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.

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9*x*A(x)^5.

Original entry on oeis.org

1, 3, 36, 585, 10935, 221697, 4740120, 105225318, 2402040420, 56029889025, 1329627118248, 31998624800220, 779102941714461, 19157195459506230, 475034438632316400, 11865382635213387504, 298265217964573747095, 7539795161286074350785, 191548870595159091038640, 4888023169106780049244275
Offset: 0

Views

Author

Paul D. Hanna, Jul 31 2014

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 36*x^2 + 585*x^3 + 10935*x^4 + 221697*x^5 +...
where A(x)^3 = 1 + 9*x*A(x)^5:
A(x)^3 = 1 + 9*x + 135*x^2 + 2430*x^3 + 48195*x^4 + 1015740*x^5 +...
A(x)^5 = 1 + 15*x + 270*x^2 + 5355*x^3 + 112860*x^4 + 2480058*x^5 +...
		

Crossrefs

Programs

  • Maple
    rec:= 2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13):
    f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=36},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Jan 30 2018
  • Mathematica
    nmax = 19; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - (1 + 9 x A[x]^5) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. HoldPattern[a[n_] -> k_] :> Set[a[n], k];
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)
  • PARI
    /* From A(x)^3 = 1 + 9*x*A(x)^5 : */
    {a(n) = local(A=1+x);for(i=1,n,A=(1 + 9*x*A^5 +x*O(x^n))^(1/3));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n+1)}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n + 1).
2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13). - Robert Israel, Jan 30 2018
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 20 2025
a(n) ~ 3^n * 5^(5*n/3-1/6) / (sqrt(Pi) * 2^(2*(n+2)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025