Take a list $L$ of roots of the $\zeta$ function, like the ones provided by Andrew Odlyzko and plug it into the Prime Counting Function $\pi(x)$ given by $$ \pi(x) \approx \operatorname{R}(x^1) - \sum_{\rho\in L}\operatorname{R}(x^{\rho}) \tag{1} $$ with $ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$ . By that we can somehow determine primes up to certain extend and accuracy.

Now, projects like GIMPS and others provide us with a bunch of exceptionally large prime numbers, see the largest here.

$\hskip.5in$ Can these large prime numbers be used to get information on the roots of $\zeta$?

And I can imagine that information might come in many ways. For example:

  • accuracy of known roots
  • bounds on the imaginary part of yet unknown $\rho$'s
  • bounds on the real part
  • ...
  • 3
    $\begingroup$ I don't believe this is possible at all. The existence of large primes doesn't tell us anything we didn't know before, as we expect about 1 out of every $\log x$ integers to be prime anyway. It is conceivable that if we knew $\pi(x)$ in a certain range, then we might be able to fish out bounds on the zeros up to a certain height - but knowing the existence of a single prime doesn't help. $\endgroup$ – Eric Naslund Jun 29 '13 at 16:35


The different zeros cancel out each others' contributions almost entirely, making it almost impossible to get information on any one root with all the noise. Even if we knew not only the large prime but its index we would not be able to do this, and since we don't have the indices of the Mersenne primes we couldn't even get started.

  • $\begingroup$ hmm, what about the Divergence of the Derivative of the Prime Counting Function. Knowing a large prime $p$ should lead to a divergent expression for $\pi'(p)$ and this depends on the roots $\rho$. So they, the roots, have to behave in such a way the $\pi'(p)$ diverges. $\endgroup$ – draks ... Jun 6 '13 at 20:53
  • $\begingroup$ and I don't see why we wouldn't be able to do this even if we knew the prime and its index. Can you explain this a little more? $\endgroup$ – draks ... Jun 6 '13 at 20:54

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