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Mats Granvik
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Mats Granvik
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Can there be more than two zeta zeros in between a Gram point and a França-LeClair point?

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 $2$.