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.

A367041 G.f. satisfies A(x) = 1 + x^2 + x*A(x)^4.

Original entry on oeis.org

1, 1, 5, 26, 168, 1195, 8988, 70318, 566388, 4665221, 39113732, 332691758, 2863778072, 24900264326, 218372530380, 1929363592870, 17157018725000, 153442147343648, 1379250344938676, 12453816724761706, 112907775890596400, 1027394297869071687
Offset: 0

Views

Author

Seiichi Manyama, Nov 03 2023

Keywords

Crossrefs

Programs

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

Formula

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

A367048 G.f. satisfies A(x) = 1 + x*A(x)^4 + x^2*A(x).

Original entry on oeis.org

1, 1, 5, 27, 177, 1270, 9645, 76206, 619913, 5156959, 43667985, 375140383, 3261467573, 28641957520, 253702185717, 2263964868768, 20334261430769, 183680693283325, 1667613040080061, 15208587941854251, 139266058402655669, 1279953660931370623
Offset: 0

Views

Author

Seiichi Manyama, Nov 03 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\2, binomial(3*n-5*k+1, k)*binomial(4*n-7*k, n-2*k)/(3*n-5*k+1));

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(3*n-5*k+1,k) * binomial(4*n-7*k,n-2*k)/(3*n-5*k+1).

A367049 G.f. satisfies A(x) = 1 + x*A(x)^4 + x^2*A(x)^2.

Original entry on oeis.org

1, 1, 5, 28, 187, 1361, 10479, 83914, 691738, 5830903, 50028259, 435454040, 3835732631, 34128555184, 306276957665, 2769050552948, 25197515469820, 230599623819217, 2121066298440282, 19597929365099640, 181814132152022195, 1692920612932871541
Offset: 0

Views

Author

Seiichi Manyama, Nov 03 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\2, binomial(3*n-4*k+1, k)*binomial(4*n-6*k, n-2*k)/(3*n-4*k+1));

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(3*n-4*k+1,k) * binomial(4*n-6*k,n-2*k)/(3*n-4*k+1).
Showing 1-3 of 3 results.