Transforming a recurrence to the product of two other recurrences This question deals with sequence:
$$a_0 = 0,\ a_1 = 7, \quad a_{n+2} = 14a_{n+1} - a_n + 6$$
The question is about establishing that $a_n$ is a composite number (except some finite cases).
In one of the comments to @Gjergji Zaimi's answer, @joro said:

So $a_n$ is the product of two lower order recurrences.

I am not clear on how this conclusion is reached. Also, I would like to see some recurrent formula in different form than original, that illustrates this.
This is interesting to me since I work on some other problems involving primality, and understanding this approach may greately help me.
I appreciate your help.
 A: From Gjergji Zaimi's answer:
$$ a_n=\left(\frac{\alpha^n-\beta^n}{2\sqrt{3}}\right)\left(\frac{\alpha^{n+1}+\beta^{n+1}}{2}\right)$$.
Both factors are integers and satisfy binary recurrences.
Experimentally the first is https://oeis.org/A001353 and the second is https://oeis.org/A001075.
Numerical data supports this.
A: Look, I want to use the most standard approach: we have 
$$
X_{n+1}=\begin{pmatrix}
a_{n+2}
\\
a_{n+1}
\end{pmatrix}
=
\begin{pmatrix}
14&-1
\\
1&0
\end{pmatrix}
\begin{pmatrix}
a_{n+1}
\\
a_{n}
\end{pmatrix}
+\begin{pmatrix}
6
\\
0
\end{pmatrix}
=\mathcal A X_n + b,\quad X_0=\begin{pmatrix}
7
\\
0
\end{pmatrix}.
$$
The matrix $\mathcal A$ is invertible and $b=\begin{pmatrix}
6
\\
0
\end{pmatrix}$. We have thus with an arbitrary vector $c\in \mathbb R^2$,
$$
X_{n+1}+c=\mathcal A \bigl(X_n + \mathcal A^{-1}(b+c)\bigr),
$$
and we may choose that $c$ such that
$
c=\mathcal A^{-1}(b+c),
$
i.e.
$
(\mathcal A-I)c= b
$
which is (uniquely) possible since $1$ is not an eigenvalue of $\mathcal A$.
With $Y_n=X_n+c$, we have thus
$$
Y_{n+1}=\mathcal A Y_n\Longrightarrow Y_{n}=\mathcal A^n Y_0.
$$
The eigenvalues of $\mathcal A$ are (real and) distinct, so we have with a diagonal 
$\mathcal D$ 
$$
\mathcal A=P\mathcal D P^{-1}\Longrightarrow\mathcal A^n=P\mathcal D^n P^{-1},
$$
making everything completely explicit.
