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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 (list; graph; refs; listen; history; text; internal format)
 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 T. D. Noe, Table of n, a(n) for n = 1..10000 Mats Granvik, Mathematics Stackexchange Andrew Guinand, A summation formula in the theory of prime numbers, page 111 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, Mathematics Stackexchange Eric W. Weisstein, MathWorld: Riemann Zeta Function Zeros Wikipedia, Riemann Zeta Function FORMULA a(n) ~ n log n / (2 * Pi). - Charles R Greathouse IV, Mar 11 2011 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*Exp(1))] + 7/8 + (im(log (zeta (1/2 + I*n))))/Pi - 1 - BigO(n^(-1)) + 1 a(n) ~ floor(n/(2*Pi)*log(n/(2*Pi*exp(1))) + 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]]]]}, 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 Cf. A002410, A013629, A092783. Sequence in context: A108955 A108956 A289133 * A176146 A171481 A230775 Adjacent sequences:  A135294 A135295 A135296 * A135298 A135299 A135300 KEYWORD nonn AUTHOR Jean-François Alcover, Mar 09 2011 STATUS approved

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Last modified June 22 14:22 EDT 2021. Contains 345380 sequences. (Running on oeis4.)