More than Nim (combinatorics problem) A two-player game is played with two piles of stones, with sizes m,n. On a player's turn, that player can remove any positive number of stones from one pile, or the same positive number of stones from each pile. A player loses when they are unable to take a stone. If 1≤m,n≤30, for how many of the 30×30=900 starting positions does the first player have a winning strategy?

As pointed out by MJD, this game is known as Wythoff's game and has a very interesting theory. As discovered by Wythoff, the losing positions are the pairs $(m_k,n_k)$ and $(n_k,m_k)$ with
$$
n_k=\left\lfloor k\phi\right\rfloor\;,\\
m_k=\left\lfloor k\phi^2\right\rfloor\;,
$$
where $k\in\mathbb N$ and $\phi=\frac{1+\sqrt5}2$ is the golden ratio. Solving $k\phi^2\lt31$ for $k$ yields $k\lt31/\phi^2\approx 11.84$, so there are $2\cdot11=22$ losing positions and thus $900-22=878$ winning positions with $1\le m,n\le30$.
