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.

A365894 Expansion of e.g.f. exp( Sum_{k>=0} x^(3*k+4) / (3*k+4)! ).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 35, 0, 1, 330, 5775, 1, 2717, 225225, 2627626, 21828, 6782490, 290990701, 2546343368, 190030590, 22939766851, 644182060203, 4514461227804, 1607617027501, 109664100094160, 2261215037103165, 13296854061626851, 15998661864449331
Offset: 0

Views

Author

Seiichi Manyama, Sep 22 2023

Keywords

Crossrefs

Programs

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

Formula

a(0)=1; a(n) = Sum_{k=0..floor((n-4)/3)} binomial(n-1,3*k+3) * a(n-3*k-4).

A365914 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(5*k+4) / (5*k+4)! ).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 0, 70, 1, 0, 0, 34650, 1430, 1, 0, 63063000, 5105100, 54740, 1, 305540235000, 40738698000, 1134117600, 1652090, 3246670537110001, 644180662125000, 33240837630000, 314389754250, 66475579247381221350, 18359921887357050001
Offset: 0

Views

Author

Seiichi Manyama, Sep 22 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+4)/(5*k+4)!))))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-4)/5)} binomial(n,5*k+4) * a(n-5*k-4).

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.