0
$\begingroup$

According to formula 163 at page 47 in the paper A theory for the zeros of Riemann Zeta and other L-functions by Guilherme França and André LeClair, the Gram points can be approximated with the formula:

$$g_n \approx \frac{2 \pi \left(n-\frac{7}{8}\right)}{W\left(\frac{n-\frac{7}{8}}{\exp (1)}\right)}$$

while the França-LeClair points are:

$$fl_n = \frac{2 \pi \left(n-\frac{11}{8}\right)}{W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)}$$

Combining them we have the formula:

$$b_n=\frac{2 \pi \left(\frac{n+1}{2}-\frac{11}{8}\right)}{W\left(\frac{\frac{n+1}{2}-\frac{11}{8}}{\exp (1)}\right)}$$

that I call the França-LeClair-Gram points.

I now downloaded the 100000 first Riemann zeta zeros from Andrew Odlyzko's site and concatenated the list of zeta zero with the list $b_n$ and sorted the resulting list from smaller to greater.

This concatenated and sorted list $d_n$ of 100000 first zeta zeros and 100000 first terms of $b_n$ starts:

$$d_n = 14.1347, 14.5213, 17.8478, 20.6557, 21.022, 23.1717, 25.0109, 25.4927,...$$ and, in the list from the program, ends at: $$d_n = ...74916.6, 74917.7, 74918.4, 74918.7, 74919.1, 74920.3, 74920.8,...$$

Question:

In the sequence $d_n$, can there be a sequence/pattern $$d_{n},d_{n+1},d_{n+2},d_{n+3},d_{n+4}$$ such that $d_{n}$ and $d_{n+4}$ are França-LeClair-Gram points (found in sequence $b_n$), while $d_{n+1}$,$d_{n+2}$,$d_{n+3}$ are imaginary parts of consecutive Riemann zeta zeros?

In other words in between $b_n$ and $b_{n+1}$, can the number of zeta zeros be greater than 2?

I looked for such a pattern of 3 zeta zeros in a row in the sequence $d_n$ but I did not find any with this program for the first 100000 zeta zeros:

(*Mathematica 8 start*)
nn = 100000;
a = Table[N[aa[[n]]], {n, 1, nn}];
(*The list aa is the downloaded list of Riemann zeta zeros from Odlyzko's site*)
(*If you want the zeta zeros from Mathematica you need to uncomment*)
(*the next line and set the variable nn equal to some*)
(*computationally reasonable number like nn=1000*)
(*a = Table[N[Im[ZetaZero[n]]], {n, 1, nn}];*)
Monitor[b = 
   Table[N[2*
      Pi*((n + 1)/2 - 11/8)/LambertW[((n + 1)/2 - 11/8)/Exp[1]]], {n, 
     1, nn}], n];
d = Sort[Flatten[{a, b}]];
Flatten[Monitor[Table[Position[d, a[[n]]], {n, 1, nn}], n]];
Flatten[Monitor[Table[Position[d, b[[n]]], {n, 1, nn}], n]];
diff = Differences[%]
Print["gaps of 3"]
Flatten[Position[diff, 3]]
Print["gaps of 4"]
Flatten[Position[diff, 4]]
(*end*)

For the pattern $d_{n},d_{n+1},d_{n+2},d_{n+3}$ such that $d_n$ and $d_{n+3}$ are Franca-LeClair-Gram points (sequence $b_n$) while $d_{n+1}$ and $d_{n+2}$ are Riemann zeta zeros, there are plenty of matches with the first starting at roughly about:

$$n=254, 269, 423, 466, 578, 736, 795, 1157, 1238, 1338,...$$

times $2$.

$\endgroup$
  • 1
    $\begingroup$ What "Combining them" means? $\endgroup$ – Matemáticos Chibchas Jan 14 '17 at 13:02
  • $\begingroup$ @MatemáticosChibchas It means dividing $n$ by $2$ in the formulas on the right hand side. Starting with the formula at the Mathworld page about Gram points we would get: $$b_n=2 \pi \exp \left(W\left(\frac{\frac{8 (n-2)}{2}+1}{8 \exp (1)}\right)+1\right)$$ What is confusing, is probably that I wrote the $\approx$ sign for Gram points and the $=$ sign for Franca-Leclair points. $\endgroup$ – Mats Granvik Jan 14 '17 at 13:07
  • $\begingroup$ @MatemáticosChibchas $\frac{7}{8}+\frac{1}{2}=\frac{11}{8}$ $\endgroup$ – Mats Granvik Jan 14 '17 at 13:17
5
$\begingroup$

It is generally believed that a positive proportion of zeros of $\zeta$ satisfy your condition. In fact, for each fixed $k$, random matrix theory predicts a distribution of the renormalized tuples $(\gamma_{n+1}-\gamma_n, \gamma_{n+2}-\gamma_n, \ldots, \gamma_{n+k}-\gamma_n)$, and the distribution is non-negative wherever it is not obviously 0 (see e.g. http://www.dam.brown.edu/people/menon/publications/notes/biane.pdf, equation (7)). Hence a positive proportion of all zeroes are expected to occur in highly concentrated clusters, which induces constellations of the type you are looking for.

Unfortunately we cannot prove much of this. In particular we do not know whether there are infinitely many roots satisfying $\gamma_{n+1}-\gamma_n\leq\frac{2\pi-\epsilon}{4\log\gamma_n}$. If there was a pair of such zeros in every interval $[T, 2T]$ for $T$ sufficiently large, then there is no Siegel zero.

Also the computational effort to find a triple of zeros between two of your points might be very large. For small $t$ the zeros of $\zeta$ do not behave as in the general range. The reason is that the sum in the Riemann-Siegel formula is essentially determined by a very short initial segment. The size of this segment increases slowly with $t$, that is, for small $t$ the first summand 1 is pretty important. So instead of looking at the first $10^6$ zeros, you might look at some interval much further out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.