login
A273061
Nearest integer to the França-Leclair approximation 2*Pi*(n - 11/8)/LambertW((n - 11/8)/exp(1)) of the Riemann zeta zeros.
9
15, 21, 25, 30, 34, 37, 41, 44, 47, 50, 53, 56, 59, 62, 64, 67, 70, 72, 75, 77, 80, 82, 85, 87, 90, 92, 94, 97, 99, 101, 103, 106, 108, 110, 112, 114, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 142, 144, 146, 148, 150, 151, 153, 155, 157, 159, 161, 163
OFFSET
1,1
COMMENTS
This sequence is also the nearest integer to the n-th point t on the critical line such that Re(zeta(1/2+i*t))=0 and such that Im(zeta(1/2+i*t)) is not equal to zero, when excluding t=0.819545... Verified for the first 10000 cases. See Mathematica program for how to verify this.
Roger Bagula pointed out that the difference between the approximation and the points t, resembles a hyperbola.
Compare this sequence to the Gram points A002505.
The first point t such that Re(zeta(1/2+i*t))=0 and Im(zeta(1/2+i*t)) is not equal to zero, is: t(1)=14.5179196282622336505419642930... while for n=1 the França-Leclair approximation is 14.5213469530656281679750582094... This gives an error of 0.0034273248033... This decreases to 0.0003990193059... by n=10.
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.
Eric Weisstein, Gram Point.
FORMULA
a(n) = round(2*Pi*(n - 11/8)/LambertW((n - 11/8)/exp(1))).
a(n) = round(2*Pi*exp(1)*exp(LambertW((n - 11/8)/exp(1)))). - Mats Granvik, Feb 27 2017
a(n) = round(2*Pi*exp(1 + LambertW((8*(n - 3/2) + 1)/(8*e)))) after the formula in MathWorld. - Mats Granvik, Feb 25 2017
For c = 1/2 the n-th complementary Gram point x is the fixed point solution to the iterative formula: x = 2*Pi*e*e^LambertW(((x/(2*Pi))*log(x/(2*Pi*e)) - c + n - 1 - RiemannSiegelTheta(x)/Pi)/e). - Mats Granvik, Jul 24 2017
MATHEMATICA
(*The nearest integer to the França-Leclair approximation*)
Round[Table[2*Pi*(n - 11/8)/ProductLog[(n - 11/8)/Exp[1]], {n, 1, 60}]]
(*The nearest integer to t such that Re(zeta(1/2+I*t))=0 while Im(zeta(1/2+I*t))=/0*)
Round[x /. Table[FindRoot[Re[Zeta[1/2 + I*x]] == 0, {x, 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]}], {n, 1, 60}]]
Clear[a, n, g]; a[n_] := g /. FindRoot[RiemannSiegelTheta[g] == Pi*(2*n - 1)/2, {g, 2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]}]; a = Table[Round[a[n]], {n, 0, 60 - 1}] (* after Jean-François Alcover in A002505 *)
PROG
(PARI) a(n)=round(2*Pi*exp(lambertw((n-11/8)/exp(1))+1)) \\ Works for n > 1 on GP 2.8.0; Charles R Greathouse IV, May 15 2016
(Sage)
R = RealField(100)
a = lambda n: R(2*pi*(n - 11/8)/lambert_w((n - 11/8)/exp(1)))
print([a(n).round() for n in (1..60)]) # Peter Luschny, May 19 2016
KEYWORD
nonn
AUTHOR
Mats Granvik, May 14 2016
STATUS
approved