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 1-2 of 2 results.

A376636 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^2)^(1/3).

Original entry on oeis.org

1, 3, 9, 18, 0, -162, -567, 0, 8019, 31590, 0, -520506, -2160756, 0, 38480265, 164549880, 0, -3072083274, -13390246485, 0, 258054995016, 1139882486490, 0, -22474826957232, -100257845970825, 0, 2011064804461548, 9039247392729582, 0, -183769714890451800
Offset: 0

Views

Author

Seiichi Manyama, Oct 23 2024

Keywords

Crossrefs

Programs

  • Mathematica
    A376636[n_] := 9^n*Binomial[(2*n + 1)/3, n]/(2*n + 1);
    Array[A376636, 35, 0] (* Paolo Xausa, Aug 04 2025 *)
  • PARI
    a(n) = 9^n*binomial(2*n/3+1/3, n)/(2*n+1);

Formula

a(n) = 9^n * binomial(2*n/3 + 1/3,n)/(2*n+1).
From Seiichi Manyama, Jun 20 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)).
a(3*n+1) = 0 for n > 0. (End)
D-finite with recurrence n*(n-2)*a(n) +54*(2*n-5)*(n-4)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A385119 G.f. A(x) satisfies A(x) = 1 + 9*x*A(x)^(5/3).

Original entry on oeis.org

1, 9, 135, 2430, 48195, 1015740, 22320522, 505692720, 11727186075, 277005649635, 6641224015140, 161193712078854, 3953072078945730, 97801207953712200, 2438092322304120720, 61182608813245896840, 1544295394480280288715, 39180450803555268621540
Offset: 0

Views

Author

Seiichi Manyama, Jun 18 2025

Keywords

Crossrefs

Programs

  • Mathematica
    A385119[n_] := 9^n*Binomial[#, n]/# & [5*n/3 + 1];
    Array[A385119, 20, 0] (* Paolo Xausa, Aug 05 2025 *)
  • PARI
    a(n) = 9^n*binomial(5*n/3+1, n)/(5*n/3+1);

Formula

a(n) = 9^n * binomial(5*n/3+1,n)/(5*n/3+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(7/3)).
G.f.: B(x)^3, where B(x) is the g.f. of A245114.
D-finite with recurrence 2*n*(n-1)*(n-2)*(2*n+3)*a(n) - 135*(5*n-9)*(5*n-3)*(5*n-12)*(5*n-6)*a(n-3) = 0. - R. J. Mathar, Jul 30 2025
a(n) ~ 3^(n+1) * 5^(5*n/3+1/2) / (sqrt(Pi) * 2^(2*(n+3)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025
Showing 1-2 of 2 results.