problem statement
Zagreb and Yerevan play a restricted version of nim as follows: They start with 2 possibly empty (but not both empty) piles of rocks, and take turns Z-Y-Z-Y-Z... until the last player who has a move wins. On a turn, each must take at least one rock out of exactly one pile, so that the total number of rocks in both piles becomes a Fibonacci number. (The starting state need not have a Fibonacci total, but all successive states must.)
An example game: $(5 \text{ rocks}, 5 \text{ rocks}) \to (5 \text{ rocks}, 3 \text{ rocks}) \to (0 \text{ rocks}, 3 \text{ rocks}) \to (0 \text{ rocks}, 0 \text{ rocks})$
A positive integer $n$ is chosen. How many starting states, with the rocks totaling at most the $n$th Fibonacci number, are winnable by Zagreb, assuming perfect play?
My try
Let's start by calculating some states. A winning state is one in which the first player can force a win, and a losing state is one in which the second player can force a win. All games are finite and predictable so all states are one of the two.
A losing state is one in which there are moves only to winning states. A winning state has a move to a losing state. We first consider only states with Fibonacci sum for clarity
We can see that the losing states form a band in the middle of each Fibonacci diagonal, and this pattern necessarily repeats forever:
Also you can see from the same diagram that all states in the triangular region above the losing segment are also losing, because they have no move to a losing state either. And, furthermore these triangular areas are the only losing cells.