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Problem 359

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Statement. Let a1<a2<⋯a_1<a_2<\cdots be an infinite sequence of integers such that a1=na_1=n and ai+1a_{i+1} is the least integer which is not a sum of consecutive earlier aja_js. What can be said about the density of this sequence?

In particular, in the case n=1n=1, can one prove that ak/k→∞a_k/k\to \infty and ak/k1+c→0a_k/k^{1+c}\to 0 for any c>0c>0?

Formulation. Read as the site words it, the rule fails for n≥2n\ge2. When a2a_2 is chosen, the only sum of consecutive earlier terms is a1=na_1=n, so the least positive integer that is not such a sum is 11, giving a2<a1a_2<a_1 against a1<a2<⋯a_1<a_2<\cdots; no sequence meets the site's wording. Erdős and Graham (1980, p. 59) use the same wording, with a1=ka_1=k. Formal-conjectures reads ai+1a_{i+1} as the least integer exceeding aia_i that is not such a sum, and the questions are recorded under that reading. For n=1n=1 the two readings agree, since every positive integer up to aia_i is already such a sum when ai+1a_{i+1} is chosen.

Status. Open.

Source. erdosproblems.com/359, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #359, https://www.erdosproblems.com/359.

References.

  • [An75] Andrews, George E., Research Problems: Mac Mahon's Prime Numbers of Measurement. Amer. Math. Monthly (1975), 922-923.
  • [Po77] Porubský, Š., On MacMahon's segmented numbers and related sequences. Nieuw Arch. Wisk. (3) 25 (1977), 403-408.

Formalization. Statement in formal-conjectures.

Current assessment

The standing recorded here targets the site's wording (page last edited 28 December 2025), whose rule for ai+1a_{i+1} fails as written for n≥2n\ge2 and is read as the Formulation above states. No independent assessment of proof coverage is recorded, and the problem has no claim page. The site's commentary credits Porubský [Po77] with two results for n=1n=1: for every ε>0\varepsilon>0 infinitely many kk have ak<(log⁡k)εklog⁡k/log⁡log⁡ka_k<(\log k)^\varepsilon k\log k/\log\log k, and lim sup⁡A(x)/π(x)≥1/log⁡2\limsup A(x)/\pi(x)\ge1/\log2, where A(x)A(x) counts the terms up to xx. Neither decides the limits the problem asks about: the first gives only lim inf⁡kak/k1+c=0\liminf_k a_k/k^{1+c}=0, and the second bounds A(x)A(x) from below only along a sequence of xx. They settle no instance, so they have no claim page. Andrews [An75] conjectures ak∼klog⁡k/log⁡log⁡ka_k\sim k\log k/\log\log k.

Search scope. As of 2026-10-07 the site's page lists no proof claim, its discussion thread holds one comment, of 29 November 2025, restating Porubský's first result in its lim inf⁡\liminf form, and the formal-conjectures statement file states both limits for n=1n=1 as open and has no solved variant.

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