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.

A367389 Expansion of g.f. A(x) = B(x^2)/(1 - 2*x*B(x^2)) where B(x) = 1 + 2*x*B(x)^3 is the g.f. of A153231.

Original entry on oeis.org

1, 2, 6, 16, 52, 152, 512, 1568, 5392, 16992, 59232, 190336, 669952, 2183680, 7742464, 25512448, 90974464, 302368256, 1083175424, 3625435136, 13036688384, 43889186816, 158323564544, 535639556096, 1937483350016, 6582584115200, 23865932414976, 81381420826624, 295661476642816
Offset: 0

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Author

Paul D. Hanna, Dec 20 2023

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 16*x^3 + 52*x^4 + 152*x^5 + 512*x^6 + 1568*x^7 + 5392*x^8 + 16992*x^9 + 59232*x^10 + ...
where
1/A(x) = 1 - 2*x - 2*x^2 - 8*x^4 - 56*x^6 - 480*x^8 - 4576*x^10 - 46592*x^12 - ... - 2^n*binomial(3*n-1,n)/(3*n-1) * x^(2*n) - ...
		

Crossrefs

Cf. A153231.

Programs

  • Mathematica
    CoefficientList[1/(-2*x + x/InverseSeries[Series[x - 2*x^3, {x, 0, 30}], x]), x] (* Vaclav Kotesovec, Dec 24 2023 *)
  • PARI
    {a(n) = my(A = 1/(-2*x + x/serreverse(x - 2*x^3 + O(x^(n+2))))); polcoeff(A,n)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A;
    A[#A] = polcoeff( (1 + (m+1)*x*Ser(A))^m / Ser(A)^m , m-1)/m - (m-1)^(m-2) ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = B(x^2)/(1 - 2*x*B(x^2)) where B(x) = 1 + 2*x*B(x)^3 is the g.f. of A153231.
(2) A(x) = 1/(1-2*x - Sum_{n>=1} 2^n * binomial(3*n-1,n)/(3*n-1) * x^(2*n) ).
(3) A(x) = 1/(-2*x + x/Series_Reversion( x - 2*x^3 )).
(4) [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n-1)^(n-2) for n > 1.
(5) [x^(n-1)] (1 + n*x*A(x))^n / A(x)^n = -n*(n-2)^(n-2) for n > 1.
(6) [x^(n-1)] (1 + (n-2)*x*A(x))^n / A(x)^n = -n*(5*n-14)*(n-4)^(n-3) for n >= 1.
a(n) ~ (15*sqrt(3/2)/2 + 9 + (15*sqrt(3/2)/2 - 9)*(-1)^n) * 3^(3*n/2) / (sqrt(Pi) * n^(3/2) * 2^(n/2)). - Vaclav Kotesovec, Dec 24 2023