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We intend to find values of the $n$-the point $t$ such that:

$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$

and

$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$

which should be what are known as the Gram points.

We start with $(1)$ the Franca-LeClair asymptotic (A slightly changed version of formula (163) page 47 in "A theory for the zeros of Riemann ζ and other L-functions" by Guilherme Franca and Andre LeClair, July 2014) - entered by Peter Bala in the OEIS, to the Riemann zeta zeros:

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

where $W(z)$ above is the Lambert W function.

We then feed the right hand side (RHS) into:

$(2)$ $$\frac{\vartheta \left(x\right)}{\pi}$$

where $\vartheta \left(x\right)$ is the Riemann–Siegel theta function and see what happens, and notice that it appears that approximately:

$(3)$ $$k-\frac{\vartheta \left(x\right)}{\pi }-\frac{3}{2} \approx 0$$

(4):

$$k-\frac{\vartheta \left(2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)\right)}{\pi }-\frac{3}{2} \approx 0$$

with the left hand side $\text{(LHS)}$ being numerically:

$\text{LHS} =-0.000456796..., -0.000321091..., -0.000260155..., -0.000223000..., -0.000197231...,...$

for $k=1,2,3,4,5,...$ a natural number.

to remedy the situation and make it approximately equal to $n$ again, we now take formula (226) at page 65 in "A theory for the zeros of Riemann ζ and other L-functions" by Guilherme Franca and Andre LeClair, July 2014, which is:

$(5)$

$$\frac{\widetilde{y}_{n}}{2\pi}\log\left(\frac{\widetilde{y}_{n}}{2\pi e}\right)=n-\frac{11}{8}$$

(This of course resembles the Riemann–von Mangoldt formula.) We then change sides of some of the terms and get:

$(6)$ $$n=\frac{\widetilde{y}_{n}}{2\pi}\log\left(\frac{\widetilde{y}_{n}}{2\pi e}\right)+\frac{11}{8}$$

The searched for remedy is now that we add the LHS of $(4)$ to the RHS of $(6)$ to make the error smaller:

$(7)$ $$n \approx \frac{x}{2 \pi }\log \left(\frac{x}{2 \pi e}\right)+ \frac{11}{8}+k-\frac{\vartheta (x)}{\pi }-\frac{3}{2}$$

where $\widetilde{y}_{n}$ is substituted with $x$. And $\frac{11}{8}-\frac{3}{2} = \frac{1}{8}$ so $(7)$ becomes:

$(8)$ $$n \approx \frac{x}{2 \pi }\log \left(\frac{x}{2 \pi e}\right)+ \frac{1}{8}+k-\frac{\vartheta (x)}{\pi }$$

For easier mouse scrolling we repeat $(1)$:

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

and guess that the RHS of $(8)$, could fit into the place of $n$ in $(1)$ so that we get $(9)$:

$(9)$ $$x=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{x}{2 \pi }\log \left(\frac{x}{2 \pi e}\right)+ \frac{1}{8}+k-\frac{\vartheta (x)}{\pi }-\frac{11}{8}}{\exp (1)}\right)\right)$$

Now: $\frac{1}{8}-\frac{11}{8}=-\frac{5}{4}$

$(10)$ $$x=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{x}{2 \pi }\log \left(\frac{x}{2 \pi e}\right) -\frac{5}{4}+k-\frac{\vartheta (x)}{\pi }}{\exp (1)}\right)\right)$$

But: This should, but does not converge to a Franca-LeClair point. As in the comment below, the definition of Franca-LeClair point is:

$$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$$ $$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$$

and the iterative formula $(11)$ converges to a Franca-LeClair point (my own wording), for $c=\frac{1}{2}$.

However, and to the main question, we do find Gram point experimentally by setting the constant $-\frac{5}{4}$ to zero $c=0$. This then gives us the iterative formula:

$(11)$ $$x=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{x}{2 \pi }\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi }}{\exp (1)}\right)\right)$$

where as said $c=0$, and where $x$ converges to a Gram point.

