Numerical calculations on the zeroes of the Riemann zeta function have reached a very high degree of refinement and sophistication and I think that the first $10^{20}$ (with positive imaginary part) or more have been calculated, all simple and all located on the critical line $\Re s=1/2$.

Question 1: what is the order of magnitude of the largest known zero?

Question 2: assuming RH and writing the $k$-th zero as $z_k=\frac12+i t_k$ with a non-decreasing sequence $t_k$ of positive numbers, is there an asymptotic formula for the size of $t_k$ in terms of $k$?

Question 3: if the Riemann hypothesis is wrong, up to a double logarithmic error, say if in fact, no better approximation than $$ \pi(x)=\text{Li}(x)+O\bigl(x^{1/2}(\ln x)( \ln \ln x)\bigr), $$ holds true, how far do the numerical computations should go to detect that "doubly logarithmic" quantity of zeroes off the critical line? If it is up to $10^{100}$, there is a chance that in the next 30 years a computer can detect a zero off line. If it is of the order of $10^{10000}$, RH won't be disproved by a computer before the sun becomes a red giant.

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    $\begingroup$ If you are not already aware of it, Odlyzko's papers here are relevant: dtc.umn.edu/~odlyzko/unpublished/index.html In view of Odlyzko's work, I'm not sure that "largest known zero" is the right concept to ask about since Odlyzko's methods allow one to "jump" to a large imaginary value and compute zeros around that point, precisely to investigate the kinds of questions you pose here. I recall that Odlyzko is agnostic about RH, and part of his reasoning is that, as you suggest here, he finds it quite possible that RH will fail at some point beyond our computational reach. $\endgroup$ Mar 8, 2017 at 16:01
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    $\begingroup$ For Question 3, doesn't that error term imply the full Riemann hypothesis? $\endgroup$
    – Will Sawin
    Aug 3, 2017 at 21:01
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    $\begingroup$ As an aside, your conditional statement that “RH won't be disproved by a computer before the sun becomes a red giant” implicitly assumes that brute force is the only approach available to a computer. $\endgroup$ Mar 6, 2021 at 17:15
  • $\begingroup$ I'm confused by question 3. Regarding question 3, that can't happen. RH is equivalent to RH is equivalent to the error term being $O(x^{\frac{1}{2} + \epsilon})$. But RH also implies a stronger result . Namely that $|\pi(x) - \operatorname{li}(x)| < \frac{1}{8\pi} \sqrt{x} \log x$ for $x \geq 2657$ (due to Schoenfeld). But the upshot is that your suggested error size would still imply RH. $\endgroup$
    – JoshuaZ
    Jul 13, 2021 at 17:24

3 Answers 3


Regarding Question 1, for their paper The zeta function on the critical line: numerical evidence for moments and random matrix theory models, Hiary and Odlyzko computed 5 billion zeros near the $10^{23}$rd zero. The last had imaginary part approximately $$ 1.30664344087959822199974045053551×10^{22} $$

See Table 2 of http://www.dtc.umn.edu/~odlyzko/doc/zeta.moments.pdf

This seems to be the current record.

Update: In “Alan Turing and the Riemann Zeta Function” by Hejhal and Odlyzko, in the book Alan Turing - His Work and Impact, Elsevier 2013, they write “It is now known that the RH is true for … some hundreds of zeros near zero number $10^{32}$” (This is $t$ near $9.04808\cdot 10^{30}$.)


For (2), according to OEIS A013629 Floor of imaginary parts of nontrivial zeros of Riemann zeta function $$ t_n \sim \frac{2\pi n}{\log{n}}$$

For (1) probably searching for "siegel z" riemann computation large will give some results.

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    $\begingroup$ This comes for 'solving' the well know asymptotic formula for the number $n$ of zeros up to height $t$, for $t$ as a function of $n$. A more precise asymptotic will use the Lambert $W$-function. $\endgroup$
    – Stopple
    Mar 8, 2017 at 17:19
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    $\begingroup$ My last comments can be illustrated by the fact that $\ln\ln 10^{100}\approx 5.4$ (quite small), $\ln\ln10^{10000}\approx 10$ (not so large), $\ln\ln10^{10^{12}}\approx 28$ (still rather small). I believe that $10^{10^{12}}$ is beyond imagination if you think that there are less than $10^{100}$ particles in the universe. Anyhow $\ln\ln$ is increasing very slowly and detecting numerically that function as not constant (!) would involve incredibly large numbers. $\endgroup$
    – Bazin
    Mar 8, 2017 at 22:45
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    $\begingroup$ $10^{10^{12}}$ is beyond imagination, but $\aleph_0$ isn't. $\endgroup$ Jul 13, 2021 at 23:48

The largest zeta zero built into Mathematica 8 and apparently also in Wolfram Alpha as of 13.7.2021 is:

0.5 + 4.99238*10^6 I

If one is happy with one significant decimal digit then this root function:

(*Mathematica start*)
Clear[f, s, n];
nn = 15;(*10^15 zeta zero*)f[x_] := Zeta[x];
(*The Franca-LeClair approximation of the zeta zeros:*)
n = 6;(*increase "n" for better precision.*)
(*The precision of N[11/8] needs to be increased too, accordingly.*)
s = 1/2 + 
   I*Table[2*Pi*Exp[1]*Exp[ProductLog[(10^n - N[11/8])/Exp[1]]], {n, 
      nn, nn}];
(*Root function for almost any function:*)
 z = Table[
   s[[j]] + 
    1/(1 - Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/
          f[k/n + s[[j]] - 1/n], {k, 1, n}]/
        Sum[((-1)^(k - 1)*Binomial[n - 1, k - 1])/f[k/n + s[[j]]], {k,
           1, n}]), {j, 1, 1}], j]

can compute double that in order of magnitude ($10^{15}$), before the built in Riemann zeta function stops evaluating.

is approximately:
{0.580063 + 2.08514*10^14 I}

The formula is based on a conjectured formula where $s$ is the value from the Franca LeClair approximation for the $j$-th Riemann zeta zero made exact by the limit:

$$s+\lim_{n \rightarrow \infty}\left(\left(1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s-\frac{1}{n}\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}}\right)^{-1}\right)$$

Mathematica code for the $10^7$ zeta zero: https://pastebin.com/tsJeGNs7


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