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%I #36 Nov 08 2017 02:33:12
%S 0,1,0,1,0,1,1,0,1,0,0,1,1,0,1,0,0,0,1,0,0,1,0,1,0,1,1,0,0,1,1,0,0,1,
%T 0,0,0,1,1,0,0,1,1,0,1,0,1,1,0,0,1,0,1,0,0,1,1,0,0,0,1,1,1,0,1,1,0,0,
%U 0,1,1,0,0,0,1,0,1,0,0,1,1,1,0,1,1,1,1,0,0,1
%N a(n) = (1 + sign(Im(ZetaZero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))))/2.
%C 2*Pi*e*exp(LambertW((n - 11/8)/e)) is the Franca-Leclair asymptotic of the nontrivial Riemann zeta zeros.
%C Positions of 0 are found in A282897. Positions of 1 are found in A282896.
%H G. C. Greubel, <a href="/A288640/b288640.txt">Table of n, a(n) for n = 1..5000</a>
%H Guilherme França and André LeClair, <a href="http://arxiv.org/abs/1407.4358">A theory for the zeros of Riemann Zeta and other L-functions</a>, arXiv:1407.4358 [math.NT], 2014, formula (163) at page 47.
%H Mats Granvik, <a href="/A288640/a288640.txt">Mathematica programs to compute the sequence.</a>
%F Let ZetaZero(k) denote the zero of the Riemann zeta function on the critical line which has the k-th smallest positive imaginary part.
%F a(n) = (1 + sign(Im(ZetaZero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))))/2.
%F a(n) ~ (floor(Im(ZetaZero(n))/(2*Pi)*log(Im(ZetaZero(n))/(2*Pi*e)) + 11/8) - n + 1).
%F a(n) ~ (1 - sign(Im(zeta(1/2 + i*2*Pi*e*exp(LambertW((n - 11/8)/e))))))/2 where i = sqrt(-1).
%F a(n) ~ floor(2*(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi - floor(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi))).
%F There is a way to compute a(n) without prior knowledge of the exact locations of the Riemann zeta zeros. Let:
%F FrancaLeclair(n) = 2*Pi*e*exp(LambertW((n - 11/8)/e)),
%F NumberOfZetaZeros(t) = RiemannSiegelTheta(t)/Pi + Im(log(zeta(1/2 + i*t)))/Pi where i = sqrt(-1),
%F Then:
%F a(n) = n - 1 - NumberOfZetaZeros(FrancaLeclair(n)).
%F Conjecture:
%F a(n) ~ (1 + sign(tan((-RiemannSiegelTheta(im(zetazero (n)))))))/2.
%t FrancaLeClair[n_] = 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]; Table[(1 + Sign[Im[ZetaZero[n]] - FrancaLeClair[n]])/2, {n, 1, 90}]
%Y Cf. A002410, A273061, A282896, A282897.
%K nonn
%O 1
%A _Mats Granvik_, Jun 17 2017
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