A192729 G.f. satisfies: A(x) = 1/(1 - x*A(x)^2/(1 - x^2*A(x)^2/(1 - x^3*A(x)^2/(1 - x^4*A(x)^2/(1 - ...))))), a recursive continued fraction.
1, 1, 3, 13, 63, 329, 1808, 10299, 60271, 360198, 2189111, 13488379, 84066176, 529037390, 3357014851, 21455604032, 137993279809, 892448240335, 5800266701499, 37864046563210, 248158092634265, 1632254493141021, 10771183395497445
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 63*x^4 + 329*x^5 + 1808*x^6 +... which satisfies A(x) = P(x)/Q(x) where P(x) = 1 - x^2*A(x)^2/(1-x) + x^6*A(x)^4/((1-x)*(1-x^2)) - x^12*A(x)^6/((1-x)*(1-x^2)*(1-x^3)) + x^20*A(x)^8/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+... Q(x) = 1 - x*A(x)^2/(1-x) + x^4*A(x)^4/((1-x)*(1-x^2)) - x^9*A(x)^6/((1-x)*(1-x^2)*(1-x^3)) + x^16*A(x)^8/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+... Explicitly, the above series begin: P(x) = 1 - x^2 - 3*x^3 - 10*x^4 - 42*x^5 - 202*x^6 - 1060*x^7 - 5862*x^8 - 33592*x^9 - 197585*x^10 - 1185867*x^11 - 7233049*x^12 +... Q(x) = 1 - x - 3*x^2 - 10*x^3 - 41*x^4 - 198*x^5 - 1041*x^6 - 5766*x^7 - 33074*x^8 - 194674*x^9 - 1168988*x^10 - 7132869*x^11 - 44097821*x^12 +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..300
- Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction.
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^2*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^2+x*O(x^n))^m); Q=sum(m=0,sqrtint(n),x^(m^2)/prod(k=1,m,1-x^k)*(-A^2+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)^2)^n / Product_{k=1..n} (1-x^k),
Q(x) = Sum_{n>=0} x^(n^2) * (-A(x)^2)^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 = 7.0656326355634513691927118582399... and c = 0.2386935555822482686868972746... - Vaclav Kotesovec, Aug 25 2017