OFFSET
1,26
COMMENTS
a(n) is not the same as A135297(n) - 1.
LINKS
André LeClair, The Riemann Hypothesis for physicists, page 30.
Raymond Manzoni, Mathematics Stack Exchange: Riemann Zeta function - number of zeros.
FORMULA
a(n) = im(LogGamma (1/4 + I*n/2))/Pi - n/(2*Pi)*log (Pi) + im(log(zeta(1/2 + I*n)))/Pi
a(n) = floor(im(LogGamma (1/4 + I*n/2))/Pi - n/(2*Pi)*log(Pi) + 1) + (sign(im(zeta (1/2 + I*n))) - 1)/2
a(n) = (RiemannSiegelTheta(n) + im(log (zeta (1/2 + I*n))))/Pi
a(n) = (floor(RiemannSiegelTheta(n)/Pi + 1)) + (sign(im (zeta(1/2 + I*n))) - 1)/2
a(n) = n/(2*Pi)*log[n/(2*Pi*Exp(1))] + 7/8 + (im(log (zeta (1/2 + I*n))))/Pi - 1 - BigO(n^(-1))
a(n) = floor(n/(2*Pi)*log(n/(2*Pi*exp(1))) + 7/8) + (sign(im(zeta (1/2 + I*n))) - 1)/2
MATHEMATICA
a = Table[Round[((Im[LogGamma[1/4 + I*t/2]]/Pi - t/(2*Pi)*Log[Pi] + Im[Log[Zeta[1/2 + I*t]]]/Pi))], {t, 1, 100}]
a = Table[Round[((Floor[Im[LogGamma[1/4 + I*t/2]]/Pi - t/(2*Pi)*Log[Pi] + 1] + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2))], {t, 1, 100}]
a = Table[Round[((RiemannSiegelTheta[t] + Im[Log[Zeta[1/2 + I*t]]])/Pi)], {t, 1, 100}]
a = Table[Round[((Floor[RiemannSiegelTheta[t]/Pi + 1]) + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2)], {t, 1, 100}]
a = Table[Round[(Floor[t/(2*Pi)*Log[t/(2*Pi*Exp[1])] + 7/8] + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2)], {t, 1, 100}]
a = Table[Round[(t/(2*Pi)*Log[t/(2*Pi*Exp[1])] + 7/8 + (Im[Log[Zeta[1/2 + I*t]]])/Pi - 1)], {t, 1, 100}]
CROSSREFS
KEYWORD
sign
AUTHOR
Mats Granvik, May 13 2017
STATUS
approved