cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

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

Views

Author

Mats Granvik, May 14 2016

Keywords

Comments

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.

Crossrefs

Programs

  • Mathematica
    (*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 *)
  • PARI
    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
    
  • Sage
    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

Formula

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

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

Views

Author

Vladimir Reshetnikov, Jan 02 2009

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Mathematica
    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))/(2*Pi)*log(Im(zetazero(k))/(2*Pi*e)) + 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

Views

Author

Mats Granvik, Feb 21 2017

Keywords

Comments

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.

Crossrefs

Programs

  • Mathematica
    (* 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)) - 2*Pi*e*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

Views

Author

Mats Granvik, Feb 24 2017

Keywords

Comments

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)

Crossrefs

Programs

  • Mathematica
    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]]

Formula

a(n) = positions where A288640 = 1.

A282897 Indices n such that sign(Im(zetazero(n)) - 2*Pi*e*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

Views

Author

Mats Granvik, Feb 24 2017

Keywords

Comments

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)

Crossrefs

Programs

  • Mathematica
    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]]

Formula

a(n) = positions where A288640 = 0.

A282794 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.

Original entry on oeis.org

136, 213, 256, 379, 399, 509, 531, 580, 639, 696, 705, 779, 795, 809, 871, 994, 1018, 1048, 1073, 1088, 1096, 1113, 1137, 1158, 1167, 1209, 1233, 1265, 1296, 1321, 1331, 1346, 1404, 1445, 1487
Offset: 1

Views

Author

Mats Granvik, Feb 21 2017

Keywords

Comments

Conjecture 1: The union of this sequence and A282793 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, undercounts the 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 - 15/8)/e))) = -1. Verified for the first 100000 zeta zeros.

Crossrefs

Programs

  • Mathematica
    (* 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]

A286707 a(n) = (RiemannSiegelTheta(n)+im(log(zeta(1/2+i*n))))/Pi.

Original entry on oeis.org

-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 12, 13, 13, 13, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 23, 24, 24, 24, 24, 25, 25, 26, 27, 27, 27, 28, 28
Offset: 1

Views

Author

Mats Granvik, May 13 2017

Keywords

Comments

a(n) is not the same as A135297(n) - 1.

Crossrefs

Cf. A135297.

Programs

  • 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}]

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
Showing 1-7 of 7 results.