Erdős problem 933 · problems.science

If $n(n+1)=2^k3^lm$ , where $(m,6)=1$ , then is it true that $\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty$ ?

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3 retained statements2415f78e850a

FormalConjectures/ErdosProblems/

933.lean

Retained formal statement3 of 3

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

FormalConjectures/ErdosProblems/933.leanErdos933.erdos_933.variants.mahler1 lineExact file

1∃ 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