The 2-adic valuation of $n(n+1)$.

The 3-adic valuation of $n(n+1)$.

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 $n$, $2^k 3^l > n\log n$.

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 $l=r+1$.