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.

A365982 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(5*k+3) / (5*k+3) ).

Original entry on oeis.org

1, 0, 0, 2, 0, 0, 80, 0, 5040, 13440, 0, 3326400, 5913600, 479001600, 3632428800, 5381376000, 1399882176000, 6586804224000, 364469833728000, 5019809832576000, 18772392038400000, 2898136435138560000, 24517466017228800000, 1203790902897623040000
Offset: 0

Views

Author

Seiichi Manyama, Sep 24 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(1/(1-sum(k=0, N\5, x^(5*k+3)/(5*k+3)))))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-3)/5)} (5*k+2)! * binomial(n,5*k+3) * a(n-5*k-3).

A365980 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(2*k+3) / (2*k+3) ).

Original entry on oeis.org

1, 0, 0, 2, 0, 24, 80, 720, 5376, 53760, 490752, 6289920, 68766720, 1024607232, 13520332800, 226177695744, 3498759290880, 65257155624960, 1153246338220032, 23793010526453760, 472374431008948224, 10686755493583257600, 235406405307208826880
Offset: 0

Views

Author

Seiichi Manyama, Sep 23 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(1/(1+x-atanh(x))))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-3)/2)} (2*k+2)! * binomial(n,2*k+3) * a(n-2*k-3).
E.g.f.: 1 / ( 1 + x - arctanh(x) ).

A365973 Expansion of e.g.f. exp( Sum_{k>=0} x^(4*k+3) / (4*k+3) ).

Original entry on oeis.org

1, 0, 0, 2, 0, 0, 40, 720, 0, 2240, 172800, 3628800, 246400, 49420800, 3531340800, 87223136000, 18450432000, 3006222336000, 225434879488000, 6411312940032000, 2744461025280000, 435228787435520000, 35074217524469760000, 1126838040745697280000
Offset: 0

Views

Author

Seiichi Manyama, Sep 23 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(sum(k=0, N\4, x^(4*k+3)/(4*k+3)))))

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=0..floor((n-3)/4)} a(n-4*k-3)/(n-4*k-3)!.
Showing 1-3 of 3 results.