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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. 5

%I #9 Nov 18 2017 04:48:35

%S 1,1,4,23,151,1075,8075,62996,505501,4145684,34594540,292794156,

%T 2507383158,21686318745,189162110341,1662142617881,14698913545378,

%U 130723572694407,1168419986539867,10490326933563842,94564400499455397,855552893388047193

%N 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.

%H Vaclav Kotesovec, <a href="/A192730/b192730.txt">Table of n, a(n) for n = 0..280</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html">Rogers-Ramanujan Continued Fraction</a>.

%F G.f. satisfies: A(x) = P(x)/Q(x) where

%F _ P(x) = Sum_{n>=0} x^(n*(n+1)) * (-A(x)^3)^n / Product(k=1..n} (1-x^k),

%F _ Q(x) = Sum_{n>=0} x^(n^2) * (-A(x)^3)^n / Product(k=1..n} (1-x^k),

%F due to Ramanujan's continued fraction identity.

%F a(n) ~ c * d^n / n^(3/2), where d = 9.72359087408044730447308019524191930733163... and c = 0.151620024312256318854728680725808488795... - _Vaclav Kotesovec_, Nov 18 2017

%e G.f.: A(x) = 1 + x + 4*x^2 + 23*x^3 + 151*x^4 + 1075*x^5 + 8075*x^6 +...

%e which satisfies A(x) = P(x)/Q(x) where

%e 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)) -+...

%e 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)) -+...

%e Explicitly, the above series begin:

%e 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 +...

%e 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 +...

%o (PARI) /* As a recursive continued fraction: */

%o {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)}

%o (PARI) /* By Ramanujan's continued fraction identity: */

%o {a(n)=local(A=1+x,P,Q);for(i=1,n,

%o P=sum(m=0,sqrtint(n),x^(m*(m+1))/prod(k=1,m,1-x^k)*(-A^3+x*O(x^n))^m);

%o 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)}

%Y Cf. A005169, A192728, A192729.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 08 2011

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