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

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

Author

Mats Granvik, Feb 24 2017

Keywords

• nonn

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)

Links

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

Crossrefs

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

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.