%I
%S 2,4,6,7,9,12,13,15,19,22,24,26,27,30,31,34,38,39,42,43,45,47,48,51,
%T 53,56,57,61,62,63,65,66,70,71,75,77,80,81,82,84,85,86,87,90,91,93,95,
%U 96,97,100,101,102,106,107
%N Indices n such that sign(Im(zetazero(n))  2*Pi*e*exp(LambertW((n  11/8)/e))) = 1.
%C 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.
%C 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.
%C From _Mats Granvik_, Jun 17 2017: (Start)
%C 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))).
%C 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 FrancaLeclair asymptotic as the argument to the zeta zero counting function. See the Mathematica program below.
%C Complement to A282897.
%C (End)
%H G. C. Greubel, <a href="/A282896/b282896.txt">Table of n, a(n) for n = 1..5500</a>
%F a(n) = positions where A288640 = 1.
%t 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]]
%Y Cf. A002505, A135297, A153815, A273061, A282793, A282794, A282896, A282897.
%K nonn
%O 1,1
%A _Mats Granvik_, Feb 24 2017
