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.

A365912 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, 1, 0, 0, 20, 0, 1, 1680, 0, 330, 369600, 1, 180180, 168168000, 13990, 163363200, 137225088001, 39041010, 232792560000, 182509367449640, 118574979600, 494730748512001, 369398970833730090, 451334037000000, 1500683270499930350, 1080492079984609149000
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+3)/(5*k+3)!))))

Formula

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

A365917 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(4*k+5) / (4*k+5)! ).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 252, 0, 0, 1, 4004, 756756, 0, 1, 65756, 69837768, 11732745024, 1, 1047508, 5772957036, 3957845988096, 623360743125121, 16781260, 475191562560, 1078063276530240, 587517500395425601, 88832646060056769732, 38604505286340
Offset: 0

Views

Author

Seiichi Manyama, Sep 23 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=40, x='x+O('x^N)); Vec(serlaplace(1/(1+x-(sinh(x)+sin(x))/2)))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-5)/4)} binomial(n,4*k+5) * a(n-4*k-5).
E.g.f.: 1 / ( 1 + x - (sinh(x) + sin(x))/2 ).

A365981 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(4*k+3) / (4*k+3) ).

Original entry on oeis.org

1, 0, 0, 2, 0, 0, 80, 720, 0, 13440, 345600, 3628800, 5913600, 296524800, 7062681600, 92559667200, 442810368000, 18037334016000, 459627769036800, 7475081822208000, 65867064606720000, 2634706112643072000, 74102151110787072000, 1464478283948359680000
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\4, x^(4*k+3)/(4*k+3)))))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-3)/4)} (4*k+2)! * binomial(n,4*k+3) * a(n-4*k-3).
Showing 1-3 of 3 results.