OPEN
This is open, and cannot be resolved with a finite computation.
If $n(n+1)=2^k3^lm$, where $(m,6)=1$, then is it true that\[\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?\]
Mahler proved (a more general result that implies in particular) that\[2^k3^l<n^{1+o(1)}.\]Erdős
[Er76d] wrote 'it is easy to see' that for infinitely many $n$\[2^k3^l>n\log n.\]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 $l=r+1$.
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Additional thanks to: Stefan Steinerberger
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T. F. Bloom, Erdős Problem #933, https://www.erdosproblems.com/933, accessed 2026-01-16
