A282794 -id:A282794 - cp's OEIS frontend

A273061 Nearest integer to the França-Leclair approximation 2Pi(n - 11/8)/LambertW((n - 11/8)/exp(1)) of the Riemann zeta zeros.

15, 21, 25, 30, 34, 37, 41, 44, 47, 50, 53, 56, 59, 62, 64, 67, 70, 72, 75, 77, 80, 82, 85, 87, 90, 92, 94, 97, 99, 101, 103, 106, 108, 110, 112, 114, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 142, 144, 146, 148, 150, 151, 153, 155, 157, 159, 161, 163

Offset: 1

Mats Granvik, May 14 2016

nonn

This sequence is also the nearest integer to the n-th point t on the critical line such that Re(zeta(1/2+i*t))=0 and such that Im(zeta(1/2+i*t)) is not equal to zero, when excluding t=0.819545... Verified for the first 10000 cases. See Mathematica program for how to verify this.
Roger Bagula pointed out that the difference between the approximation and the points t, resembles a hyperbola.
Compare this sequence to the Gram points A002505.
The first point t such that Re(zeta(1/2+i*t))=0 and Im(zeta(1/2+i*t)) is not equal to zero, is: t(1)=14.5179196282622336505419642930... while for n=1 the França-Leclair approximation is 14.5213469530656281679750582094... This gives an error of 0.0034273248033... This decreases to 0.0003990193059... by n=10.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Guilherme França and André LeClair, A theory for the zeros of Riemann Zeta and other L-functions, arXiv:1407.4358 [math.NT], 2014, formula (163) at page 47.
Mats Granvik, Mathematica program for the iterative formula.
Eric Weisstein, Gram Point.

Cf. A002505, A177885, A135297, A273061, A153815, A282793, A282794, A282896, A282897.

(*The nearest integer to the França-Leclair approximation*)
Round[Table[2*Pi*(n - 11/8)/ProductLog[(n - 11/8)/Exp[1]], {n, 1, 60}]]
(*The nearest integer to t such that Re(zeta(1/2+I*t))=0 while Im(zeta(1/2+I*t))=/0*)
Round[x /. Table[FindRoot[Re[Zeta[1/2 + I*x]] == 0, {x, 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]}], {n, 1, 60}]]
Clear[a, n, g]; a[n_] := g /. FindRoot[RiemannSiegelTheta[g] == Pi*(2*n - 1)/2, {g, 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]}]; a = Table[Round[a[n]], {n, 0, 60 - 1}] (* after Jean-François Alcover in A002505 *)

a(n)=round(2*Pi*exp(lambertw((n-11/8)/exp(1))+1)) \\ Works for n > 1 on GP 2.8.0; Charles R Greathouse IV, May 15 2016

R = RealField(100)
a = lambda n: R(2*pi*(n - 11/8)/lambert_w((n - 11/8)/exp(1)))
print([a(n).round() for n in (1..60)]) # Peter Luschny, May 19 2016

a(n) = round(2*Pi*(n - 11/8)/LambertW((n - 11/8)/exp(1))).
a(n) = round(2*Pi*exp(1)*exp(LambertW((n - 11/8)/exp(1)))). - Mats Granvik, Feb 27 2017
a(n) = round(2*Pi*exp(1 + LambertW((8*(n - 3/2) + 1)/(8*e)))) after the formula in MathWorld. - Mats Granvik, Feb 25 2017
For c = 1/2 the n-th complementary Gram point x is the fixed point solution to the iterative formula: x = 2*Pi*e*e^LambertW(((x/(2*Pi))*log(x/(2*Pi*e)) - c + n - 1 - RiemannSiegelTheta(x)/Pi)/e). - Mats Granvik, Jul 24 2017

A135297 Number of Riemann zeta function zeros on the critical line, less than n.

Original entry on oeis.org

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

Jean-François Alcover, Mar 09 2011

nonn

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

			The first nontrivial zero is 1/2 + 14.1347...*i; hence, a(15)=1.

H. M. Edwards, Riemann's Zeta Function, Dover Publications, New York, 1974 (ISBN 978-0-486-41740-0)

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, Riemann Zeta function - number of zeros, Mathematics Stackexchange, 2013.
Eric W. Weisstein, MathWorld: Riemann Zeta Function Zeros
Wikipedia, Riemann Zeta Function
Index entries for sequences related to zeta function

Cf. A002410, A013629, A092783.

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 *)

a(n) = #lfunzeros(L, n) \\ Felix Fröhlich, Jun 10 2019

# 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

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)

A153815 Indices of nontrivial zeros of the Riemann zeta function where the real part of zeta'(s) becomes negative.

