Hello,
the problem is following quadratic recurrence equation:
$u_0=1$
$u_1=2$
$u_n=u_{n-2}^2u_{n-1}, \ n \geq 2$

I know how to solve most of the linear homogeneous and non-homogeneous equations, but I haven't encountered quadratic equations before. I tried googling and using forums search function, but I couldn't found any information, how to solve these. So, any help is appreciated. Thanks!

WolframAlpha gives following solution, but that doesn't help me either.

2. Setting $v_{n} = \log_{2} u_{n}$ the 'recursive relation' becomes...

$\displaystyle v_{n}= v_{n-1} + 2\ v_{n-2}\\ , \\ v_{0}=0 \\, v_{1}= 1$ (1)

... which is linear...

Kind regards

$\chi$ $\sigma$

3. Very interesting is the case of the 'recursive relation'...

$\displaystyle u_{n}= u_{n-1}\ u_{n-2}\\,\\ u_{0}=1\\,\\ u_{1}=2$ (1)

Setting $v_{n}= \log_{2} u_{n}$ we obtain the solution...

$\displaystyle u_{n}= 2^{f_{n}}$ (2)

... where the $f_{n}$ are the terms of the Fibonacci's sequence...

Kind regards

$\chi$ $\sigma$