12
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – François Brunault
    Jan 11, 2012 at 22:34
  • 2
    $\begingroup$ 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... $\endgroup$
    – François Brunault
    Jan 11, 2012 at 23:07

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.