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This question is inspired by Tao’s answer in this post. I have thought about this occasionally for several months without anything concrete.

Question:

Given $\delta>0$ and $k\ge 3$, let $N= N_k(\delta)$ be such that for $N’>N$, and $A\subset G:=\Bbb{Z}/N’\Bbb{Z}$ with $|A| > \delta N’$, we have that $A$ contains an arithmetic progression of length $k$ (i.e., there exists $x,d \in G, d\neq 0$ where $P:= \{x,x+d,x+2d,\dots,x+(k-1)d\} \subset A$).

Does there exist some absolute constant $C$, such that $N_k(1-10/k) \le k^{C+o(1)}$ for all large $k$?

Comments: The construction from this post can be applied to show that: fixing any $a\ge 1,\epsilon > 0$, we will have $N_k(1-a/k) \ge k^{(1-\epsilon)a}$ for all sufficiently large $k$. It sounds plausible to me that this may be sharp, in the sense that, for fixed $a\ge 1,\epsilon > 0$, we have $N_k(1-a/k)\le k^{(1+\epsilon)a}$ for all sufficiently large $k$.

I have defined $N_k(\cdot)$ in terms of progression-free subsets of groups, rather than subsets of the interval $\{1,\dots,N\}$. This was simply to make a bit easier to get better upper bounds. I would of course also be interested if anything can be said about the interval variant.

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    $\begingroup$ What is the meaning of $1 - 10/k$ and $C+o(1)$ instead of simply $C$? The question seems to be equivalent to asking for bound $N_k\le k^C$ for large $k$ (possibly with different $C$). $\endgroup$
    – Fedor Petrov
    Commented May 11, 2023 at 8:01
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    $\begingroup$ $N_k(\cdot)$ is function, and 10 is just some arbitrary constant a bit bigger than 1. you are right about $C$ of course, but I find more intuitive to write it this way. if someone proves (say) $N_k\le k^5$ for all $k\ge 10^{1000}$, I find it more helpful to view this as $k^{5+o(1)}$ rather than $k^C$ for some abstract $C$. $\endgroup$
    – Zach Hunter
    Commented May 11, 2023 at 8:22

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