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Table[Limit[ Zeta[s]*Sum[(1 - If[Mod[k, n] == 0, n, 0])/k^(s - 1), {k, 1, n}], s -> 1], {n, 1, 12}]

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

https://oeis.org/A177885

Plot[Im[LogGamma[1/4 + I*t/2]/Pi - I*t/(2*Pi)*Log[Pi] + Log[Zeta[1/2 + I*t]]/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 + I*t]]]/Pi + 1, {t, 0, 100}]]]]

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

by Raymond Manzoni.

This Excel Spreadsheet formula uses Andre LeClaire's formula to approximate the Riemann zeta zeros:

=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)

you need to divide the result with: /2/PI()/EXP(1) and take the reciprocal. tetration this is.

The von Mangoldt function matrix:

=IF(OR(ROW()=1; COLUMN()=1); 1; IF(ROW()>=COLUMN();-SUM(INDIRECT(ADDRESS(ROW()-COLUMN()+1;COLUMN(); 4)&":"&ADDRESS(ROW()-1; COLUMN(); 4); 4));-SUM(INDIRECT(ADDRESS(COLUMN()-ROW()+1;ROW(); 4)&":"&ADDRESS(COLUMN()-1; ROW(); 4); 4))))

The number of divisors of n:

Table[Sum[ Limit[((s + 1)*(-1)^(n/k*2) + s - 1)/s/2, s -> 0]^-1, {k, 1, n}], {n, 1, 32}]


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