Please solve this recurrence relation question for $8a_na_{n+1}-16a_{n+1}+2a_n+5=0$ Suppose $a_1=1$ and $$8a_na_{n+1}-16a_{n+1}+2a_n+5=0,\forall n\geq1,$$Please help to sort out the general form of $a_n$.
Here are the first a few values of the series. Not sure if they are useful as they seem quite random to me.
$$1,{7\over8},{3\over4},{13\over20},{7\over12},\dots$$
Thanks.

Let $p_n$, $q_n$ be two sequences to be determined such that $\displaystyle\;a_n = \frac{p_n}{q_n}\;$. In terms of $p_n$, $q_n$, the recurrence relation for $a_n$ can be rewritten as
$$a_{n+1} = \frac{2a_n+5}{-8a_n+16} 
\quad\iff\quad\frac{p_{n+1}}{q_{n+1}} = \frac{2p_n+5q_n}{-8p_n+16q_n}\tag{*1}
$$
If we scale $p_n, q_n$ to make them satisfy the linear recurrence relation
$$\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix}
= \begin{bmatrix}2 & 5\\-8 & 16\end{bmatrix} \begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}
\tag{*2}
$$
then any solution of $(*2)$ will lead to a solution of $(*1)$. It is easy to check the square matrix in $(*2)$ has eigenvalues $6$ and $12$ with corresponding eigenvectors
$\begin{bmatrix}5\\4\end{bmatrix}$ and $\begin{bmatrix}1\\2\end{bmatrix}$. The general
solution of $(*2)$ has the form:
$$
\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}
= A \times 6^n \begin{bmatrix}5\\4\end{bmatrix} + B \times 12^n \begin{bmatrix}1\\2\end{bmatrix}$$
for suitable chosen constants $A$ and $B$. By inspection, we can reproduce $a_1 = 1$ by setting $A = \frac16$ and $B = \frac{1}{12}$.
This leads to  a solution of the original recurrence relation subject to $a_1 = 1$:
$$a_n = \frac{p_n}{q_n} = \frac{5 \times 6^{n-1} + 12^{n-1}}{4 \times 6^{n-1} + 2\times 12^{n-1}} = \frac{2^{n-1} + 5}{2^n + 4}$$
