We consider the following combinatorial game (with two players alternatively playing optimally). Posititions are given by heaps containing $b\geq 0$ black and $w\geq 0$ white stones and are encoded by $(b,w)$ in $\mathbb N^2$ where $\mathbb N=\lbrace 0,1,\ldots\rbrace$.

Players are allowed to remove either a unique black stone or a set of two stones containing at least one white stone. (Moves starting at $(b,w)$ correspond therefore to steps in $\{(-1,0),(-1,-1),(0,-2)\rbrace$ which do not leave the first quadrant.)

Final positions with no move left are $(0,0)$ and $(0,1)$ and are either winning or losing independently (there are therefore $4=2^2$ possibilities).

There seems to be no easy obvious pattern in the set of winning positions for these four games but my computer seems to be convinced of the following fact (for each possible game):

Fix $(\alpha,\beta)$ in $\mathbb N^2$ and consider for every $(a,b)$ in $\mathbb N^2$ the characteristic sequence $W_{(\alpha,\beta)}(a,b)$ in $\lbrace 0,1\rbrace^{\mathbb N}$ encoding winning positions in the arithmetic progression $(\alpha,\beta)+\mathbb N(a,b)$. All sequences $W_{(\alpha,\beta)}(a,b)$ with $a,b\geq 1$ are seemingly ultimately periodic and moreover the union $\cup_{a,b\geq 1}W_{(\alpha,\beta)}(a,b)$ of sequences corresponding to all arithmetic progressions (with integral steps in the open first quadrant) starting at $(\alpha,\beta)$ seems to be a finite (and generally fairly small) set.

Should I throw away my computer?

Upgrade: The characteristic sequence for winning positions $(0,3)+\mathbb N(1,1)$ (with respect to the rule that the last player with moves win) has a short non-periodic prefix since it is seemingly given by $110\overline{101}$.