Let $(a_n)$ be the A001921 sequence
$$ a_0 := 0,\ a_1 := 7, \quad a_{n+2} = 14a_{n+1} - a_n + 6. $$
Is it true that $a_n$ is always a composite integer for any $n\geq 2$ ?
UPDATE : I now make a much stronger conjecture : if we define $b_k$ as the gcd of all the integers $a_{2^kn+2^{k-1}-1}(n\geq 0)$, then $(b_k)_{k\geq 1}$ is increasing (numerical values suggest it grows very very fast, see below). For example :
$a_{2n}$ is always divisible by $b_0=2$.
$a_{4n+1}$ is always divisible by $b_1=7$.
$a_{8n+3}$ is always divisible by $b_2=97$.
$a_{16n+7}$ is always divisible by $b_3=607$.
$a_{32n+15}$ is always divisible by $b_4=708158977$.
$a_{64n+31}$ is always divisible by $b_5=1002978273411373057$.
$a_{128n+63}$ is always divisible by $b_6=2011930833870518011412817828051050497$.
This question has been answered completely on MO.
To summarize, the conjecture is true because if we put $\alpha=2-\sqrt{3}$ and $\beta=2+\sqrt{3}$, then $a_n=u_nv_{n+1}$ where $(u_n)$ and $(v_n)$ are the integer sequences defined by
$$ u_n=\frac{\beta^n-\alpha^n}{\beta-\alpha}, \ v_n=\frac{\alpha^{n}+\beta^{n}}{2} $$
and $b_n=v_{2^n}$.