OFFSET

1,1

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)

LINKS

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

FORMULA

a(n) = positions where A288640 = 1.

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]]

CROSSREFS

KEYWORD

nonn

AUTHOR

Mats Granvik, Feb 24 2017

STATUS

approved