Consider a 2-player game where the players take turns claiming an integer from $1$ to $n$. The same integer cannot be claimed twice. Player 1 wins if the set of integers he claims contains an arithmetic progression of $\ge k$ terms; player 2 wins if this is prevented by the time all available numbers have been claimed.
Unless I'm missing something fundamental, for every $k$ there is $s(k)$ such that P2 wins for every $n<s(k)$ and P1 wins for every $n\ge s(k)$. I wonder: are the explicit values of $s(k)$, or at least any good asymptotics, known?
What I have so far:
Trivially, $s(1)=1$ and $s(2)=3$. I was also able to show that $s(3)\le12$, $s(4)\le30$ and $s(5)\le225$, by comparison to $m,n,k$-games. I expect $s(5)$ to be significantly smaller, though. For $k\ge6$, comparisons to $m,n,k$-games cease to be helpful.
In general, the number of winning sets is equal to: $w:=tn-\frac{t^2+t}2(k-1)$, where $t=\lfloor\frac n{k-1}\rfloor$. As proven by Erdős and Selfridge, if $w<2^{k-1}$ then P2 has a winning strategy. Conversely, if $w>2^{k-4} \cdot(k^2-k) \cdot n$ then P1 has a winning strategy (however, this bound can possibly be improved).
Those bounds aren't particularly heplful, though. For instance, they yield $101\le s(10)\le103690$.
As a side note:
Unless some clever general strategy can be found, I expect the game to actually be quite fun for $100 \le n \le 200$, especially if balanced by choosing a good value of $k$ (no less than $10$ to be sure) and introducing some kind of pie rule!
EDIT: $s(k)>2^\frac k2 \sqrt{k-1}$ will hold for large enough $k$s.
EDIT2: $s(3)=5$
EDIT3: a friend of mine was able to prove $s(4)=15$ with a computer search.
EDIT4: the upper bounds I initially provided were wrong.