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Claim. Samuel Korsky's proof claim on the minimum $g(n)=\min_{\pi\in S_n}S(\pi)$, submitted on 15 September 2026 to the proof-claim tab of Problem 34, claims an elementary construction giving, for every integer $k\ge1$, $g(2\cdot6^k)\le31^k+35^k+2\cdot6^k\le4\cdot35^k$. Hence $g(n)=O(n^{\log_635})$ along an infinite sequence, with $\log_635=1.9842775\ldots<2$, which would refute the two expectations of the erdosproblems.com commentary, $g(n)\ge n^{2-o(1)}$ and $g(n)\gg S(\iota)$ for the identity $\iota$ (Konieczny's Questions 3 and 4). The proof is an external document linked from the tab, and erdosproblems.com lists the entry as a partial proof claim.

Submission note. Posted to erdosproblems.com as a proof claim by Samuel Korsky (account SamKorsky) on 15 September 2026, giving "GPT Astra" as the AI used:

Let $g(n)$ be the minimum number of distinct sums of nonempty consecutive intervals in a permutation of $\{1,\ldots,n\}$. We give an elementary construction showing that, for every integer $k\geq 1$,

$$g(2\cdot 6^k)\leq > 31^k+35^k+2\cdot 6^k\leq 4\cdot 35^k.$$

Thus $g(n)=O(n^{\log_6 35})$ along an

infinite sequence, where $\log_6 35=1.9842775\ldots<2$, disproving the last two conjectures listed in the problem statement. Notes: The exponent is non-optimized; I pushed Astra to search for sharper constructions and was able to get the exponent down to $\sim1.86$, though I didn't try too hard. It would be very interesting to find the correct exponent - maybe it's even $3/2$!

Why it is rejected. It answers a variant, not the Statement. Problem 34 asks whether every permutation $\pi\in S_n$ has $S(\pi)=o(n^2)$, a question about the largest values of $S(\pi)$. The claim answers instead the question the erdosproblems.com commentary raises about the smallest value $g(n)$, whether $g(n)\ge n^{2-o(1)}$ or even $g(n)\gg S(\iota)$. An upper bound on $g(n)$ says nothing about whether some permutation has $cn^2$ distinct consecutive sums, so its mathematics answers the variant and does not settle this problem; the value disproved is its answer to that variant.

Standing. Rejected, as answering a variant. The entry carries no comments and no review on erdosproblems.com, which warns that listing on the tab is not a check of the proof; this corpus has not read the external document.

Depends on. No page of this wiki.