Erdős problem 933 · problems.science

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

Sources

Browse retained paths and inspect the exact material available for this Problem.

3 retained statements2415f78e850a

FormalConjectures/ErdosProblems/

933.lean

Retained formal statement2 of 3

Erdős [Er76d] wrote 'it is easy to see' that for infinitely many $n$ , $2^k 3^l > n\log n$ .

Steinerberger has noted a simple proof of this fact follows from taking $n=2^{3^r}$ for any integer $r\geq 1$ , when $k=3^r$ and $l=r+1$ .

FormalConjectures/ErdosProblems/933.leanErdos933.erdos_933.variants.lower_bound1 lineExact file

1{n | ↑(2 ^ Erdos933.k n * 3 ^ Erdos933.l n) > ↑n * Real.log ↑n}.Infinite

SolvedStatement only, no proof