How to solve this recurrence relation: T(n) = T(n-1) * T(n-2)

Question

Consider this recurrence relation:

T(n) = T(n-1) * T(n-2)    n>2
T(1) = 1, T(2) = 2

How i can solve it? And finally: T(n) = O(?)

I think we should take log of both sides or something like. But i have no idea to continue.

Answer 1

You start by taking the logarithm of both parts to get:

log(T(n)) = log(T(n-1)) + log(T(n-2))

Now you substitute it with log(T(n)) to K(n) so you have to solve the problem K(n) = K(n-1) + K(n-2) . Doing similar thing you will get the solution

K(n) = c1 * F(n) + c2 * L(n), where F(n) is a Fibonacci number and L(n) is a Lucas number and c1, c2 are just some constants. So now to get your answer you just have to revert the logarithm.

So your solution is e^(c1 * F(n) + c2 * L(n)) . Following my previous explanation the complexity of this is O(e^(phi^n))