A demonstration with Mathematica shows that formula $(11)$ indeed converges to a Gram point:

Clear[x, k, c, numberOfIterations, i, t]
numberOfIterations = 20;
(* Interesting values of c are: c = 0, c = 1/2, c = 1/4, c = 3/4 *)
(* c = 0 gives Gram points *)
(* c = 1/2 gives Franca-LeClair points *)
(* c = 1/4 gives non-zero self intersections: Re[Zeta[1/2+I*t]] = Im[Zeta[1/2+I*t]] *)
(* c = 3/4 gives: Re[Zeta[1/2+I*t]] = -Im[Zeta[1/2+I*t]] *)
c = 0;
t = Table[x = 1; 
  Do[x = N[Round[
       2*Pi*E^1*
        E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + k - 
             RiemannSiegelTheta[x]/Pi)/E^1], 10^-20], 20];, {i, 1, 
    numberOfIterations}]; x, {k, 0, 12}]
Zeta[1/2 + I*t]

So my question is, what is the correct way to handle the constant $c$ so that one arrives at $c=0$ and thereby the Gram points in the iterative formula $(11)$?

$$x=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{x}{2 \pi}\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi}}{\exp (1)}\right)\right)$$

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  • $\begingroup$ I maybe should add that I started by trying to find the Franca-LeClair points which I define as: $$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$$ $$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$$ which by this derivation is what should be found, but I did not find them either. Franca-Leclair points correspond to $c=\frac{1}{2}$ but for marketing purposes I wrote about "trying to find Gram Points" because Gram points are what is known. Franca-LeClair points is my own wording. $\endgroup$ – Mats Granvik Jun 20 '17 at 14:06
  • $\begingroup$ determine the Julia and Fatou sets of the maps you mention. if you do that you should discover where and why it converges or diverges $\endgroup$ – crow Jul 27 '17 at 3:51
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This is not a complete answer either, but it might be worth noticing that the equation in the iterative formula can be put in a different form as follows:

When $x=$ the $k$-th Gram point: $$x=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{x}{2 \pi}\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi}}{\exp (1)}\right)\right) $$ the equation can be written: $$\frac{x}{2 \pi \exp (1)}= \exp \left(W\left(\frac{\frac{x}{2 \pi}\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi}}{\exp (1)}\right)\right) $$

$$\log\left(\frac{x}{2 \pi \exp (1)}\right)= W\left(\frac{\frac{x}{2 \pi}\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi}}{\exp (1)}\right) $$

$$\frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi \exp (1)}\right)= \frac{\frac{x}{2 \pi}\log \left(\frac{x}{2 \pi e}\right) -c+k-\frac{\vartheta (x)}{\pi}}{\exp (1)}$$

$$\frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi e}\right) = \frac{x}{2 \pi e}\log \left(\frac{x}{2 \pi e}\right) + \frac{-c+k-\frac{\vartheta (x)}{\pi}}{e}$$

$$\frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi e}\right) = \frac{x}{2 \pi e}\log \left(\frac{x}{2 \pi e}\right) + \frac{-c+k}{e}-\frac{\vartheta (x)}{\pi e}$$

Series expansion in Mathematica of RiemannSiegelTheta[x] at infinity;

Clear[x];
FullSimplify[Series[RiemannSiegelTheta[x]/Pi/E, {x, Infinity, 6}]]

gives:

$$\frac{\vartheta (x)}{\pi e} \approx -\frac{x \left(\log \left(\frac{2 \pi }{x}\right)+1\right)}{2 (e \pi )}-\frac{1}{8 e}+\frac{1}{48 e \pi x}+\frac{7}{5760 e \pi x^3}+\frac{31}{80640 e \pi x^5}+O\left(\left(\frac{1}{x}\right)^7\right)$$

which is:

$$\frac{\vartheta (x)}{\pi e} \approx \frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi e}\right)-\frac{1}{8 e}+\frac{1}{48 e \pi x}+\frac{7}{5760 e \pi x^3}+\frac{31}{80640 e \pi x^5}+O\left(\left(\frac{1}{x}\right)^7\right)$$

at $x= \infty$ this simplifies to:

$$\frac{\vartheta (x)}{\pi e} \approx \frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi e}\right)-\frac{1}{8 e}$$

But that is as far as I can get.

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