If $a_{n+2}=2a_{n+1}a_{n}$ where $a_0=a_1=1$.Find closed formula 
It is given that $a_{n+2}=2a_{n+1}a_{n}$ where $a_0=a_1=1$.Find closed formula for this recurrence relation using generating functions.

My thought:
I am familiar with working generating functions when the operation between variables are summation or subtraction such that $a_{n+2}=2a_{n+1}+a_{n}$. However , there are multiplication in my question , so i could not apply my classic solution way. I thought to use exponenial generating functions but the multiplication is a hindrance again. Then , how can i solve it using generating functions ? Is there any special generating function for multiplied variables ?
 A: HINT:
You can write
$$2 a_{n+2} = (2 a_{n+1}) \cdot (2 a_n)$$
so consider $b_n = 2 a_n$, $\ \ b_{n+2} = b_{n+1} \cdot b_n$, $\ b_0 = b_1= 2$.
Recall the Fibonacci sequence $F_0=0$, $F_1=1$, $F_2=1$ &c. So try to prove by induction that
$$b_n = 2^{F_{n+1}}$$
A: Hint
There is no general theory for nonlinear recurrence relations, but in this case there is a trick that lets us reduce things to the linear case. Namely, to convert multiplication to addition, you can take logarithms. Specifically, looking at a couple of terms of the series, you can see all of the terms are powers of $2$, so taking the log base $2$ is convenient.
Let $b_n=\log_2 a_n$. From $a_n=2a_{n-1}a_{n-2}$, we get
$$
2^{b_n}=2\cdot 2^{b_{n-1}}\cdot 2^{b_{n-2}}
$$
Take the $\log_2$ of both sides, to get
$$
b_n=1+b_{n-1}+b_{n-2}\tag{$*$}
$$
which is a nice additive equation that can be solved using generating functions in the usual way. That is, letting $B(x)=\sum_{n\ge 0}b_nx^n$,  you multiply both sides of ($*$) by $x^n$, and take the sum over all $n\ge 2$. This gives an equation involving $B(x)$ that lets you solve for $B(x)$. You will of course need the base cases $b_0=\log_2 a_0=0$ and $b_1=\log_2a_1=0$.
A: One good approach, suggested in the comment by
@lhf, is to examine the first few terms, guess the general answer, and verify it by induction. Another approach is to define
$b_n=\log_2 a_n$ (taking logs to base 2 is the most natural here) so that
$b_{n+2}=1+b_{n+1}+b_n$. This almost looks like the Fibonacci recursion. Indeed, $g_n=b_{n-1}+1$ satisfy $g_0=g_1=1$ and $g_{n+2}=g_{n+1}+g_n$,
so $g_n=F_{n+1}$ according to the current convention [1]  for Fibonacci numbers. We conclude that
$$a_n=2^{b_n}=2^{F_{n+1}-1} \,.$$
https://en.wikipedia.org/wiki/Fibonacci_number
