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Erdős problem 933

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If n(n+1)=2k3lmn(n+1)=2^k3^lm, where (m,6)=1(m,6)=1, then is it true that lim sup⁡n→∞2k3lnlog⁡n=∞\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?

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3 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

933.lean

Retained formal statement3 of 3

Mahler proved (a more general result that implies in particular) that 2k3l<n1+o(1)2^k3^l<n^{1+o(1)}.

FormalConjectures/ErdosProblems/933.leanErdos933.erdos_933.variants.mahler1 lineExact file
∃ c, c =o[Filter.atTop] 1 ∧ ∀ᶠ (n : ℕ) in Filter.atTop, ↑(2 ^ Erdos933.k n * 3 ^ Erdos933.l n) < ↑n ^ (1 + c n)
SolvedStatement only, no proof