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.

Showing 1-3 of 3 results.

A276592 Numerator of the rational part of the sum of reciprocals of even powers of odd numbers, i.e., Sum_{k>=1} 1/(2*k-1)^(2*n).

Original entry on oeis.org

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 56963745931, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961, 14129659550745551130667441, 16103843159579478297227731
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Comments

Apart from signs, same as A089171 and A279370. - Peter Bala, Feb 07 2019

Crossrefs

Programs

  • Maple
    seq(numer(sum(1/(2*k-1)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..22);
  • Mathematica
    a[n_]:=Numerator[Pi^(-2 n) (1-2^(-2 n)) Zeta[2 n]]  (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Numerator[(-1)^n SeriesCoefficient[1/(E^x+1),{x,0,2 n-1}]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Numerator[(-1)^n Residue[Zeta[s] Gamma[s] (1-2^(1-s)),{s,1-2 n}]] (* Steven Foster Clark, Mar 11 2023 *)

Formula

a(n)/A276593(n) + A276594(n)/A276595(n) = A046988(n)/A002432(n).
a(n)/A276593(n) = (-1)^(n+1) * B_{2*n} * (2^(2*n) - 1) / (2 * (2*n)!), where B_n is the Bernoulli number. - Seiichi Manyama, Sep 03 2018

A276593 Denominator of the rational part of the sum of reciprocals of even powers of odd numbers, i.e., Sum_{k>=1} 1/(2*k-1)^(2*n).

Original entry on oeis.org

8, 96, 960, 161280, 2903040, 638668800, 49816166400, 83691159552000, 2845499424768000, 1946321606541312000, 408727537373675520000, 48662619743783485440000, 124089680346647887872000000, 174221911206693634572288000000, 70734095949917615636348928000000
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Comments

A276592(n)/a(n) * Pi^(2*n) = Sum_{k>=1} 1/(2*k-1)^(2*n) > 1. So Pi^(2*n) > a(n)/A276592(n). - Seiichi Manyama, Sep 03 2018

Examples

			From _Seiichi Manyama_, Sep 03 2018: (Start)
n |    Pi^(2*n)   |   a(n)/A276592(n)
--+---------------+------------------------------------
1 |        9.8... |           8
2 |       97.4... |          96
3 |      961.3... |         960
4 |     9488.5... |      161280/17     =     9487.0...
5 |    93648.0... |     2903040/31     =    93646.4...
6 |   924269.1... |   638668800/691    =   924267.4...
7 |  9122171.1... | 49816166400/5461   =  9122169.2... (End)
		

Crossrefs

Programs

  • Maple
    seq(denom(sum(1/(2*k-1)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..22);
  • Mathematica
    a[n_]:=Denominator[(1-2^(-2 n)) Zeta[2 n]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Denominator[1/2 SeriesCoefficient[1/(E^x+1),{x,0,2 n-1}]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Denominator[1/2 Residue[Zeta[s] Gamma[s] (1-2^(1-s)) x^(-s),{s,1-2 n}]] (* Steven Foster Clark, Mar 11 2023 *)

Formula

A276592(n)/a(n) + A276594(n)/A276595(n) = A046988(n)/A002432(n).
A276592(n)/a(n) = (-1)^(n+1) * B_{2*n} * (2^(2*n) - 1) / (2 * (2*n)!), where B_n is the Bernoulli number. - Seiichi Manyama, Sep 03 2018

A276594 Numerator of the rational part of the sum of reciprocals of even powers of even numbers, i.e., Sum_{k>=1} 1/(2*k)^(2*n).

Original entry on oeis.org

1, 1, 1, 1, 1, 691, 1, 3617, 43867, 174611, 77683, 236364091, 657931, 3392780147, 1723168255201, 7709321041217, 151628697551, 26315271553053477373, 154210205991661, 261082718496449122051, 1520097643918070802691, 2530297234481911294093
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Crossrefs

Programs

  • Maple
    seq(numer(sum(1/(2*k)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..24);

Formula

A276592(n)/A276593(n) + a(n)/A276595(n) = A046988(n)/A002432(n).
Showing 1-3 of 3 results.