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A135297
Number of Riemann zeta function zeros on the critical line, less than n.
8
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 9, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 13, 14, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 25, 25, 25, 25, 26, 26, 27, 28, 28, 28, 29, 29
OFFSET
1,22
COMMENTS
This sequence is just the cumulative distribution of the zeros.
Apart from differing singularities, the beginning of this sequence agrees with the zeta zero counting functions (RiemannSiegelTheta(n) + im(log(zeta(1/2 + i*n))))/Pi + 1 and (sign(im(zeta(1/2 + i*n))) - 1)/2 + floor(n/(2*Pi)*log(n/(2*Pi*e)) + 7/8) + 1, but disagrees later. The first deviations are seen in the continuous counting function at locations of zeta zeros with indices A153815. See also A282793 and A282794. - Mats Granvik, Feb 21 2017
REFERENCES
H. M. Edwards, Riemann's Zeta Function, Dover Publications, New York, 1974 (ISBN 978-0-486-41740-0)
LINKS
Andrew Guinand, A summation formula in the theory of prime numbers, Proc. London Math. Soc. (1948) s2-50 (1): 107-119, see page 111.
Raymond Manzoni, Riemann Zeta function - number of zeros, Mathematics Stackexchange, 2013.
FORMULA
a(n) ~ n log (n/(2*Pi*e)) / (2*Pi). - Charles R Greathouse IV, Mar 11 2011, corrected by Hal M. Switkay, Oct 03 2021
From Mats Granvik, May 13 2017: (Start)
a(n) ~ im(LogGamma(1/4 + i*n/2))/Pi - n/(2*Pi)*log(Pi) + im(log(zeta(1/2 + i*n)))/Pi + 1.
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 + 1.
a(n) ~ (RiemannSiegelTheta(n) + im(log(zeta(1/2 + i*n))))/Pi + 1.
a(n) ~ (floor(RiemannSiegelTheta(n)/Pi + 1)) + (sign(im(zeta(1/2 + i*n))) - 1)/2 + 1.
a(n) ~ n/(2*Pi)*log(n/(2*Pi*e)) + 7/8 + (im(log(zeta(1/2 + i*n))))/Pi - 1 - O(n^(-1)) + 1.
a(n) ~ floor(n/(2*Pi)*log(n/(2*Pi*e)) + 7/8) + (sign(im(zeta(1/2 + i*n))) - 1)/2 + 1.
See A286707 for exact relations.
(End)
EXAMPLE
The first nontrivial zero is 1/2 + 14.1347...*i; hence, a(15)=1.
MATHEMATICA
nn = 100; t = Table[0, {nn}]; k = 1; While[z = Im[ZetaZero[k]]; z < nn, k++; t[[Ceiling[z] ;; nn]]++]
With[{zz=Ceiling[Im[N[ZetaZero[Range[30]]]]]}, Table[If[MemberQ[zz, n], 1, 0], {n, Max[zz]}]]//Accumulate (* Harvey P. Dale, Aug 15 2017 *)
PROG
(Sage)
# This function makes sure no zeros are missed.
def A135297_list(n):
Z = lcalc.zeros(n)
R = []; pos = 1; count = 0
for z in Z:
while pos < z:
R.append(count)
pos += 1
count += 1
return R
A135297_list(30) # Peter Luschny, May 02 2014
(PARI) a(n) = #lfunzeros(L, n) \\ Felix Fröhlich, Jun 10 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved