According to formula 163 at page 47 in the paper [A theory for the zeros of Riemann Zeta and other L-functions](https://arxiv.org/abs/1407.4358) 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](http://www.dtc.umn.edu/~odlyZko/zeta_tables/index.html) 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 $2$.