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.

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A377267 G.f. A(x) satisfies A(x) = 1/(1 - 9*x*A(x))^(2/3).

Original entry on oeis.org

1, 6, 81, 1386, 26676, 551124, 11939967, 267647490, 6155681103, 144442697256, 3444310087155, 83222570083068, 2033143391152440, 50136558534943776, 1246312401751305306, 31197886757177379570, 785732617740674763684, 19896044539519213482690
Offset: 0

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Author

Seiichi Manyama, Oct 22 2024

Keywords

Crossrefs

Programs

  • Mathematica
    A377267[n_] := 9^n*Binomial[(5*n - 1)/3, n]/(n + 1);
    Array[A377267, 20, 0] (* Paolo Xausa, Aug 05 2025 *)
  • PARI
    a(n) = 9^n*binomial(5*n/3-1/3, n)/(n+1);

Formula

G.f.: (1/x) * Series_Reversion( x*(1-9*x)^(2/3) ).
a(n) = 9^n * binomial(5*n/3 - 1/3,n)/(n+1).
G.f.: B(x)^2, where B(x) is the g.f. of A245114.
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(7/2)). - Seiichi Manyama, Jun 22 2025
E.g.f.: (2*hypergeom([2/15, 8/15, 11/15, 14/15], [1/3, 2/3, 2/3, 5/6, 1, 4/3], 5^5*x^3/4) + 12*x*hypergeom([7/15, 13/15, 16/15, 19/15], [2/3, 1, 7/6, 4/3, 4/3, 5/3], 5^5*x^3/4) + 81*x^2*hypergeom([4/5, 6/5, 7/5, 8/5], [4/3, 4/3, 3/2, 5/3, 5/3, 2], 5^5*x^3/4))/2. - Stefano Spezia, Jun 22 2025
From Karol A. Penson, Aug 23 2025: (Start)
a(n) = Integral_{x=0..15*5^(2/3)*2^(1/3)/2} x^n*W(x)*dx, where W(x) can be represented in terms of 4 different generalized hypergeometric functions of type 4F3 of argument 4*x^3/84375, W(x) = W1(x) + W2(x) + W3(x) + W4(x):
W1(x) = sqrt(15)*3^(7/10)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([-1/5, 2/15, 3/10, 7/15], [1/5, 2/5, 3/5], (4*x^3)/84375)/(18*Pi*x^(3/5)),
W2(x) = -sqrt(15)*3^(3/10)*sec(Pi/10)*sin((3*Pi)/10)*x^(3/5)*hypergeom([1/5, 8/15, 7/10, 13/15], [3/5, 4/5, 7/5], (4*x^3)/84375)/(1350*Pi),
W3(x) = -sqrt(15)*3^(1/10)*sec((3*Pi)/10)*sin(Pi/10)*x^(6/5)*hypergeom([2/5, 11/15, 9/10, 16/15], [4/5, 6/5, 8/5], (4*x^3)/84375)/(30375*Pi),
W4(x) = -sqrt(15)*3^(9/10)*csc(Pi/5)*sin(Pi/10)*x^(9/5)*hypergeom([3/5, 14/15, 11/10, 19/15], [6/5, 7/5, 9/5], (4*x^3)/84375)/(911250*Pi).
W(x) is a positive function on x = (0,15*5^(2/3)*2^(1/3)/2), is singular at x=0 with the singularity x^(-3/5), and tends to zero at x = 15*5^(2/3)*2^(1/3)/2.
Thus a(n) is a positive definite sequence. This representation is unique as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) ~ 3^n * 5^(5*n/3+1/6) / (sqrt(Pi) * 2^(2*(n+1)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025
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