# Solving the recurrence $T(n)=T(n-1)*T(n-2)$

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!

• Please use MathJax for typesetting equations. Jan 24, 2016 at 14:55

Your reasoning is perfect. To wit, here is what you get if you don't make any mistakes in your algebra. As you mention, $S(n)$ satisfies the Fibonacci recurrence, and furthermore $S(1) = 2 = F_3$, $S(0) = 1 = F_2$. We conclude that $S(n) = F_{n+2}$. Therefore $T(n) = 2^{F_{n+2}}$. The first few values are $$2, 4, 8, 32, 256, \ldots$$ To get a closed formula, note that $$F_n = \frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^n.$$ Therefore \begin{align*} S(n) &= \frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^{n+2} - \frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^{n+2} \\ &= \frac{3+\sqrt{5}}{2\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{3-\sqrt{5}}{2\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^n \\ &= \frac{5+3\sqrt{5}}{10} \left(\frac{1+\sqrt{5}}{2}\right)^n + \frac{5-3\sqrt{5}}{10} \left(\frac{1-\sqrt{5}}{2}\right)^n, \end{align*} using the fact that $(1\pm\sqrt{5})/2$ are roots of $x^2=x+1$.