cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

Original entry on oeis.org

1, 2, 10, 244, 554, 202084, 2162212, 1594887848, 7756604858, 9619518701764, 59259390118004, 554790995145103208, 954740563911205348, 32696580074344991138888, 105453443486621462355224, 7064702291984369672858925136, 4176926860695042104392112698
Offset: 0

Views

Author

Artur Jasinski, Feb 26 2023

Keywords

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

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)!.

A360966 a(n) = denominator of (Zeta(2*n+1,1/4) - Zeta(2*n+1,3/4))/Pi^(2*n+1) where Zeta is the Hurwitz zeta function.

Original entry on oeis.org

1, 1, 3, 45, 63, 14175, 93555, 42567525, 127702575, 97692469875, 371231385525, 2143861251406875, 2275791174570375, 48076088562799171875, 95646113035463615625, 3952575621190533915703125, 1441527579493018251609375, 68739242628124575327993046875, 333120945043988326589504765625
Offset: 0

Views

Author

Artur Jasinski, Apr 09 2023

Keywords

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 numerators see A360945.

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) = 45 because (Zeta(2*3+1,1/4) - Zeta(2*3+1,3/4))/Pi^(2*3+1) = 244/45.
		

Crossrefs

Programs

  • Mathematica
    Table[(Zeta[2*n + 1, 1/4] - Zeta[2*n + 1, 3/4]) / Pi^(2*n + 1), {n, 0, 25}] // FunctionExpand // Denominator
    (* Second program: *)
    a[n_] := SeriesCoefficient[Tan[x + Pi/4], {x, 0, 2n}] // Denominator;
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 16 2023 *)
  • PARI
    a(n) = denominator(abs(eulerfrac(2*n))*(2*n + 1)*2^(2*n)/(2*n + 1)!); \\ Michel Marcus, Apr 11 2023

Formula

a(n) = A046983(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)!.
Showing 1-2 of 2 results.