A360945 a(n) = numerator of (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) where Zeta is the Hurwitz zeta function.

1, 2, 10, 244, 554, 202084, 2162212, 1594887848, 7756604858, 9619518701764, 59259390118004, 554790995145103208, 954740563911205348, 32696580074344991138888, 105453443486621462355224, 7064702291984369672858925136, 4176926860695042104392112698

Offset: 0

Author

Artur Jasinski, Feb 26 2023

Keywords

• nonn
• frac

Comments

The function (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) is rational for every positive integer n.

For denominators see A360966.

(Zeta(2*n+1,1/4) + Zeta(2*n+1,3/4))/Zeta(2*n+1) = 4*16^n - 2*4^n; see A193475.

For numerators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A361007.

For denominators of the function (Zeta(2*n,1/4) + Zeta(2*n,3/4))/Pi^(2*n) see A036279.

(Zeta(2*n,1/4) - Zeta(2*n,3/4))/beta(2*n) = 16^n (see A001025) where beta is the Dirichlet beta function.

From the above formulas we can express Zeta(k,1/4) and Zeta(k,3/4) for every positive integer k.

Examples

			a(0) = 1 because lim_{n->0} (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) = 1.
a(3) = 244 because (Zeta(2*3+1,1/4) - Zeta(2*3+1,3/4))/Pi^(2*3+1) = 244/45.
		

Crossrefs

Cf. A000364, A046982, A173945, A173947, A173948, A173949, A173953, A173954, A173955, A173982, A173983, A173984, A173987, A360966, A361007, A361007.

Programs

Mathematica

Table[(Zeta[2*n + 1, 1/4] - Zeta[2*n + 1, 3/4]) / Pi^(2*n + 1), {n, 1, 25}] // FunctionExpand // Numerator (* Vaclav Kotesovec, Feb 27 2023 *)
t[0, 1] = 1; t[0, _] = 0;
t[n_, k_] := t[n, k] = (k-1) t[n-1, k-1] + (k+1) t[n-1, k+1];
a[n_] := Sum[t[2n, k]/(2n)!, {k, 0, 2n+1}] // Numerator;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Mar 15 2023 *)
a[n_] := SeriesCoefficient[Tan[x+Pi/4], {x, 0, 2n}] // Numerator;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Apr 15 2023 *)

PARI

a(n) = numerator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023

Formula

a(n) = A046982(2*n).

(Zeta(2*n + 1, 1/4) - Zeta(2*n + 1, 3/4))/(Pi^(2*n + 1)) = A000364(n)*(2*n + 1)*2^(2*n)/(2*n + 1)!.