Erdős problem 933If n(n+1)=2k3lmn(n+1)=2^k3^lm, where (m,6)=1(m,6)=1, then is it true that lim supn→∞2k3lnlogn=∞\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?WorkspaceContinue locallyOpen sourceSign in to contributeOpen this exact Problem, source revision, and authority Repository in Workbench. This handoff does not clone, switch, upload, or execute anything.FilesErdős problem 9334 retained source recordsCanvaspublic previewSource#933→ResultNone→Checks0