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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. Write F(n)F(n) for the largest size of A⊆{1,…,n}A\subseteq\{1,\ldots,n\} whose nonzero subset sums form a primitive set, the quantity the problem asks for. Samuel Korsky's manuscript "Near-Exact Bounds for Primitive Subset Sums" (dated 26 July 2026, 8 pages) proves a lower bound on nn for every kk-element set A⊆{1,…,n}A\subseteq\{1,\ldots,n\} under the weaker hypothesis that no nonempty subset sum equals twice another nonempty subset sum, and derives from it, with the witness set of Erdős, Lev, Rauzy, Sándor and Sárközy, that

F(n)∈{m, m+1},m=⌊log⁡2(n+1)⌋,F(n)\in\{m,\,m+1\},\qquad m=\lfloor\log_2(n+1)\rfloor,

for every n≥1n\ge1 (its Corollary 1.2), and that F(n)=mF(n)=m whenever 2m−1≤n<2m+(2m−1+⌊m2/4⌋)/(m+1)2^m-1\le n<2^m+(2^m-1+\lfloor m^2/4\rfloor)/(m+1) (its Corollary 1.3), so in particular F(2m−1)=F(2m)=mF(2^m-1)=F(2^m)=m and the lower value is exact on an initial interval of length (1+o(1))2m/m(1+o(1))2^m/m in each dyadic block. The manuscript conjectures that F(n)=mF(n)=m for every nn, that is, that every kk-element set with primitive positive subset sums has largest element at least 2k−12^k-1, and says that its method, which uses only the excluded ratio 22, leaves a factor of about 22 in the lower bound on nn that the full divisibility condition would have to close.

Submission note. Posted to erdosproblems.com as a proof claim by Samuel Korsky (account SamKorsky) on 27 July 2026, giving "GPT 5.6-Pro" as the AI used:

While this problem is already marked as solved, the current state of the art seems to still have wide error bars on the size of a maximal AA:

>log⁡2(n)−1<∣A∣≤log⁡2(n)+12⋅log⁡2(log⁡(n))+O(1)>> \log_2(n) - 1 < |A| \le \log_2(n) + \frac{1}{2}\cdot\log_2(\log(n)) + O(1) >

By adapting some of the ideas from my work on #817, one can actually show

that the largest AA satisfying the weaker condition that no nonempty subset sum is exactly twice another has size at most $\lfloor\log_2(n + 1)\rfloor + 1$. This confirms and sharpens the claim at the bottom of the problem description and shows that a maximal ∣A∣|A| must take one of two consecutive values; GPT was then able to refine the argument further and show for infinitely many nn that the construction of Erdos, Lev, Rauzy, Sandor, and Sarkozy is optimal. Notes: Is this a valid proof claim, given that the problem is already marked as solved?

Covers. On the stronger exact-value question that the problem page's Formulation names: the upper bound F(n)≤⌊log⁡2(n+1)⌋+1F(n)\le\lfloor\log_2(n+1)\rfloor+1 for every nn, which replaces the O(log⁡log⁡n)O(\log\log n) error of the accepted bound by a single unit, and the exact value F(n)=⌊log⁡2(n+1)⌋F(n)=\lfloor\log_2(n+1)\rfloor for the nn in the initial interval of each dyadic block stated above, hence for infinitely many nn. It does not determine F(n)F(n) for every nn.

Depends on. [[problems/divisors/E0882/claims/1999_04_01_erdos_lev_rauzy_sandor_sarkozy|The accepted two-sided bound]] supplies the lower bound F(n)≥mF(n)\ge m through its witness set; the manuscript proves the upper bound.

Authorship and tools. The author's acknowledgments say that the author's initial argument restricted F(n)F(n) to three possible values and that GPT-5.6 Pro assisted in identifying the refinements that gave two values and the exact value for infinitely many nn, and that the author verified all arguments and takes responsibility for the content; the forum entry names the system as GPT 5.6-Pro. The method adapts ideas from the author's work on Problem 817.

Standing. Posted on the problem's forum as a proof claim on 27 July 2026 with the manuscript as its external link; the author asks there whether a claim on a problem already marked solved is valid; the entry had no comments on 2026-10-07. The manuscript is not refereed, and no one has recorded accepting it, so the claim is claimed. The site's remarks do not mention it.