ISL 2000 A4

The function $F$ is defined on the set of nonnegative integers and takes nonnegative integer values satisfying the following conditions: for every $n \geq 0,$

(i) $F(4n) = F(2n) + F(n),$ (ii) $F(4n + 2) = F(4n) + 1,$ (iii) $F(2n + 1) = F(2n) + 1.$

Prove that for each positive integer $m,$ the number of integers $n$ with $0 \leq n < 2^m$ and $F(4n) = F(3n)$ is $F(2^{m + 1}).$