Solving the recurrence F(n) = 3F(n - 12).

Question

I'm very much stuck and don't even know where to begin here, any help would be much appreciated. Thanks.

Answer 1

Note that the chains $F(0),F(12),F(24)\dots $ and $F(1),F(13),F(25)\dots $ and $\dots$ are independent. There is nothing to link their values. Let us concentrate on $F(0),F(12),F(24)\dots $ and assume you are given $F(0)$ Define $G(n)=F(12n)$ so the recurrence becomes $G(n)=3G(n-1)$. Clearly the solution is $G(n)=G(0)3^n$ as you multiply by $3$ each step, so $F(12n)=F(0)3^n$ and $F(12n+k)=F(k)3^n$ for $ \le k \lt 12$. You need starting values for $F(0), F(1)\dots F(11)$

Answer 2

Let F(n+13)=3F(n) then F(13)=3F(0),F(14)=3F(1),...,then recurrence is clear.