I have to solve given recurrence relation: $ a_{n+2} = 2a_{n+1}^2 a_{n}^3$, where: $ a_{1} = a_{0} = 1.$
I have no idea how to do it honestly, I have calculated that first values of $a_{n}$ are (1, 1, 2, 8, 1024...). $a_{5}$ is so big that I will not put it down here.
I've tried using generating functions to solve it but with no success. I don't know how to split given sequence in order to receive couple of generating functions so that I can solve them. Maybe we have to use auxiliary sequence for that?
Thanks in advance! Any help is appreciated.
First transform your recurrence: Let $b_n = \log a_n$, and:
$\begin{align*} \log a_{n + 2} &= \log 2 + 2 \log a_{n + 1} + 3 \log a_n \\ b_{n + 2} &= 2 b_{n + 1} + 3 b_n + \log 2 \end{align*}$
Now use generating functions. Define $B(z) = \sum_{n \ge 0} b_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognize the resulting sums:
$\begin{align*} \sum_{n \ge 0} b_{n + 2} z^n &= 2 \sum_{n \ge 0} b_{n + 1} z^n + 3 \sum_{n \ge 0} b_n z^n + \log 2 \sum_{n \ge 0} z^n \\ \frac{B(z) - b_0 - b_1 z}{z^2} &= 2 \frac{B(z) - b_0}{z} + 3 B(z) + \frac{\log 2}{1 - z} \\ \end{align*}$
Use the initial values $b_0 = b_1 = 0$, solve for $B(z)$ and express as partial fractions:
$\begin{align*} B(z) &= \frac{z^2}{1 - 3 z - z^2 + 3 z^2} \log 2 \\ &= \frac{\log 2}{8 (1 - 3 z)} + \frac{\log 2}{8 (1 + z)} - \frac{\log 2}{4 (1 - z)} \end{align*}$
Everything in sight is just geometric series:
$\begin{align*} b_n &= \frac{\log 2}{8} \cdot 3^n + \frac{\log 2}{8} \cdot (-1)^n - \frac{\log 2}{4} \\ a_n &= 2^{3^n / 8 + (-1)^n / 8 - 1 / 4} \end{align*}$