OIMU 2005 Problem 5

Arnaldo and Bernaldo play a game where they alternate saying natural numbers, and the winner is the one who says $0$ . In each turn except the first the possible moves are determined from the previous number $n$ in the following way: write $$n =\sum_{m\in O_n}2^m;$$ the valid numbers are the elements $m$ of $O_n$ . That way, for example, after Arnaldo says $42= 2^5 + 2^3 + 2^1$ , Bernaldo must respond with $5$ , $3$ or $1$ .

We define the sets $A,B\subset \mathbb{N}$ in the following way. We have $n\in A$ iff Arnaldo, saying $n$ in his first turn, has a winning strategy; analogously, we have $n\in B$ iff Bernaldo has a winning strategy if Arnaldo says $n$ during his first turn. This way, $$A =\{0, 2, 8, 10,\cdots\}, B = \{1, 3, 4, 5, 6, 7, 9,\cdots\}$$ Define $f:\mathbb{N}\to \mathbb{N}$ by $f(n)=|A\cap \{0,1,\cdots,n-1\}|$ . For example, $f(8) = 2$ and $f(11)=4$ . Find $$\lim_{n\to\infty}\frac{f(n)\log(n)^{2005}}{n}$$