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It is known that for $p$ a prime, the following conditions are equivalent:

  1. ${2p-1\choose p-1}\equiv 1\mod{p^4}$
  2. $p^3$ divides the numerator of $\sum_{n=1}^{p-1}\frac{1}{n}$
  3. $p$ divides the numerator of the Bernoulli number $B_{p-3}$ (i.e., $(p,p-3)$ is an irregular pair)

A prime satisfying these conditions is called a Wolstenholme prime; only two are known. It is conjectured that there are infinitely many, and that their density is 0.

Are there known to be infinitely-many primes which are NOT Wolstenholme primes?

This would follow from the conjecture that their are infinitely many regular primes, but it seems that the property of not being a Wolstenholme prime is much weaker than being regular.

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Are some upper bounds already known for the $p$-adic valuation of the quantities you mentioned ? By comparison, I'm not aware of decent upper bounds on the $p$-adic valuation of $2^p-2$ or $(p-1)!+1$, for example. – François Brunault Jan 11 '12 at 22:34
Silverman proved that the abc-conjecture implies that there are infinitely many primes $p$ such that $2^{p-1} \not\equiv 1 \pmod{p^2}$ (Wieferich's criterion and the abc-conjecture, Journal of Number Theory, 1988). This suggests that your question might be difficult... – François Brunault Jan 11 '12 at 23:07

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