Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$ 
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.

I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,5741$… So I ran out of ideas.

FYI, 
I noticed that 
$\dfrac{a_{n+1}+a_{n-1}}{a_n}=6 \hspace{1cm}\forall n \gt1\tag1$ 
(unfortunately  I could not prove it so far).
Using this recursion formula easy to determine the general formula of $a_{n+1}$.
Searching the general formula in the form  
$a_n=c_1 r_1^n+c_2 r_2^n\tag2$
where $r_1, r_2$ are the roots of the following equation: 
$r^2-6r+1=0\tag3$
So $\hspace{1cm}$ $r_1=3+2\sqrt2$ $\hspace{0,5cm}$ and $\hspace{0.5cm}$$r_2=3-2\sqrt2. $
$c_1, c_2$  can be determined from the initial values: 
$a_1=1=c_1 (3+2\sqrt2)+c_2 (3-2\sqrt2)\tag4$
$a_2=5=c_1 (3+2\sqrt2)^2+c_2 (3-2\sqrt2)^2\tag5$ 
From (4) and (5) we get: 
$c_1=\frac {1}{4+2\sqrt2}$ and $c_2=\frac {3+2\sqrt2}{4+2\sqrt2}$
Finally, based on (2): 
$a_{n+1}=\Big(\frac {3+2\sqrt2}{4+2\sqrt2}\Big)^{n+1}+ \Big(\frac {3-2\sqrt2}{4+2\sqrt2}\Big)^{n}\tag5$ 
I have checked the first 15 items by Excel; the results are matches.
