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## Mats Granvik $(function() {$(".top-badge").addSpinner().load("/users/rank?userId=8530"); });

14,134725141734693790457251983562 = ?

s = ZetaZero[1]; Limit[Zeta[s*c]/Zeta[c + s - 1], c -> 1]

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

https://oeis.org/A191898

The Franca-Leclair zeta zeros asymptotic:

http://i.stack.imgur.com/CmZmV.png

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 + It]] + IPi]/Pi, {t, 0, 60}, ImageSize -> Large]

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

https://oeis.org/A177885

Table[Im[LogGamma[ZetaZero[n]/2]/Pi - I*Im[ZetaZero[n]]/(2*Pi)*Log[Pi] + Log[Zeta[ZetaZero[n]]]/Pi + I], {n, 1, 12}]

Plot[Im[LogGamma[1/4 + It/2]/Pi - It/(2*Pi)Log[Pi] + Log[Zeta[1/2 + It]]/Pi + I], {t, 0, 60}, ImageSize -> Large]

Round[Chop[ N[Table[Im[LogGamma[1/4 + I*t/2]]/Pi - t/(2*Pi)Log[Pi] + Im[Log[Zeta[1/2 + It]]]/Pi + 1, {t, 0, 100}]]]]

From this answer: http://math.stackexchange.com/a/442686/8530

by Raymond Manzoni.

=IF(OR(ROW()=1; COLUMN()=1);0; IF(ROW()>=COLUMN();EXP(-(1-11/8/(COLUMN()-1))/EXP(1)*SUM(INDIRECT(ADDRESS(ROW()-COLUMN()+1; COLUMN(); 4)&":"&ADDRESS(ROW()-1; COLUMN(); 4); 4)));0))

(European dot-comma)

divide the result with: /2/PI()/EXP(1) and take the reciprocal.

The von Mangoldt function matrix:

=REPLACE(A1;FIND(".";A1);1;",")

http://pastebin.com/u/MatsGranvik

Clear[x]

x = x /. FindRoot[ 2*Pi/ProductLog[Log[x^x]] == Im[ZetaZero[1]], {x, 1.5}, WorkingPrecision -> 100]

Log[x]

(2*Pi)/Log[x]

Divisibility recurrence:

Clear[nn];

nn = 12

f[n_, s_] = ((s + 1)^(n - 1) + s - 1)/s;

TableForm[ FullSimplify[ Table[Integrate[Integrate[f[n, s], {n, 1, 2}], {s, 0, k}], {k, 0, nn}]]]

Table[Limit[f[n, s], s -> 0], {n, 1, nn}]

Table[Limit[D[f[n, s], s], s -> 0], {n, 1, nn}]

Table[Limit[Integrate[f[-n, s], s], s -> 0], {n, 1, nn}]

FullSimplify[ Differences[Table[Limit[Sum[f[-n, s], s], s -> 0], {n, -1, nn}]]]

Table[Limit[(-1 + n s (1 + s) + (2 + s)^n)/((1 + s)^2), s -> -1], {n, 1, nn}]

z = Integrate[((s + 1)^(-n - 1) + s - 1)/s, s]; a = Limit[z, s -> 0]

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