Let $H(j)=1+\frac 12+\cdots+\frac 1j$ and let $N(j)$ be the numerator of the reduced fraction form of $H(j)$. If $p$ is a prime greater than $3$, it is known that $p^2$ divides $N(p-1)$.
My question is: For what primes $p$, if any, do powers like $p^3$ or higher divide $N(p-1)$?
These are exactly the Wolstenholme primes; as Wikipedia notes, heuristic arguments suggest that there should be infinitely many (approximately $\log\log n$ of them $\leq n$, IIRC), but only two are known.