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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

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)!.

A361007 a(n) = numerator of (zeta(2*n,1/4) + zeta(2*n,3/4))/Pi^(2*n) where zeta is the Hurwitz zeta function.

Original entry on oeis.org

0, 2, 8, 64, 2176, 31744, 2830336, 178946048, 30460116992, 839461371904, 232711080902656, 39611984424992768, 955693069653508096, 1975371841521663868928, 1124025625663103358205952, 369906947004953565463576576, 278846808228005417477465964544
Offset: 0

Views

Author

Artur Jasinski, Mar 15 2023

Keywords

Comments

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

Examples

			tan(2*x) = 2*x + (8/3)*x^3 + (64/15)*x^5 + (2176/315)*x^7 + (31744/2835)*x^9 + ...

Crossrefs

Cf. A036279 (denominators).
Cf. A000364, A046982, A173945, A173947, A173948, A173949, A173953, A173954, A173955, A173982, A173983, A173984, A173987, A360945, A360966.

Programs

  • Mathematica
    Table[(Zeta[2*n, 1/4] + Zeta[2*n, 3/4])/Pi^(2*n), {n, 0, 25}] //
  FunctionExpand // Numerator
Table[4^(2 k) (2^(2 k) - 1) (-1)^(k + 1) BernoulliB[2 k]/(2 (2 k)!), {k, 0, 25}] // Numerator
  • PARI
    my(x='x+O('x^50), v = Vec(tan(2*x)/x)); apply(numerator, vector(#v\2, k, v[2*k-1])) \\ Michel Marcus, Apr 09 2023

Formula

a(n) = numerator( [x^(2*n-1)] tan(2*x) ).
a(n) = numerator( (-1)^(n + 1)*4^(2*n)*(2^(2*n) - 1)*B(2*n)/(2*(2*n)!) ) where B(2*n) are Bernoulli numbers.
Showing 1-2 of 2 results.

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