Non-linear recurrence with square root: $a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$ How should I approach this problem:
$a_{n+2}=\sqrt{a_{n+1}\cdot  a_{n}}$
where $a_0 = 2, a_1=8$
 A: An alternative approach to the one given by Oliver Oloa:
$$a_{n+2}=\sqrt{a_{n+1}\cdot  a_{n}}$$
$$\implies a^2_{n+2}={a_{n+1}\cdot  a_{n}}$$
$$\implies \frac{a_{n+2}}{a_{n+1}}=\frac{a_{n}}{a_{n+2}} \tag1$$
Similarly we can write that
$$\frac{a_{n+1}}{a_{n}}=\frac{a_{n-1}}{a_{n+1}} \tag2$$
$$\frac{a_{n}}{a_{n-1}}=\frac{a_{n-2}}{a_{n}} \tag3$$
$$\ldots$$
$$\frac{a_{2}}{a_{1}}=\frac{a_{0}}{a_{2}}  \tag{n+1}$$
Multiplying these $n+1$ relations, we get 
$$\frac{a_{n+2}}{a_{n+1}}\cdot\frac{a_{n+1}}{a_{n}}\cdot\frac{a_{n}}{a_{n-1}}\ldots \frac{a_{2}}{a_{1}}=\frac{a_{n}}{a_{n+2}}\cdot\frac{a_{n-1}}{a_{n+1}}\cdot\frac{a_{n-2}}{a_{n}}\ldots\frac{a_{0}}{a_{2}}$$
$$\implies \frac{a_{n+2}}{a_{1}}=\frac{a_{1}\cdot a_{0}}{a_{n+2}\cdot a_{n+1}}$$
$$\implies \frac{a_{n+2}}{8}=\frac{16}{a_{n+2}\cdot a_{n+1}}$$
$$\implies a_{n+2}=\frac{12}{\sqrt{a_{n+1}}}$$
$$\implies a_{n+2}=\sqrt{12}\cdot \sqrt[4]{a_{n}}$$
$$\implies \boxed{a_{n}=\sqrt{12}\cdot \sqrt[4]{a_{n-2}}}$$
So we can write that $$a_{n}=12^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots + \frac{(-1)^{r-1}}{2^{r-1}}}\cdot (a_{n-r})^{\frac{(-1)^r}{2^r}} \,\,\,\,\,\,\, \text{where} \,\,\, r \ge 1$$
For $r=n$, we get
$$a_{n}=12^{\frac{1-(-\frac{1}{2})^n}{1-(-\frac{1}{2})}}\cdot (a_0)^{\frac{(-1)^n}{2^n}}$$
$$\boxed{a_{n}=12^{\frac{1}{3}\cdot\frac{2^n-(-1)^n}{2^{n-1}}}\cdot (2)^{\frac{(-1)^n}{2^n}}} \,\,\,\,\,\,\,\,\, \forall \,\,\,\,\,\, n\ge0$$
This is the closed form.
Hope this helps you.
A: Hint. One may notice that $a_n>0$, for $n\ge0$, and that
$$
\log(a_{n+2})=\frac12\log(a_{n+1})+\frac12\log(a_{n}).
$$
