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%I #10 Mar 30 2012 18:37:31
%S 1,1,4,23,158,1212,10058,88811,826982,8085950,82922624,893003234,
%T 10129641140,121552747370,1550460365622,21115793548491,
%U 308004022741254,4817224946243142,80703099826887368,1444218797390453282,27501426760092853796,554910390616969332656
%N G.f. satisfies: A(x) = Sum_{n>=0} n!*x^n*A(x)^(2*n).
%F G.f.: A(x) = sqrt((1/x)*Series_Reversion( x/[Sum_{n>=0} n!*x^n]^2 )).
%F G.f. satisfies: A(x) = 1/(1 - x*A(x)^2/(1 - x*A(x)^2/(1 - 2*x*A(x)^2/(1 - 2*x*A(x)^2/(1 - 3*x*A(x)^2/(1 - 3*x*A(x)^2/(1 - 4*x*A(x)^2/(1 - ...)))))))), a recursive continued fraction.
%F G.f. satisfies: A(x/F(x)^2) = F(x) where F(x) = Sum_{n>=0} n!*x^n.
%F a(n) = [x^n] (Sum_{k>=0} k!*x^k)^(2*n+1) / (2*n+1).
%e G.f.: A(x) = 1 + x + 4*x^2 + 23*x^3 + 158*x^4 + 1212*x^5 + 10058*x^6 +...
%e where
%e A(x) = 1 + x*A(x)^2 + 2!*x^2*A(x)^4 + 3!*x^3*A(x)^6 + 4!*x^4*A(x)^8 +...
%o (PARI) {a(n)=polcoeff((1/x*serreverse(x/sum(m=0, n, m!*x^m+x^2*O(x^n))^2))^(1/2), n)}
%o (PARI) /* Recursive continued fraction: */
%o {a(n)=local(A=1+x,CF=1+x*O(x^(n+2))); for(i=1,n, for(k=1, n+1, CF=1/(1-((n-k+1)\2+1)*x*A^2*CF));A=CF); polcoeff(A,n)}
%o (PARI) {a(n)=polcoeff(sum(k=0,n,k!*x^k +x*O(x^n))^(2*n+1)/(2*n+1),n)}
%Y Cf. A088368.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 31 2011
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