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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}).

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