General solution for the series $a_n = \sqrt{(a_{n-1} \cdot a_{n-2})}$

Question

Hey I'm searching a general solution for this recursive series:

$a_n = \sqrt{(a_{n-1}\cdot a_{n-2})}$

$\forall n \geq 2$

$a_0 = 1$, $a_1 = 2$

Answer 1

Squaring both sides, $a_n^2=a_{n-1}a_{n-2}$. Then, $2\log(a_n)=\log(a_{n-1})+\log(a_{n-2})$. Set $b_n=\log(a_n)$. Thus $2b_n=b_{n-1}+b_{n-2}$ with $b_0=0$ and $b_1=\log2$.

Can you coninue?

Answer 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, $$a_n=\begin{cases}\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_1}}}}}} & \text{if n is odd} \\\sqrt[2]{2\sqrt[4]{2{\sqrt[4]{2\sqrt[4]{\ldots \sqrt{a_0}}}}}} & \text{if n is even} \end{cases}$$