OFFSET
1,2
COMMENTS
Between the n-th and (n+1)-th nontrivial Riemann zeros there is exactly one extremum of the Riemann-Siegel Z function.
If n is odd then Z(t) > 0 else Z(t) < 0, where z(n) is the imaginary part of the n-th Riemann zero, z(n) < t < z(n+1), and Z'(t) = 0.
Successive records occur when gaps between two successive zeros are large.
LINKS
Tadej Kotnik, Computational estimation of the order of zeta(1/2 + i t), Mathematics of Computation, Vol. 73, No. 246 (2004), pp. 949-956.
EXAMPLE
n | a(n) | max Z(t) | t
---+------+------------+------------
1 | 1 | 2.340551 | 17.882582
2 | 3 | 2.847472 | 27.735883
3 | 5 | 2.942394 | 35.392730
4 | 8 | -3.664836 | 45.636113
5 | 14 | -4.166936 | 63.060427
6 | 25 | 4.477140 | 90.723857
7 | 33 | 5.193289 | 108.986790
8 | 64 | -5.980169 | 171.759106
9 | 79 | 6.062599 | 199.651794
MATHEMATICA
aa = {}; prec = 50; d = 30; e = 1/10^d; max = 0; Do[
p = N[Im[ZetaZero[t]], prec]; k = N[Im[ZetaZero[t + 1]], prec];
f = N[RiemannSiegelZ[(p + k)/2], prec];
g = N[RiemannSiegelZ[(p + k)/2 + e], prec];
Do[If[Abs[f - g] < 10^-40, Break[]];
If[f < g, p = (p + k)/2 + e; f = N[RiemannSiegelZ[(p + k)/2], prec];
g = N[RiemannSiegelZ[(p + k)/2 + e], prec], k = (p + k)/2;
f = N[RiemannSiegelZ[(p + k)/2], prec];
g = N[RiemannSiegelZ[(p + k)/2 + e], prec]], {m, 1, 1000}];
If[Abs[g] > max, max = Abs[g]; AppendTo[aa, t]], {t, 1, 1000}]; aa
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 22 2019
STATUS
approved