Erdős Problem #933

declared status 'open'. Formalized: no. If $n(n+1)=2^k3^lm$, where $(m,6)=1$, then is it true that$$\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?$$ Current best: Mahler proved (a more general result that implies in particular) that$$2^k3^l<n^{1+o(1)}.$$Erd\H{o}s [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 $l=r+1$. Prize: no. Tags: number theory.

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