Problem solving episode---WalkTilings(3)

I have shown that f(1) = 1, f(2) = 2, f(n) = f(n-1) + f(n-2) for n >= 3. So it seems that the solution is a sequence of Fabonacci numbers with f(n) = F(n+1), where F is Fabonacci numbers. Now I will prove this using complete induction.

Case1: When n = 1, f(1) = 1 = F(2).
Case2: When n = 2, f(2) = 2 = F(3).
Case3: When n >=3, suppose f(n-1) and f(n-2) and ... and f(1) is true. f(n) = f(n-1) + f(n-2) = F(n) + F(n-1) = F(n+1). So f(n) is true.

Conclude: for any natural number n >=1, f(n) = F(n+1).