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.

A291656 Square array T(n,k), n>=0, k>=0, read by antidiagonals: T(n,k) = ((2n-1)!!)^k * Sum_{i=1..n} 1/(2*i-1)^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 10, 23, 4, 0, 1, 28, 259, 176, 5, 0, 1, 82, 3527, 12916, 1689, 6, 0, 1, 244, 51331, 1213136, 1057221, 19524, 7, 0, 1, 730, 762743, 123296356, 885533769, 128816766, 264207, 8, 0, 1, 2188, 11406979, 12820180976, 809068942341, 1179489355164, 21878089479, 4098240, 9
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2017

Keywords

Examples

			Square array begins:
  0,   0,     0,       0,         0, ...
  1,   1,     1,       1,         1, ...
  2,   4,    10,      28,        82, ...
  3,  23,   259,    3527,     51331, ...
  4, 176, 12916, 1213136, 123296356, ...
		

Crossrefs

Columns k=0-5 give: A001477, A004041(n+1), A001824(n+1), A291585, A291586, A291587.
Rows n=0-2 give: A000004, A000012, A034472.
Main diagonal gives A291676.
Cf. A291556.

Formula

T(0,k) = 0, T(1,k) = 1 and T(n+1, k) = ((2*n-1)^k+(2*n+1)^k) * T(n, k) - (2*n-1)^(2*k) * T(n-1, k).

A335091 a(n) = ((2*n+1)!!)^3 * (Sum_{k=1..n} 1/(2*k+1)^3).

Original entry on oeis.org

0, 1, 152, 55511, 41625144, 56246975289, 124697847089808, 423322997436687375, 2088114588247920714000, 14363296872939657999716625, 133299155158711610547152961000, 1624450039177408057102079622846375, 25413656551949715361011431877529125000, 500711137690193661025654228810320074015625
Offset: 0

Views

Author

Seiichi Manyama, Sep 11 2020

Keywords

Crossrefs

Column k=3 of A335095.

Programs

  • Mathematica
    a[n_] := ((2*n + 1)!!)^3 * Sum[1/(2*k + 1)^3, {k, 1, n}]; Array[a, 14, 0] (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    {a(n) = prod(k=1, n, 2*k+1)^3*sum(k=1, n, 1/(2*k+1)^3)}
    
  • PARI
    {a(n) = if(n<2, n, ((2*n-1)^3+(2*n+1)^3)*a(n-1)-(2*n-1)^6*a(n-2))}

Formula

a(n) = ((2*n-1)^3+(2*n+1)^3) * a(n-1) - (2*n-1)^6 * a(n-2) for n>1.
a(n) ~ (7*zeta(3)/8 - 1) * 2^(3*n + 9/2) * n^(3*n + 3) / exp(3*n). - Vaclav Kotesovec, Sep 25 2020

A291587 a(n) = ((2n-1)!!)^5 * Sum_{i=1..n} 1/(2*i-1)^5.

Original entry on oeis.org

0, 1, 244, 762743, 12820180976, 757031629267449, 121921454556651769524, 45268703999809586294371407, 34375967164840303438628549400000, 48808991831991566280900452880679940625, 120855944455445379138034328603009420077012500
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Table[(2*n-1)!!^5 * Sum[1/(2*i-1)^5, {i, 1, n}], {n, 0, 12}] (* Vaclav Kotesovec, Aug 27 2017 *)

Formula

a(0) = 0, a(1) = 1, a(n+1) = ((2*n-1)^5+(2*n+1)^5)*a(n) - (2*n-1)^10*a(n-1) for n > 0.
a(n) ~ 31*Zeta(5) * 2^(5*n-5/2) * n^(5*n) / exp(5*n). - Vaclav Kotesovec, Aug 27 2017
Showing 1-3 of 3 results.