Erdős problem 933 · problems.science

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

No current result

No reviewed Result is current in Vela Mathematics Program. Retained source material is shown below.

FormalConjectures/ErdosProblems/933.lean

Formal Conjectures

FormalConjectures/ErdosProblems/933.leanErdos933.erdos_9331 lineExact file

1True ↔ Filter.limsup (fun n => ↑↑(2 ^ Erdos933.k n * 3 ^ Erdos933.l n) / (↑↑n * ↑(Real.log ↑n))) Filter.atTop = ⊤

OpenStatement only, no proof