General solution for the series $a_n = \sqrt{(a_{n-1} \cdot a_{n-2})}$ Hey I'm searching a general solution for this recursive series:
$a_n = \sqrt{(a_{n-1}\cdot a_{n-2})}$
$\forall n \geq 2$
$a_0 = 1$,
$a_1 = 2$

Elaborating on what Wojowu has mentioned,
$$a_n^2=a_{n-1}\cdot a_{n-2}$$
$$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}$$
That is, $a_n^2\cdot a_{n-1}=$ constant is invariant. 
Hence $$a_n^2\cdot a_{n-1}=a_{n-1}^2\cdot a_{n-2}= \ldots = a_1^2a_0 = 4$$
or, $$a_n^2=\frac{4}{a_{n-1}}=\frac{4}{\frac{2}{\sqrt{a_{n-2}}}}=2\sqrt{a_{n-2}}$$
or, $$a_n=\sqrt{2\sqrt{a_{n-2}}}=\sqrt{2\sqrt{\sqrt{2\sqrt{a_{n-4}}}}}$$
or,$$a_n=\sqrt[2]{2\sqrt[4]{2\sqrt{a_{n-4}}}}$$
or,$$a_n=\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt{a_{n-6}}}}}}$$
Therefore, $$a_n=\begin{cases}\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_1}}}}}}  & \text{if n is odd} \\\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_0}}}}}}  & \text{if n is even} \end{cases}$$
