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.

A381574 G.f. A(x) satisfies A(x) = 1/(1 - x*A(x*A(x)))^3.

Original entry on oeis.org

1, 3, 15, 118, 1206, 14712, 204385, 3143826, 52580328, 944416084, 18056415144, 365065244238, 7765839784508, 173123253590079, 4031536347783786, 97807655876704029, 2466489368705170539, 64527021089110890192, 1748298996924574135699, 48982266056400514509660
Offset: 0

Views

Author

Seiichi Manyama, Feb 28 2025

Keywords

Crossrefs

Column k=1 of A381573.

Programs

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

Formula

See A381573.
G.f.: B(x)^3, where B(x) is the g.f. of A381615.

A381029 G.f. A(x) satisfies A(x) = 1/(1 - x * A(x*A(x)^2)^2).

Original entry on oeis.org

1, 1, 3, 16, 113, 955, 9178, 97427, 1121705, 13836694, 181295019, 2507119320, 36416096984, 553461581406, 8774534872463, 144744539399484, 2479088917439527, 44004108702467428, 808171916050540308, 15335535608825061803, 300272362335527090277, 6059534345675248667550
Offset: 0

Views

Author

Seiichi Manyama, Mar 01 2025

Keywords

Crossrefs

Programs

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

Formula

Let a(n,k) = [x^n] A(x)^k.
a(n,0) = 0^n; a(n,k) = k * Sum_{j=0..n} binomial(2*n-j+k,j)/(2*n-j+k) * a(n-j,2*j).

A381649 G.f. A(x) satisfies A(x) = 1 + x * A(x)^2 * A(x*A(x)^3)^3.

Original entry on oeis.org

1, 1, 5, 44, 510, 7024, 109362, 1871530, 34590180, 682396379, 14251399805, 313170119013, 7207845252630, 173129413258492, 4327373963163746, 112289379643018983, 3018922654575996866, 83951253980821314446, 2411137697712963195801, 71427857356498491780290
Offset: 0

Views

Author

Seiichi Manyama, Mar 03 2025

Keywords

Crossrefs

Column k=1 of A381648.

Programs

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

Formula

See A381648.
Showing 1-3 of 3 results.