OFFSET
1,2
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
FORMULA
a(n) = positions where A288640 = 0.
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