As I found in wikipedia Riemann Hypothesys has been verified numerically by X. Gourdon (2004) up to 10000000000000 ($10^{13}$) zeroes.

I have a few question about how they did it. I tried to read on their official website (numbers.computation.free.fr), but could not find all answers. if I use notation $\zeta(s)$.

  1. Do I correctly understand, that they did not just find zeroes, but they verified Riemann Hypothesis? In other words, if they claim that they do it up to some number (for example $10^{13}$), from scientific point of view it does not make sense to verify it below this value and it does make sense only to go above this number?

  2. Do I correctly understand that they verified up to $10^{13}$ zeros (or up to zero with index $10^{13}$), but not up to Im(s)=$10^{13}$? If yes, how to understand up to which Im(s) they did it?

  3. Is there an explicit formula for zero with index n? What does it mean when they mention "The first column contains a zero index n"?

  4. Do I correctly understand, that to verify it, they first calculated number of zeros like described here and then tried to find all these zeroes and then they checked if Zeta function change sign in Re(s)=1/2. If Zeta function change sign in Re(s)=1/2 with specific Im(s), that means zero can иe only in the point where Re(s)=1/2.

  • $\begingroup$ I think they proved that the first $10^{13}$ non-trivial zeros of $\zeta(s)$ have real part $1/2$. Here the zeros of $\zeta(s)$ are ordered by the absolute value of $s$. $\endgroup$ Commented Aug 20, 2017 at 21:08
  • 3
    $\begingroup$ recommend Edwards store.doverpublications.com/0486417409.html inexpensive $\endgroup$
    – Will Jagy
    Commented Aug 20, 2017 at 21:09
  • $\begingroup$ @principal-ideal-domain Maybe you forgot to put link for "Here"? $\endgroup$
    – Zlelik
    Commented Aug 20, 2017 at 21:18
  • $\begingroup$ @Zlelik With "here" I meant my first sentence where I talked about "first" which implies that they are ordered in some way. $\endgroup$ Commented Aug 20, 2017 at 21:20
  • $\begingroup$ Yes for 1. 2. 4. Do you have any question then ? There is no explicit formula for the $n$th zero except the obvious one in term of the inverse of $t \mapsto \vartheta(t)+ \text{arg}(\zeta(1/2+it)$ which can be approximated with $\vartheta^{-1}(t)$ (see this) $\endgroup$
    – reuns
    Commented Aug 20, 2017 at 22:12

1 Answer 1


1,4. You can't talk about sign changes of $\zeta$, as it is complex-valued. There's a normalized $\zeta$, sometimes denoted $\Xi$, that is real-valued for real inputs, and $\Xi(\gamma)=0$ if and only if $\zeta(1/2+i\gamma)=0$. One finds zeros of $\zeta$ on the critical line by finding sign changes of $\Xi$ on the real line. A sign change indicates a zero, but conceivably a triple zero (or even higher multiplicity). And a double-zero would not have a sign change at all.

There is an integral that gives the number of zeros (with multiplicity) $\rho=\beta+i \gamma$ with $|\gamma|\leq H$. Since this must be a whole number, one can numerically evaluate the integral to within $1/2$ and get the exact count of zeros. Combining, the two types of information, Gourdon knows that he found all of the zeros and they are all on the critical line and are all simple zeros (single).

2,3. Let $N(T)$ be the number of zeros with height at most $H$. We know (from work of Trudgian) that $$\left| N(T) - \left(\frac{T}{\pi}\log \frac{T}{2\pi e} +\frac 74 \right) \right| < 0.34 \log(T) + 4$$ for $T>100$. Gourdon worked to $N(T)=2\cdot 10^{13}$ (they come in pairs), and solving that for $T$ gives $2445999556028 \leq T \leq 2445999556032$.


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