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.

A366088 Expansion of (1/x) * Series_Reversion( x*(1-x)^2/(1-x-x^4) ).

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 96, 264, 719, 1913, 4875, 11478, 22860, 26044, -77216, -793820, -4394125, -20304455, -85805571, -343282020, -1321898694, -4943906064, -18052305410, -64551823869, -226418611750, -779487689870, -2633172840764, -8717790419014
Offset: 0

Views

Author

Seiichi Manyama, Sep 28 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, (-1)^k*binomial(n+1, k)*binomial(2*n-3*k, n-4*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} (-1)^k * binomial(n+1,k) * binomial(2*n-3*k,n-4*k).

A366055 Expansion of (1/x) * Series_Reversion( x*(1-x)^3/(1-x+x^4) ).

Original entry on oeis.org

1, 2, 7, 30, 144, 741, 3996, 22287, 127494, 743919, 4410255, 26489073, 160843708, 985729010, 6089215057, 37875775533, 237021929322, 1491204370335, 9426547131330, 59843910602283, 381378377720469, 2438954925930558, 15646857920046108
Offset: 0

Views

Author

Seiichi Manyama, Sep 27 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n+1, k)*binomial(3*n-3*k+1, n-4*k))/(n+1);

Formula

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

A366056 Expansion of (1/x) * Series_Reversion( x*(1-x)^4/(1-x+x^4) ).

Original entry on oeis.org

1, 3, 15, 91, 613, 4408, 33143, 257400, 2048825, 16625940, 137033316, 1144010387, 9653706723, 82208879366, 705587243802, 6097408839400, 53007770199641, 463269048213536, 4067950092964440, 35871913838983980, 317533385082542404, 2820492099258807887
Offset: 0

Views

Author

Seiichi Manyama, Sep 27 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n+1, k)*binomial(4*n-3*k+2, n-4*k))/(n+1);

Formula

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