Question

Suppose $A \subseteq \{1,\dots,n\}$ does not contain any arithmetic progressions of length $k+1$. What is the largest number of $k$-term arithmetic progressions that $A$ can have? (one may also wish to put some lower or upper on the size of $A$) We can work over $\mathbb{Z}_p$ if it makes the answer any easier. The "degenerate" case $k=2$ asks for the largest size of the set without arithmetic progressions and it is known that there exist $A$'s with this property of almost linear size.

Answer 1

Let $B\geq2k$ and let $$A=\left\{\sum_{i=0}^na_iB^i:n=0,1,...;a_i=0,1,...,k-1\right\}$$ It's not hard to show that $A$ has no $k+1$-long arithmetic progression. Using the density Hales-Jewett theorem we get that any subset $B\subset A$ with positive relative density has a $k$-long arithmetic progression.

I don't know the best bounds on the density Hales-Jewett, but I think there are some from the polymath proof, so in principle this would give an answer to your question.