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Given $\delta>0$ and positive integers $k$, write $h(\delta;k)$ for the smallest $N$ such that for any $S\subset [N]:=\{1,\dots,N\}$ of size $\ge \delta N$, there exists non-zero integers $n_0,d_1,\dots,d_k$ such that $n_0+\sum_{i\in I}d_i\in S$ for all $I\subset [k]$ (these sums need not all be distinct, in fact, they may be any $k$-term arithmetic progression).

It occurs to me that there is a short proof that for any fixed $\delta$, that $h(\delta;k)$ grows superexponentially as $k\to \infty$ (i.e., for any $C$, we have $h(\delta;k)>C^k$ for all large $k$).

Is such a bound recorded in the literature? I am aware of the work of Gunderson, Rodl, and Siderenko (e.g., "Extremal problems for sets forming Boolean algebras and complete partite hypergraphs"), but their bounds fixate on when $k$ is fixed and $\delta\to 0$.

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If I read Theorem 1.6 of

Sándor, Csaba, Non-degenerate Hilbert cubes in random sets, J. Théor. Nombres Bordx. 19, No. 1, 249-261 (2007). ZBL1126.11014.

correctly, I think he shows (in your language) that for $\delta=1/2$ one has

$$ k \leq (1+\varepsilon)\log_2 \log_2 h(1/2;k)$$

for $k$ sufficiently large depending on $\varepsilon$, or equivalently $$ h(1/2;k) \gg_\varepsilon 2^{2^{k/(1+\varepsilon)}},$$

by computing the largest cube that can be located inside a random subset of $[n]$ (which has density at least $1/2$ at least half of the time). This is sharp up to the epsilon loss, as noted in that paper. In fact Sándor shows a more precise estimate with some lower order terms that I will not detail here.

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    $\begingroup$ due the presence of the term “non-degenerate”, I fear this is not what what I seek. the only restriction on the integers $d_1,\dots,d_k$ i meant to impose was that they are non-zero (i.e., the subset sums need not be distinct). I will update the question accordingly. $\endgroup$ Commented Feb 7, 2023 at 18:59
  • $\begingroup$ Fair enough. I guess a random density 1/2 set in $[n]$ will already contain an arithmetic progression of length about $\log_2 n$ or so, so some other construction will be needed to do better than exponential. $\endgroup$
    – Terry Tao
    Commented Feb 7, 2023 at 22:22
  • $\begingroup$ indeed. a minor tweak to this is the solution. basically you take a random dense subset of the construction in this post (which was known before Salem-Spencer and Behrend constructions). to my knowledge, said construction is the only way we know how to make dense subset of $[n]$ where all APs have length $o(\log n)$. $\endgroup$ Commented Feb 7, 2023 at 22:34

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