TITLE: Tic-tac-toe with one mark type
QUESTION [18 upvotes]: Parameters $a,b,c$ are given such that $c\leq\max(a,b)$. In an $a\times b$ board, two players take turns putting a mark on an empty square. Whoever gets $c$ consecutive marks horizontally, vertically, or diagonally first wins. (Someone must win because we use only one mark type.) For each triple $(a,b,c)$, who has a winning strategy?
For $a=b=c=3$ (tic-tac-toe size), the first player can win by first going on the middle square and winning in the next turn.
In the one-dimensional case ($a=1$), this may well be a known game, but I also cannot find a reference. I asked the question on math.SE but it has not been solved.

REPLY [4 votes]: All I have are three small observations.
For all $a,c\in \mathbb N^+$, $(2a,1,2c)$ is a second player win. Player two's winning strategy is to divide the board into adjacent pairs, and to respond with the square paired with player one's last move.
I tried to find a similar pairing strategy for a $(2a,2b,2c)$ board, but failed. In the game where only orthogonal rows count as a win, there is a sort of pairing strategy for the second player: divide the board into $2\times 2$ blocks, and make a winning move if it exists, otherwise respond diagonally opposite in the same block. Unfortunately, this fails when diagonals are wins.
The game $(a,1,3)$ is the octal game  0.11337, which is equivalent to 0.007 with an offset, as Timothy Chow said. That is, when $n\ge 2$, an empty $(n,1,3)$-board  is equivalent to a 0.11337-heap of size $n$, which it turn equivalent to a 0.007-heap of size $n-2$ .