The Pawn March Problem is something I just came up with, and I thought it was interesting enough to share.
White has a line of $n$ pawns on an infinite chess board, all on rank 0 (negative positions are legal, and pawns may never make double moves). Black has a single rook on $ωω$. Assuming optimal play and that Black cannot move to a position where it is stalemate, what is the furthest rank to which White can get a pawn?
The first couple positions are rather easy. Let the first pawn start on $A0$, and each additional pawn start to the right $B0$, $C0$, $D0$,...
White's only move is to $A1$. Black moves to $Aω$. White moves to $A2$. Black takes $A2$, and we're done. $PM(1)=2$.
White's moves are symmetric, so wlog White moves $A1$. There are a couple of possibilities here, but Black's best move is to $Aω$. If White pushes the A-pawn, Black takes on $A2$ and eventually takes the B-pawn on $B2$. If White pushes the B-pawn, Black takes $A1$, White moves $B2$, Black moves $B1$, and the White pawn is eventually taken on $B3$. $PM(2)=3$.
It's trickier here. It's important to note that $1.A1$ is different than $1.C1$, because Black can move to $ω0$ but not $-ω0$. I might write a program later to attempt to solve it, but figured that the lovely folks here might be able to solve it or know of something like it that would give some insight.