Original entry on oeis.org

127, 136, 196, 213, 233, 256, 289, 368, 379, 380, 399, 401, 462, 509, 519, 531, 568, 580, 596, 619, 627, 639, 655, 669, 693, 696, 705, 716, 729, 767, 779, 795, 796, 809, 820, 849, 858, 871, 888, 965, 994, 996

Offset: 1

Vladimir Reshetnikov, Jan 02 2009

nonn

From Mats Granvik, Feb 21 2017: (Start)
Conjecture 1: Indices n of nontrivial zeros of the Riemann zeta function such that: abs(floor(im(zetazero(n))/(2*Pi)*log(im(zetazero(n))/(2*Pi*e)) + 7/8) - n + 1) = 1.
Conjecture 2: The zeta zeros with these indices are also the locations where the zeta zero counting sequence A135297 disagrees with the zeta zero counting function: (RiemannSiegelTheta(t) + im(log(zeta(1/2 + I*t))))/Pi + 1. The locations where the counting function overcounts are given by A282793, and the locations where the counting function undercounts are given by A282794.
Conjecture 3: Union of A282793 and A282794.
(End)
Floor(im(zetazero(n))/(2*Pi)*log(im(zetazero(n))/(2*Pi*e)) + 7/8) - n + 1 is the branch of the argument of zeta at the n-th zero on the critical line (conjectured). - Stephen Crowley, Mar 09 2017

			Re(zeta'(zetazero(127))) < 0.

Stephen Crowley, An Expression For The Argument of zeta at Zeros on the Critical Line, arXiv:1703.03490 [math.NT], 2017.

Cf. A002505, A135297, A153815, A273061, A282793, A282794, A282896, A282897.

Select[Range[1000], N[Re[Zeta'[ZetaZero[ # ]]] < 0] &]
(* Conjecture 1: *) Monitor[Flatten[Position[Table[Abs[Floor[Im[ZetaZero[n]]/(2*Pi)*Log[Im[ZetaZero[n]]/(2*Pi*Exp[1])] + 7/8] - n + 1], {n, 1, 1000}], 1]], n] (* Mats Granvik, Feb 21 2017 *)

A282793 Indices k of nontrivial Riemann zeta zeros such that floor(Im(zetazero(k))/(2Pi)log(Im(zetazero(k))/(2Pie)) + 7/8) - k + 1 = 1.

Original entry on oeis.org

127, 196, 233, 289, 368, 380, 401, 462, 519, 568, 596, 619, 627, 655, 669, 693, 716, 729, 767, 796, 820, 849, 858, 888, 965, 996, 1029, 1035, 1044, 1114, 1179, 1210, 1251, 1277, 1291, 1308, 1332, 1343, 1431, 1457, 1488, 1496, 1499

Offset: 1

Mats Granvik, Feb 21 2017

nonn

Conjecture 1: The union of this sequence and A282794 is A153815.
Conjecture 2: The zeta zeros whose indices are terms of this sequence are the locations where the zeta zero counting function, (RiemannSiegelTheta(t) + Im(log(zeta(1/2 + i*t))))/Pi + 1 overcounts the number of zeta zeros on the critical line.
Conjecture 3: This sequence consists of the numbers k such that sign(Im(zetazero(k)) - 2*Pi*e*exp(LambertW((k - 7/8)/e))) = 1. Verified for the first 100000 zeta zeros.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000

Cf. A002505, A135297, A153815, A273061, A282794, A282896, A282897.

(* Definition: *)
fQ[n_] := Block[{a = N[Im@ ZetaZero@ n, 32]}, Floor[a (Log[a] - Log[2Pi] - 1)/(2Pi) + 7/8] == n]; Select[ Range@ 1550, fQ] (* Robert G. Wilson v, Feb 21 2017 *)
(* Definition: *)
Monitor[Flatten[Position[Table[Floor[Im[ZetaZero[n]]/(2*Pi)*Log[Im[ZetaZero[n]]/(2*Pi*Exp[1])] + 7/8] - n + 1, {n, 1, 1500}], 1]], n]
(* Conjecture 3: *)
Monitor[Flatten[Position[Table[Sign[Im[ZetaZero[n]] - 2*Pi*E*Exp[LambertW[(n - 7/8)/E]]], {n, 1, 1500}], 1]], n]

A282896 Indices n such that sign(Im(zetazero(n)) - 2Pie*exp(LambertW((n - 11/8)/e))) = 1.

Original entry on oeis.org

2, 4, 6, 7, 9, 12, 13, 15, 19, 22, 24, 26, 27, 30, 31, 34, 38, 39, 42, 43, 45, 47, 48, 51, 53, 56, 57, 61, 62, 63, 65, 66, 70, 71, 75, 77, 80, 81, 82, 84, 85, 86, 87, 90, 91, 93, 95, 96, 97, 100, 101, 102, 106, 107

