Let $(a\_n)$ be the [A001921](http://oeis.org/A001921) sequence 


$$
 a\_0 = 0,\ a\_1 = 7, \quad a\_{n+2} = 14a\_{n+1} - a\_n + 6.
$$


Let $(b\_k)$ be the (almost)"tower-of-squares" sequence defined by


$$
 b\_0=2, \quad b\_{k+1}=2b\_k^2-1
$$


Is it true that $a\_{2^kn+2^{k-1}-1}$ is always divisible by
 $b\_k$, for any $k,n\geq 0$ ?


I have checked this up to $k=6$. 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 $b\_2=97$.
* Etc. Up to : $a\_{128n+65}$ is always divisible by $b\_6=2011930833870518011412817828051050497$.


This is a cross-post from a MSE [question](https://math.stackexchange.com/questions/1002792/does-the-a001921-linear-recurrent-integer-sequence-always-yield-composite-number).



Tower-of-squares sequence divides linear recurrent A001921 sequence?
The elements of your sequence are
$$a\_n=\left(\frac{\alpha^n-\beta^n}{2\sqrt{3}}\right)\left(\frac{\alpha^{n+1}+\beta^{n+1}}{2}\right)$$
where $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$. Notice that both factors are integers. We can also compute that
$$b\_n=\frac{\alpha^{2^n}+\beta^{2^n}}{2}.$$
Now your statement that $b\_k$ always divides $a\_{2^{k+1}n+2^k-1}$ follows from the fact that
$\frac{\alpha^{2^k}+\beta^{2^k}}{2}$ divides $\frac{\alpha^{2^k(2n+1)}+\beta^{2^k(2n+1)}}{2}$, which is very easy to check (the ratio is a polynomial in $\alpha, \beta$ and symmetric in both).


