I have been trying to solve the following recurrence:
$$T(n)=T(n-1)*T(n-2)$$ The initial conditions are $n \ge 2$ and $T(0) = 2$ and $T(1) = 4$.
I started by taking the $\log_{2}$ of both sides to get
$$\log(T(n)) = \log(T(n-1)) + \log(T(n-2))$$
which looks an awful lot like a Fibonacci recurrence.
So if I set $S(k) = \log(T(n))$, then I have a recurrence of the form $$S(k) = S(k-1) + S(k-2)$$ with $S(0) = 1$ and $S(1) = 2$.
I can solve this easily using a characteristic polynomial and get
$$S(k) = A*(0.5*(1+\sqrt{5}))^k + B*(0.5*(1-\sqrt{5}))^k$$ where $A$ and $B$ are constants. To get $A$ and $B$ I use the initial conditions for $S$, namely $S(0) = 1$ and $S(1) = 2$.
The problematic part is that I think to get $T(n)$ I just simply raise $2$ to the power of the entire bunch above to get
$$T(n) = 2^{n*A*0.5*(1+\sqrt{5})+n*B*0.5(1-\sqrt{5})}$$.
However, when I plug in some values for n to check I get some different integer values from the original recurrence.
Is my reasoning flawed somewhere in the process or am I just making an algebraic mistake somewhere?
Thanks!