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.

A365972 Expansion of e.g.f. exp( Sum_{k>=0} x^(5*k+2) / (5*k+2) ).

Original entry on oeis.org

1, 0, 1, 0, 3, 0, 15, 720, 105, 25920, 945, 712800, 39927195, 18532800, 4522138335, 486486000, 324696683025, 20936022307200, 19281559579425, 5026330152043200, 1045172736969075, 704618287875216000, 51145202778981886575, 76432873129004106000
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\5, x^(5*k+2)/(5*k+2)))))

Formula

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

A365978 Expansion of e.g.f. 1 / ( 1 - Sum_{k>=0} x^(3*k+2) / (3*k+2) ).

Original entry on oeis.org

1, 0, 1, 0, 6, 24, 90, 1008, 7560, 54432, 712152, 7620480, 81130896, 1266632640, 17587441872, 246734377344, 4527397929600, 77238618702336, 1340945212763520, 28407941067018240, 574640938744314624, 11868502219930137600, 285787326567523173120
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-sum(k=0, N\3, x^(3*k+2)/(3*k+2)))))

Formula

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

A365990 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, 6, 0, 0, 0, 2520, 40320, 0, 0, 7484400, 345945600, 6227020800, 0, 81729648000, 7410154752000, 307697854464000, 6402373705728000, 2375880867360000, 354798209525760000, 25460995321681920000, 1090665702016450560000, 26003493399464380800000
Offset: 0

Views

Author

Seiichi Manyama, Sep 25 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)} (5*k+3)! * binomial(n,5*k+4) * a(n-5*k-4).
Showing 1-3 of 3 results.