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## Mats Granvik

Engineer.

Set $$s=14.000000000000000000000000000000...i$$ with 1000 zeros after the decimal point. Set $$n=21$$ and set $$c = 1 + 1/10^{40}$$; With those parameters compute this formula:

$$s-\frac{n \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}$$

What you will get is the first 25 digits of the first Riemann zeta zero.

Of course ideally: $$n \rightarrow \infty$$ and $$c \rightarrow 1$$

$$T(n,k)$$ = https://oeis.org/A191898 $$(n,k)$$

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

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

$$\mu(n) = \underset{n = 1}{1} - \underset{a = n}{\sum_{a \geq 2}} 1 + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots$$

$$1/a^{b+i c}=1/a^b (\cos (c \log (1/a))+i \sin (c \log (1/a)))$$

1/a^(b + I*c) = 1/a^b*(Cos[c*Log[1/a]] + I*Sin[c*Log[1/a]])


N[Table[2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]], {n, 1, 12}]]

Plot[RiemannSiegelTheta[t]/Pi +
Im[Log[Zeta[1/2 + I*t]] + I*Pi]/Pi, {t, 0, 60}, ImageSize -> Large]

Table[Limit[
Zeta[s] Total[1/Divisors[n]^(s - 1)*MoebiusMu[Divisors[n]]],
s -> 1], {n, 1, 32}]


divided with: /2/PI()/EXP(1) gives reciprocal.

von Mangoldt function matrix:

http://pastebin.com/u/MatsGranvik

Divisibility:

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