $(a_n)_{n\geq 1}$ be a sequence defined by $a_{n+1}=(2n^2+2n+1)a_n-(n^4+1 )a_{n-1} $.

$a_1=1$, $a_2=3$.

I have to show that $a_n $ is a positive integer for any $n\in \mathbb {N}, n\geq 1$.

I tried to prove it by induction but it doesn't work.

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    $\begingroup$ for $n=1$ we get $$a_2=(2+2+1)a_1-(1+1)a_0$$ but $a_0$ is not given! $\endgroup$ – Dr. Sonnhard Graubner Feb 4 '18 at 18:16
  • $\begingroup$ @Lord Shark the Unknown I tried to prove that $a_n> \frac {n^4+1}{2n^2+2n+1}a_{n-1}$ but the induction hypothesis falls. $\endgroup$ – rafa Feb 4 '18 at 18:24
  • $\begingroup$ It feels like this should not be difficult, given that the sequence grows superexponentially, but somehow I can't make it work. $\endgroup$ – Patrick Stevens Feb 4 '18 at 21:51
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    $\begingroup$ @PatrickStevens One difficulty here is that the recurrence is very sensitive on the initial conditions. Changing $a_2=3$ to $a_2=2.5$ for example makes it go negative pretty quickly. $\endgroup$ – dxiv Feb 4 '18 at 22:39
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    $\begingroup$ @dxiv: Using the inequality $(1)$ in my answer, we see that, in order for $a_n$ to be a positive integer for all $n\ge 1$, it is sufficient that $a_1,a_2$ are positive integers such that $a_3\gt 0$ and $a_4\gt 0$ and $22\lt\frac{a_5}{a_4}\lt 28$, i.e. $\frac{a_2}{a_1}\gt \frac{1853}{638}\approx 2.904$. $\endgroup$ – mathlove May 22 at 15:59
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It can be proven that for $n\ge 4$, $$n^2+\frac 32n\lt \frac{a_{n+1}}{a_n}\lt n^2+3n\tag1$$

From $(1)$, we have, for $n\ge 4$, $$\frac{a_{n+1}}{a_n}\gt 0$$

Since we have $$a_1=1,\quad a_2=3,\quad a_3=22,\quad a_4=304,\quad a_5=6810$$ it follows that $a_n$ is a positive integer for all $n\ge 1$.


Now, let us prove that $(1)$ holds for $n\ge 4$.

Proof :

Let us prove $(1)$ by induction on $n$.

We see that $(1)$ holds for $n=4$ since $$n^2+\frac 32n=22,\quad \frac{a_{n+1}}{a_n}=22+\frac{61}{152},\quad n^2+3n=28$$

Suppose that $(1)$ holds for some $n\ (\ge 4)$.

Then, we have $$\begin{align}&\frac{a_{n+2}}{a_{n+1}}-\left((n+1)^2+\frac 32(n+1)\right) \\\\&=2(n+1)^2+2(n+1)+1-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)-\left((n+1)^2+\frac 32(n+1)\right) \\\\&\gt 2(n+1)^2+2(n+1)+1-\frac{(n+1)^4+1}{n^2+\frac 32n}-\left((n+1)^2+\frac 32(n+1)\right) \\\\&=\frac{n^2 - n - 8}{2 n (2 n + 3)} \\\\&\gt 0\end{align}$$

and

$$\begin{align}&(n+1)^2+3(n+1)-\frac{a_{n+2}}{a_{n+1}} \\\\&=(n+1)^2+3(n+1)-\left((2(n+1)^2+2(n+1)+1)-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)\right) \\\\&\gt (n+1)^2+3(n+1)-(2(n+1)^2+2(n+1)+1)+\frac{(n+1)^4+1}{n^2+3n} \\\\&=\frac{2 n^2 + n + 2}{n (n + 3)} \\\\&\gt 0\qquad\blacksquare\end{align}$$

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This answer provides partial results.

For each $n$ put $b_n=\frac {a_{n+1}}{a_n}$. Then $b_1=3$ and

$$b_n=2n^2+2n+1-\frac{n^4+1}{b_{n-1}}.$$

It suffices to prove that $b_n>0$. Computer evaluation suggests that a sequence $\{c_n=b_n-n^2-2n\}$ decreases and converges to about $-5.78734$. We have $c_1=0$ and

$$c_n=n^2+1-\frac{n^4+1}{c_{n-1}+n^2-1}= \frac{c_{n-1}(n^2+1)-2}{c_{n-1}+n^2-1}.$$

It seems that we can show that $c_n<c_{n-1}$ provided the denominator $ c_{n-1}+n^2-1>0$.

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