## Interesting iterative formula that fails to converge to Riemann zeta zeros for c close to 1

Let the matrix $A$ be the lower triangular matrix defined as:
$$A=\text{If } n \bmod k=0 \text{ then } \frac{1}{n^c} \text{ else } 0$$
and let the matrix $B$ be the upper triangular matrix defined as:
$$B=\text{If } k \bmod n=0 \text{ then } \frac{n \cdot \mu (n)}{k^s} \text{ else } 0$$
where $\mu(n)$ is the Möbius function.

Then the matrix product of the matrices $A$ and $B$ is the symmetric matrix $T$ starting:

$$T=A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which has the Dirichlet generating function:

$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s – 1)}$$

For a while consider the three limits of the generating function:

$(1)$
$$\lim_{c\to \infty } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=\zeta (s)$$

$(2)$
For $s$ equal to a Riemann zeta zero:
$$\lim_{c\to 1 } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=0$$
$(3)$
while for $s$ not equal to a Riemann zeta zero:
$$\lim_{c\to 1 } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=\infty$$

So thereby there is/exists a continuous transformation through the value of the variable $c$ between points $t$ such that:

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

and points $t$ such that (both):

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

or in other words we can push the Franca-LeClair points (or complementary Gram points) to coincide with the Riemann zeta zeros by letting $c\to 1$.

The plot for:
$c= 1+\frac{1}{100}$
$$f(t)=\frac{\zeta (\frac{1}{2}+it) \zeta (c)}{\zeta (\frac{1}{2}+it+c-1)}$$

[![zeta zero spectrum][1]][1]

illustrates the accentuation of the zeta zeros.

There is a Fourier like series of this spectrum [over here](https://mathoverflow.net/questions/162076/is-this-riemann-zeta-function-product-equal-to-the-fourier-transform-of-the-von)
For $c=1$:
$$f(t)=\sum\limits_{n=1}^{\infty} \frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(\frac{1}{2}+it-1)}}}{n^c}$$

[![Fourierlikespectrum][2]][2]

and that works better for finding points $t$ such that the real part of the Riemann zeta function is zero, iteratively, but I find the Dirichlet generating function more interesting because it is smoother and accentuates the Riemann zeta zeros more.

The Franca-LeClair equation (131) at page 37 in their [arXiv paper](https://arxiv.org/abs/1407.4358) is:

$(4)$
$$\frac{y_n}{2\pi}\log\left(\frac{y_n}{2\pi e}\right)+\frac{1}{\pi}\lim_{\delta\to 0^{+}} \, \arg \left(\zeta \left(\delta+i y_n+\frac{1}{2}\right)\right)=n-\frac{11}{8}$$

The Franca-LeClair formula (163) at page 47:

$(5)$
$$y_n = 2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)$$

We then solve $(4)$ for $n$ and input the resulting expression into $(5)$ which gives as at mathematics stack exchange at [this question](https://math.stackexchange.com/questions/2329867/how-to-prove-that-this-iterative-formula-converges-to-the-gram-points-for-the-co) for $k=\frac{1}{2}$ the relationship:

$(6)$
$$y_n=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{y_n}{2 \pi }\log \left(\frac{y_n}{2 \pi e}\right) -k+n-\frac{\vartheta (y_n)}{\pi }}{\exp (1)}\right)\right)$$

The above can be iterated and converges to the Franca-LeClair points. With Franca-LeClair points (or complementary Gram points) as already said I mean points $t$ such that:$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$ and:$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$.

What remains is to include the Dirichlet generating function $\frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}$ to push the Franca-LeClair points closer to the Riemann zeta zeros. Since the Franca-LeClair equation (131) already corresponds to $\zeta(s)$, we only need to add the part $\frac{\zeta (c)}{\zeta (c+s-1)}$ to $(6)$. This we include as:

$$\arg \left(\frac{\zeta (c)}{\zeta (c+s-1)}\right)$$

or on the critical line:

$$\arg \left(\frac{\zeta (c)}{\zeta (\frac{1}{2}+it+c-1)}\right)$$

The right hand side in $(6)$ for $k=\frac{1}{2}$ then becomes:
$(7)$

$$t = 2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$

Plots of the family of functions $f(t,n,k,c)$ for $k=\frac{1}{2}$, $c = 1 + 1/40$ and $n=1,2,3…12$:
$$f(t,n,k,c)=2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$

[![family of functions passing through the line x at zeta zeros][3]][3]

Black dots are at $t$ equal to imaginary parts of zeta zeros.

