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A288640
a(n) = (1 + sign(Im(ZetaZero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))))/2.
3
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, 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, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1
OFFSET
1
COMMENTS
2*Pi*e*exp(LambertW((n - 11/8)/e)) is the Franca-Leclair asymptotic of the nontrivial Riemann zeta zeros.
Positions of 0 are found in A282897. Positions of 1 are found in A282896.
LINKS
Guilherme França and André LeClair, A theory for the zeros of Riemann Zeta and other L-functions, arXiv:1407.4358 [math.NT], 2014, formula (163) at page 47.
FORMULA
Let ZetaZero(k) denote the zero of the Riemann zeta function on the critical line which has the k-th smallest positive imaginary part.
a(n) = (1 + sign(Im(ZetaZero(n)) - 2*Pi*e*exp(LambertW((n - 11/8)/e))))/2.
a(n) ~ (floor(Im(ZetaZero(n))/(2*Pi)*log(Im(ZetaZero(n))/(2*Pi*e)) + 11/8) - n + 1).
a(n) ~ (1 - sign(Im(zeta(1/2 + i*2*Pi*e*exp(LambertW((n - 11/8)/e))))))/2 where i = sqrt(-1).
a(n) ~ floor(2*(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi - floor(RiemannSiegelTheta(Im(ZetaZero(n)))/Pi))).
There is a way to compute a(n) without prior knowledge of the exact locations of the Riemann zeta zeros. Let:
FrancaLeclair(n) = 2*Pi*e*exp(LambertW((n - 11/8)/e)),
NumberOfZetaZeros(t) = RiemannSiegelTheta(t)/Pi + Im(log(zeta(1/2 + i*t)))/Pi where i = sqrt(-1),
Then:
a(n) = n - 1 - NumberOfZetaZeros(FrancaLeclair(n)).
Conjecture:
a(n) ~ (1 + sign(tan((-RiemannSiegelTheta(im(zetazero (n)))))))/2.
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Mats Granvik, Jun 17 2017
STATUS
approved