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).

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)!.

A365989 Expansion of e.g.f. exp( Sum_{k>=0} x^(5*k+4) / (5*k+4) ).

Original entry on oeis.org

1, 0, 0, 0, 6, 0, 0, 0, 1260, 40320, 0, 0, 1247400, 172972800, 6227020800, 0, 3405402000, 1235025792000, 153848927232000, 6402373705728000, 19799007228000, 14783258730240000, 4243499220280320000, 545332851008225280000, 25852227123135781368000
Offset: 0

Views

Author

Seiichi Manyama, Sep 25 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(sum(k=0, N\5, x^(5*k+4)/(5*k+4)))))

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=0..floor((n-4)/5)} a(n-5*k-4)/(n-5*k-4)!.
Showing 1-3 of 3 results.