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Erdős problem 933

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If n(n+1)=2k3lmn(n+1)=2^k3^lm, where (m,6)=1(m,6)=1, then is it true that lim sup⁡n→∞2k3lnlog⁡n=∞\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty?

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Problem row
sha256:0771c950c614dd2a4246b02a747a0da9b81a4a75fd58ca0ac48dffd64634a130
Metadata
sha256:2356c384e6b59ff6510dd34d88686eb67041c2f19f861593da4e6c33d1a9b4d6
Observation
sha256:9c8326e865b6017b103af7611c852a06776d81964979b8ab5a1b800cd550dff6
Content
sha256:1ad05b090a1a2c70cee5ff1267479711dc9824c823f65de41d8b3932178980f8
Repository
sha256:a956b84c437202e5a02cc9e036a621bd14a302b34a75758115730bdbb77c52a4
Projection
sha256:a4a96441668a9f8b33afbcb3f696a2639f5fae145f36ddc063d7f67d5fb87ce8
Source commit
36c6d0fef71a3fec84d9dcd5eeca4e22e378f7cb
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