If for some reason you really wanted to compute Fib(n) for ridiculously large numbers of n, you would probably use that [Fib(n), Fib(n-1)] = A [Fib(n-1), Fib(n-2)] for the transition matrix A = [[1, 1], [1, 0]] and thus [Fib(n+1), Fib(n)] = A^n [Fib(1), Fib(0)] and then use exponentiation by squaring to compute A^n directly and thus Fib(n) in log_2(n) steps.