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.
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...