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.
 A: 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}$$r_2=3-2\sqrt2. $
$c_1, c_2$  can be determined from the initial values: 
$a_1=1=c_1 (3+2\sqrt2)+c_2 (3-2\sqrt2)\tag4$
$a_2=5=c_1 (3+2\sqrt2)^2+c_2 (3-2\sqrt2)^2\tag5$ 
From (4) and (5) we get: 
$c_1=\frac {1}{4+2\sqrt2}$ and $c_2=\frac {3+2\sqrt2}{4+2\sqrt2}$
Finally, based on (2): 
$a_{n+1}=\Big(\frac {3+2\sqrt2}{4+2\sqrt2}\Big)^{n+1}+ \Big(\frac {3-2\sqrt2}{4+2\sqrt2}\Big)^{n}\tag5$ 
I have checked the first 15 items by Excel; the results are matches.
A: If we start from $a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)$ we will get:
$$a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)\\ \Leftrightarrow\\ a_{n+1}^2+a_{n-1}a_{n+1} = a_n^2+a_{n}a_{n+2}\\ \Leftrightarrow\\ \frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{n}+a_{n+2}}{a_{n+1}}$$
Therefore we have $\frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{0}+a_{2}}{a_1} = 6$. After that, we can follow the solution of JV.Stalker:
$$a_{n+2}-6a_{n+1}+a_n = 0 \Rightarrow r^2-6r+1=0 \Rightarrow r_{1,2} = 3\pm2\sqrt{2}\Rightarrow a_n = c_1(3+2\sqrt{2})^n+c_2(3-2\sqrt{2})^n$$
We can define $a_0$ as $a_0a_2=a_1^2+4\Rightarrow a_0 = 1$. Therefore we have $c_1+c_2=1$ and $3(c_1+c_2)+2\sqrt{2}(c_1-c_2)=1$ and hence $c_1=\frac{2-\sqrt{2}}{4}$ and $c_2=\frac{2+\sqrt{2}}{4}$. If we plug this into the equation for $a_n$ we receive $$a_n = \frac{2-\sqrt{2}}{4}(3+2\sqrt{2})^n+\frac{2+\sqrt{2}}{4}(3-2\sqrt{2})^n\\ \Leftrightarrow\\ a_n = \frac{2-\sqrt{2}}{4}(\sqrt{2}+1)^{2n}+\frac{2+\sqrt{2}}{4}(\sqrt{2}-1)^{2n}\\ \Leftrightarrow\\ a_n = \frac{\sqrt{2}}{4}(\sqrt{2}+1)^{2n-1}+\frac{\sqrt{2}}{4}(\sqrt{2}-1)^{2n-1}$$
