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Problem 357

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claims/: The 4 claim pages of Problem 357, one per claimant's result; the problem's standing derives from them.


Statement. Let 1≤a1<⋯<ak≤n1\leq a_1<\cdots <a_k\leq n be integers such that all sums of the shape ∑u≤i≤vai\sum_{u\leq i\leq v}a_i are distinct. Let f(n)f(n) be the maximal such kk.

How does f(n)f(n) grow? Is f(n)=o(n)f(n)=o(n)?

Status. Open. Two accepted partial claims give refereed upper bounds: Hegyvári's f(n)≤(2/3+o(1))nf(n)\le(2/3+o(1))n (Acta Math. Hungar. 48 (1986); claim page) and Coppersmith and Phillips's f(n)≤(2/3−1/512)n+O(log⁡n)f(n)\le(2/3-1/512)n+O(\log n) (SIAM J. Discrete Math. 9 (1996); claim page). Two partial claims are pending, neither reviewed: a claim of 27 July 2026 by Lenthall-Cleary (using GPT-5.6 Sol, as the proof-claims tab names it), the upper bound f(n)≤n/2+O(n2/3)f(n)\le n/2+O(n^{2/3}) with its finite inequality in Lean 4 (claim page); and Pickhardt's manuscript, written with the Paratelligent Research Agent and linked in the discussion thread on 31 August 2026, the lower bound f(n)≥(4/3−o(1))nf(n)\ge(4/\sqrt3-o(1))\sqrt n and an upper bound of the same shape with a smaller second-order constant (claim page). None of the four touches the question f(n)=o(n)f(n)=o(n). The site's label was OPEN on 2026-10-07 (page last edited 12 January 2026).

Source. erdosproblems.com/357, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #357, https://www.erdosproblems.com/357.

References.

  • [Er77c] Erdős, Paul, Problems and results on combinatorial number theory. III. Number theory day (Proc. Conf., Rockefeller Univ., New York, 1976) (1977), 43-72.
  • [He86] Hegyvári, N., On consecutive sums in sequences. Acta Math. Hungar. 48 (1986), no. 1--2, 193--200, DOI 10.1007/BF01949064. The introduction (printed p. 193) records the Erdős--Harzheim question in the unrestricted form and their conjecture "that this is not true if a1<a2<a…<aka_1<a_2<a\ldots<a_k [sic] is also assumed", this problem's question, which the paper leaves open; Theorem 1 (p. 193) gives (1/3+o(1))n≤f(n)≤(2/3+o(1))n(1/3+o(1))n\le f(n)\le(2/3+o(1))n for the unrestricted form, whose upper bound applies to this problem's f(n)f(n) while the lower bound's sequence is not increasing. Library home: hegyvari_1986_consecutive_sums_sequences; result page theorem_1.

Formalization. Statement in formal-conjectures.

Current assessment

No independent assessment of proof coverage is recorded. The frontmatter standing is derived from the four claim pages, two accepted partial claims with refereed upper bounds and two pending partial claims, so the problem is open. Search scope, 2026-10-07: the site's page, its discussion thread and its proof-claims tab, Lenthall-Cleary's repository and Pickhardt's manuscript; no literature search beyond the page's References was made here.

Known Results

Upper bounds. Hegyvári's Theorem 1 ([He86], refereed; claim page) gives f(n)≤(2/3+o(1))nf(n)\le(2/3+o(1))n through the unrestricted form, as the References record; a thread comment of 9 December 2025 derives f(n)≤(2/3−1/512)n+log⁡nf(n)\le(2/3-1/512)n+\log n for all large nn from the Coppersmith-Phillips bound of Problem 867 (claim page); the site's commentary states that bound for the unrestricted g(n)g(n) instead, which a comment of 9 April 2026 disputes, saying that the bound applies only to f(n)f(n), and the commentary was unchanged on 2026-10-07. Lenthall-Cleary's paper states the f(n)f(n) bound as f(n)≤(2/3−1/512)n+O(log⁡n)f(n)\le(2/3-1/512)n+O(\log n) (its eq. (1.2)), and Pickhardt's repeats the site's form f(n)≤g(n)≤(2/3−1/512+o(1))nf(n)\le g(n)\le(2/3-1/512+o(1))n. Lenthall-Cleary's preprint draft of 26 July 2026, a partial proof claim of 27 July 2026 on the proof-claims tab (claim page), proves f(n)≤n/2+9⋅2−7/3n2/3+O(n1/3)f(n)\le n/2+9\cdot2^{-7/3}n^{2/3}+O(n^{1/3}) by attaching to each start aia_i a block of ⌈n/(2ai)⌉\lceil n/(2a_i)\rceil consecutive terms, whose sums are distinct and mostly lie in an interval of length about n/2n/2; its finite inequality is formalized in Lean 4, neither built nor audited here, and its own remark says that the framework cannot reach f(n)=o(n)f(n)=o(n). Neither the preprint nor the formalization has a refereed version, site acceptance or independent review. Pickhardt's manuscript (dated 14 July 2026, published 31 August 2026 with the Paratelligent Research Agent named as co-author; claim page) proves f(n)≤⌈n/2⌉+⌊n/(2R)⌋+R(R−1)/2f(n)\le\lceil n/2\rceil+\lfloor n/(2R)\rfloor+R(R-1)/2 for every R≥1R\ge1, so f(n)≤n/2+((3/2)2−2/3+o(1))n2/3f(n)\le n/2+((3/2)2^{-2/3}+o(1))n^{2/3}, by packing the sums of rr consecutive terms taken from (n/(2r),n/r](n/(2r),n/r] into (n/2,n](n/2,n]; it is a pending claim with no Lean development, no refereed version and no independent review. Lower bound. The site records f(n)≥(2+o(1))n1/2f(n)\ge(2+o(1))n^{1/2} from Straus's construction for the stronger property of Problem 874, an observation in a thread comment of 24 August 2025, which as a thread post has no claim page; Pickhardt's manuscript improves this to f(n)≥(4/3−o(1))nf(n)\ge(4/\sqrt3-o(1))\sqrt n by the sequence ai=B+yia_i=B+y_i, where the yiy_i are the first kk nonnegative integers outside one residue class modulo 33 and B>3k2/16+k/8+1B>3k^2/16+k/8+1, also pending and unreviewed.

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