# $a_n$ is a positive integer for any $n\in \mathbb {N}$.

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.

• for $n=1$ we get $$a_2=(2+2+1)a_1-(1+1)a_0$$ but $a_0$ is not given! – Dr. Sonnhard Graubner Feb 4 '18 at 18:16
• @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. – rafa Feb 4 '18 at 18:24
• It feels like this should not be difficult, given that the sequence grows superexponentially, but somehow I can't make it work. – Patrick Stevens Feb 4 '18 at 21:51
• @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. – dxiv Feb 4 '18 at 22:39
• @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. $\frac{a_2}{a_1}\gt \frac{1853}{638}\approx 2.904$. – mathlove May 22 at 15:59

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}

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 $$c_n provided the denominator $$c_{n-1}+n^2-1>0$$.