Plots of the sharpened family of functions $f(t,n,k,c)$ for $k=\frac{1}{2}$, $c = 1 + 1/10^{11}$ and $n=1,2,3…12$:

[![sharpened family of functions passing through the line x at zeta zeros][4]][4]

> Is this proof enough that for $k=\frac{1}{2}$ the imaginary part of the $n$-th Riemann zeta zero is the point $t$ such that: $$t = \lim_{c\to 1} \, 2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$
> ?

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

Mathematica program to verify the claim:

(* start *)
Clear[k, c, n, t];
k = 1/2;
c = 1 + 1/10^100;
n = Range[50];
t = Im[ZetaZero[n]]
N[2*Pi*E*E^
LambertW[((t/(2*Pi))*Log[t/(2*Pi*E)] +
Arg[Zeta[c]/Zeta[1/2 + I*t + c – 1]]/Pi – k + n – 1 –
RiemannSiegelTheta[t]/Pi)/E], 110]
% – t
Print[“The differences above are zero which verifies the conjecture.”]
(* end *)

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## Polynomial root find algorithm

http://pastebin.com/kk9fHAaC

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## Dirichlet series for a symmetric matrix

Let $\mu(n)$ be the Möbius function

$a(n) = \sum\limits_{d|n} d \cdot \mu(d)$

$T(n,k)=a(GCD(n,k))$

$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$

$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$

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## Train of thought leading from the zeta function to the Möbius function

(*start Mathematica 8*) (*Start with Riemann zeta:*) Zeta[s] (*Take the logarithm:*) Log[Zeta[s]] (*Take the derivative:*) D[Log[Zeta[s]], s] Clear[s, c] (*Generalize it:*) Limit[Zeta[c] - Zeta[s]*Zeta[c]/Zeta[s + c - 1], c -> 1] (*See that Zeta[s]*Zeta[c]/Zeta[s+c-1] is the Dirichlet generating \ function of:*) Table[Limit[ Zeta[s]*Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)]/n^c, s -> 1], {n, 1, 12}] (*Which in turn is the Dirichlet generating function of the rows or \ columns of the symmetric matrix:*) nn = 32; A = Table[ Table[If[Mod[n, k] == 0, k^ZetaZero[k], 0], {k, 1, nn}], {n, 1, nn}]; B = Table[ Table[If[Mod[k, n] == 0, MoebiusMu[n]*n^ZetaZero[-n], 0], {k, 1, nn}], {n, 1, nn}]; MatrixForm[N[A.B]] (*For comparison,here is a plot of the von Mangoldt function from the \ matrix:*) ListLinePlot[Total[N[A.B]/Range[nn]]] (*end*)

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## The Möbius function times n

1, -2, -3, 0, -5, 6, -7, 0, 0, 10, -11, 0, -13, 14, 15, 0, -17, 0, -19, 0, 21, 22, -23, 0, 0, 26, 0, 0, -29, -30, -31, 0, 33, 34, 35, 0, -37, 38, 39, 0, -41, -42, -43, 0, 0, 46, -47, 0, 0, 0, 51, 0, -53, 0, 55, 0, 57, 58, -59, 0, -61, 62, 0, 0, 65, -66, -67, 0, 69, -70, -71, 0, -73, 74, 0, 0, 77, -78, -79, 0, 0, 82, -83, 0, 85, 86, 87, 0, -89, 0, 91, 0, 93, 94, 95, 0, -97, 0, 0, 0

Mathematica: MoebiusMu[Range[100]]*Range[100]

Keywords: core, sign, number theory

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## Arne Bergstroms paper 26 6 2013

$u=2 i \pi c_2+\log \left(i \left(2 \pi c_1+\pi \right)\right)$

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## Arne Bergstroms paper

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