Offset: 1

Mats Granvik, Feb 24 2017

nonn

The beginning of this sequence agrees with the sequence of numbers n such that floor(Im(zetazero(n))/(2*Pi)*log(Im(zetazero(n))/(2*Pi*e)) + 11/8 - n + 1) = 1, but disagrees later. The first disagreements are at n = 28813, 30264, 36720, 45925, 46590, 50513, 55258, 63925, 64573, 73615, 78374, 82247, 94463, ... and these numbers are in a(n) but not in the sequence that uses the floor function.
The beginning of this sequence also agrees with numbers n such that sign(Im(zeta(1/2 + I*2*Pi*e*exp(LambertW((n - 11/8)/e))))) = -1, but disagrees later. The first numbers that are in a(n) but not in the sequence that uses the sign function are n = 28814, 30265, 36721, 45926, 46591, ... The first numbers that are in the sequence that uses the sign function but not in a(n) are n = 39325, 44468, ... Compare this to the sequences in Remark 2 in A282897.
From Mats Granvik, Jun 17 2017: (Start)
There is at least an initial agreement between a(n) and the positions of 1 in: floor(2*(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi - floor(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi))).
There is at least an initial agreement between a(n) and the positions of 1 in the sequence computed without prior knowledge of the exact locations of the Riemann zeta zeros, that instead uses the Franca-Leclair asymptotic as the argument to the zeta zero counting function. See the Mathematica program below.
Complement to A282897.
(End)

G. C. Greubel, Table of n, a(n) for n = 1..5500

Cf. A002505, A135297, A153815, A273061, A282793, A282794, A282896, A282897.

FrancaLeClair[n_] = 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]; f = Table[Sign[ Im[ZetaZero[n]] - FrancaLeClair[n]], {n, 1, 110}]; Flatten[Position[f, 1]]

a(n) = positions where A288640 = 1.

A282897 Indices n such that sign(Im(zetazero(n)) - 2Pie*exp(LambertW((n - 11/8)/e))) = -1.

Original entry on oeis.org

1, 3, 5, 8, 10, 11, 14, 16, 17, 18, 20, 21, 23, 25, 28, 29, 32, 33, 35, 36, 37, 40, 41, 44, 46, 49, 50, 52, 54, 55, 58, 59, 60, 64, 67, 68, 69, 72, 73, 74, 76, 78, 79, 83, 88, 89, 92, 94, 98, 99, 103, 104, 105, 108, 109, 110

Offset: 1

Mats Granvik, Feb 24 2017

nonn

The beginning of a(n) agrees with the sequence of numbers n such that floor(Im(zetazero(n))/(2*Pi)*log(Im(zetazero(n))/(2*Pi*e)) + 11/8 - n + 1) = 0, but disagrees later. The first disagreements are at n = 39326, 44469, 64258, 68867, 74401, 90053, 94352, 96239, ... and these numbers are in a(n) but not in the sequence that uses the floor function.
The beginning of a(n) also agrees with numbers n such that sign(Im(zeta(1/2 + i*2*Pi*e*exp(LambertW((n - 11/8)/e))))) = 1, but disagrees later. The first numbers that are in a(n) but not in the sequence that uses the sign function are n = 39325, 44468, ... The first numbers that are in the sequence that uses the sign function but not in a(n) are n = 28814, 30265, 36721, 45926, 46591, ... Compare this to the sequences in Remark 2 in A282896.
From Mats Granvik, Jun 17 2017: (Start)
There is at least an initial agreement between a(n) and the positions of zeros in: floor(2*(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi - floor(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi))). - Mats Granvik, Jun 17 2017
There is at least an initial agreement between a(n) and the positions of -1 in the sequence computed without prior knowledge of the exact locations of the Riemann zeta zeros, that instead uses the Franca-Leclair asymptotic as the argument to the zeta zero counting function. See the Mathematica program below.
Complement to A282896.
(End)

G. C. Greubel, Table of n, a(n) for n = 1..5400
Mats Granvik, Mathematica programs to compute the sequence

Cf. A002505, A135297, A153815, A273061, A282793, A282794, A282896, A282897.

FrancaLeClair[n_] = 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]; f = Table[Sign[Im[ZetaZero[n]] - FrancaLeClair[n]], {n, 1, 110}]; Flatten[Position[f, -1]]

a(n) = positions where A288640 = 0.

cp's OEIS Frontend

A273061 Nearest integer to the França-Leclair approximation 2*Pi*(n - 11/8)/LambertW((n - 11/8)/exp(1)) of the Riemann zeta zeros.

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A135297 Number of Riemann zeta function zeros on the critical line, less than n.

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A153815 Indices of nontrivial zeros of the Riemann zeta function where the real part of zeta'(s) becomes negative.

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A282793 Indices k of nontrivial Riemann zeta zeros such that floor(Im(zetazero(k))/(2*Pi)*log(Im(zetazero(k))/(2*Pi*e)) + 7/8) - k + 1 = 1.

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A282896 Indices n such that sign(Im(zetazero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))) = 1.

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A282897 Indices n such that sign(Im(zetazero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))) = -1.

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A273061 Nearest integer to the França-Leclair approximation 2Pi(n - 11/8)/LambertW((n - 11/8)/exp(1)) of the Riemann zeta zeros.

A282793 Indices k of nontrivial Riemann zeta zeros such that floor(Im(zetazero(k))/(2Pi)log(Im(zetazero(k))/(2Pie)) + 7/8) - k + 1 = 1.

A282896 Indices n such that sign(Im(zetazero(n)) - 2Pie*exp(LambertW((n - 11/8)/e))) = 1.

A282897 Indices n such that sign(Im(zetazero(n)) - 2Pie*exp(LambertW((n - 11/8)/e))) = -1.