2000 IMO Shortlist S08

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 $F(2^{m + 1}).$

函数 $F$ 定义在非负整数集合上，并采用满足以下条件的非负整数值：对于每个 $n \geq 0,$

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

证明对于每个正整数 $m,$ 满足 $0 \leq n < 2^m$ 且 $F(4n) = F(3n)$ 的整数 $n$ 的个数为 $F(2^{m + 1})。$