Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$? Numerical evidence suggests the following.
For $c \in \mathbb{N}, c > 2$ define the sequence $a_n$ by
$a_0=0,a_1=1, \; a_n=c a_{n-1} - a_{n-2}$
For $ 5 < n < 500, \; 2 < c < 100$ there are no primes in $a_n$ though
semiprimes exist.

Is it true that $a_n$ is always composite for $n > 5$
If yes is there explicit partial factorization?

Searching OEIS solved the case $c=6$
with a Pell equation.
Counterexamples are welcome.

Put $u = (c + \sqrt{c^2-4})/2$. We have
$$a_{2n} = \frac{u^{2n}-u^{-2n}}{u-u^{-1}} = \left( \frac{ u^n-u^{-n}}{u-u^{-1}} \right) \left( \vphantom{\frac{ u^n-u^{-n}}{u-u^{-1}}} u^n + u^{-n} \right)$$
$$a_{2n+1} = \frac{u^{2n+1}-u^{-2n-1}}{u-u^{-1}} = \left( \frac{u^{n+1/2}-u^{-2n-1}}{u^{1/2}-u^{-1/2}} \right) \left( \frac{u^{n+1/2}+u^{-n-1/2}}{u^{1/2}+u^{-1/2}} \right) =$$
$$\left( \frac{u^{n+1/2}-u^{-n-1/2}}{u^{1/2}-u^{-1/2}} \right) \left( \frac{u^{n+1/2}+u^{-n-1/2}}{u^{1/2}+u^{-1/2}} \right)=\left( \frac{u^{n+1}-u^{-n}}{u-1} \right) \left( \frac{u^{n+1}+u^{-n}}{u+1} \right).$$
Put  $v_{n} = u^n + u^{-n}$, $x_{n} =  \frac{u^{n+1}-u^{-n}}{u-1} $, $y_{n}=\frac{u^{n+1}+u^{-n}}{u+1} $ so $a_{2n} = a_n v_n$ and $a_{2n+1} = x_n y_n$. I claim that each of  $v$, $x$ and $y$ are integer valued sequences which are greater than $1$ for $n \geq 2$, thus proving the claim.
The fastest way to see this is to note the recursions:
$$v_n = c v_{n-1} - v_{n-2} \quad v_0=2 \quad v_1 =c $$
$$x_n = c x_{n-1} - x_{n-2} \quad x_0=1 \quad x_1 =c+1 $$
$$y_n = c y_{n-1} - y_{n-2} \quad y_0=1 \quad y_1 =c-1 $$
A more conceptual way to see the rationality is to note that the Galois symmetry $\sqrt{c^2-4} \mapsto - \sqrt{c^2-4}$ takes $u$ to $u^{-1}$ and takes each of $v_n$, $x_n$ and $y_n$ to themselves.
