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Updated Oct 8

Problem 933

claims/: The 1 claim page of Problem 933, one per claimant's result; the problem's standing derives from them.

Statement. If $n(n+1)=2^k3^lm$ , where $(m,6)=1$ , then is it true that

\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?

Status. Open. A claimed negative answer of 7 February 2026 by Mohamed Amine Belachhab, that the limsup equals $3/\log 2$ , was refuted in the site's discussion thread by the counterexample $n=3^{14}\cdot 311$ and is recorded as rejected on its claim page.

Source. erdosproblems.com/933, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #933, https://www.erdosproblems.com/933.

References.

[Er76d] Erdős, P., Problems and results on number theoretic properties of consecutive integers and related questions. Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1975) (1976), 25-44.

Formalization. Statement in formal-conjectures.

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Linked library material

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Linked from (6)

Belachhab's claimed bound 3/log 2 on the {2,3}-part of n(n+1)problems/diophantine_problems/E0933/claims Diophantine Problems and Powers diophantine_problems/erdos_1976_problems_results_number_theoretic_properties_consecutive Conjecture (16), p. 31: limsup A(n(n+1); 2, 3)/(n log n) = ∞Inequality (15): Mahler's bound A((n+1)...(n+k); p_1, ..., p_r) < n^{1+ε}

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