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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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


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


This answer provides partial results.

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


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$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.