Proving that the sequence $a_{n + 2} = -2 \left( 2n^2 + 1 \right) a_n - a_{n - 2}$ is unbounded I came across this recurrence relation
$a_{n + 2} = -2 \left( 2n^2 + 1 \right) a_n - a_{n - 2}$
with $a_0, a_1 \in \mathbb{R} \setminus \left\lbrace 0 \right\rbrace$ to be taken arbitrary and $a_2 = - a_0, a_3 = -7a_1$. Then, from the very first terms of the recurrence relation, I can say that the sequence defined will be unbounded. However, I am unable to find a analytic proof for the same.
My idea is to just check the even subsequence and prove it to be unbounded so that the overall sequence also becomes unbounded.
 A: The odd and the even terms constitute independent sequences deserving a separate analysis.
Let us consider the even case.
The given relationship with $n=2m$ becomes 
$$a_{2m+2}=-2(8m^2+1)a_{2m}-a_{2m-2}$$
Setting $u_m:=a_{2m}$, we get :
$$u_{m+1}=-2(8m^2+1)u_{m}-u_{m-1}\tag{1}$$
Said otherwise :
$$\dfrac12 (u_{m+1}+u_{m-1})=-(8m^2+1)u_{m}\tag{2}$$
As the LHS of (2) is the midpoint of $u_{m+1}$ and $u_{m-1}$, by an immediate recurrence, we see the sequence $u_m$ is alternate. Taking $v_n:=(-1)^n u_n=|u_n|$, we get the equivalent recursion with all positive terms :
$$v_{m+1}=2(8m^2+1)v_{m}-v_{m-1}, \ \text{with} \ v_0=v_1=a_0$$
or
$$v_{m+1}=(16m^2+1)v_{m}+(v_m-v_{m-1})\tag{3}$$
Set apart the first term, (3) shows, here also by an immediate recurrence, that sequence $(v_n)$ is increasing. 
Two cases : either this sequence converges to a limit $L<+\infty$, either it is unbounded. The first hypothesis cannot hold when we take $n\to \infty$ in (3) : we would get $L$ on the LHS and $\infty$ on the RHS...
A: What I can think is to express $a_n$ in terms of $a_0$. It should be the form $a_n = f(n)\cdot a_0$. If $f(n)$ is unbounded, then we are done.
The most complicated thing is to derive $f(n)$, though...
