Solve the recurrence relation
\begin{align}
a_n &= a_{n-1} ^ 3 \cdot a_{n-2} ^ 2 \\
a_0 &= a_1 = 2
\end{align}
Taking the logarithm of both sides seem helpful, but I could not continue. Is it valid to write characteristic equation like in the ordinary recurrence relations in that case?
Best Answer
Your idea was good $$a_n = a_{n-1} ^ 3 \, a_{n-2} ^ 2 \implies\log(a_n)=3\log(a_{n-1})+2\log(a_{n-2})$$ Let $b_n=\log(a_n)$ to make $$b_n=3b_{n-1}+2b_{n-2}$$ The characteristic equation being $r^2=3r+2$ gives for the roots $r_\pm=\frac{1}{2} \left(3\pm\sqrt{17}\right)$. So $$b_n=c_1 \left(\frac{3-\sqrt{17}}2 \right)^n+c_2 \left(\frac{3+\sqrt{17}}2 \right)^n$$ Apply the conditions $b_0=b_1=\log(2)$ and go back to $a_n$ to get $$a_n=2^{c_n}$$ where $$ c_n=\frac 1{34}\Bigg[\left(17+\sqrt{17} \right)\left(\frac{3-\sqrt{17}}2 \right)^n+\left(17-\sqrt{17} \right)\left(\frac{3+\sqrt{17}}2 \right)^n\Bigg]$$ and the first terms of the sequence are $$\{2,2,32,131072,2305843009213693952,\cdots\}$$