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.

A192730 G.f. satisfies: A(x) = 1/(1 - x*A(x)^3/(1 - x^2*A(x)^3/(1 - x^3*A(x)^3/(1 - x^4*A(x)^3/(1 - ...))))), a recursive continued fraction.

Original entry on oeis.org

1, 1, 4, 23, 151, 1075, 8075, 62996, 505501, 4145684, 34594540, 292794156, 2507383158, 21686318745, 189162110341, 1662142617881, 14698913545378, 130723572694407, 1168419986539867, 10490326933563842, 94564400499455397, 855552893388047193
Offset: 0

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Author

Paul D. Hanna, Jul 08 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 23*x^3 + 151*x^4 + 1075*x^5 + 8075*x^6 +...
which satisfies A(x) = P(x)/Q(x) where
P(x) = 1 - x^2*A(x)^3/(1-x) + x^6*A(x)^6/((1-x)*(1-x^2)) - x^12*A(x)^9/((1-x)*(1-x^2)*(1-x^3)) + x^20*A(x)^12/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Q(x) = 1 - x*A(x)^3/(1-x) + x^4*A(x)^6/((1-x)*(1-x^2)) - x^9*A(x)^9/((1-x)*(1-x^2)*(1-x^3)) + x^16*A(x)^12/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Explicitly, the above series begin:
P(x) = 1 - x^2 - 4*x^3 - 19*x^4 - 113*x^5 - 763*x^6 - 5557*x^7 - 42472*x^8 - 335804*x^9 - 2723164*x^10 - 22523476*x^11 - 189267247*x^12 +...
Q(x) = 1 - x - 4*x^2 - 19*x^3 - 112*x^4 - 757*x^5 - 5517*x^6 - 42188*x^7 - 333673*x^8 - 2706555*x^9 - 22390279*x^10 - 188175369*x^11 - 1602132261*x^12 +...
		

Crossrefs

Programs

  • PARI
    /* As a recursive continued fraction: */
    {a(n)=local(A=1+x,CF);for(i=1,n,CF=1+x;for(k=0,n,CF=1/(1-x^(n-k+1)*A^3*CF+x*O(x^n)));A=CF);polcoeff(A,n)}
    
  • PARI
    /* By Ramanujan's continued fraction identity: */
    {a(n)=local(A=1+x,P,Q);for(i=1,n,
    P=sum(m=0,sqrtint(n),x^(m*(m+1))/prod(k=1,m,1-x^k)*(-A^3+x*O(x^n))^m);
    Q=sum(m=0,sqrtint(n),x^(m^2)/prod(k=1,m,1-x^k)*(-A^3+x*O(x^n))^m);A=P/Q);polcoeff(A,n)}

Formula

G.f. satisfies: A(x) = P(x)/Q(x) where
P(x) = Sum_{n>=0} x^(n*(n+1)) * (-A(x)^3)^n / Product_{k=1..n} (1-x^k),
Q(x) = Sum_{n>=0} x^(n^2) * (-A(x)^3)^n / Product_{k=1..n} (1-x^k),
due to Ramanujan's continued fraction identity.
a(n) ~ c * d^n / n^(3/2), where d = 9.72359087408044730447308019524191930733163... and c = 0.151620024312256318854728680725808488795... - Vaclav Kotesovec, Nov 18 2017