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%I #7 Mar 06 2014 12:37:36
%S 1,1,2,7,29,133,658,3471,19400,114417,709815,4619048,31446579,
%T 223419752,1652599036,12698380493,101151995810,833740791381,
%U 7098646227614,62335051895044,563749889969108,5244173616702347,50117689766439784
%N G.f.: A(x) = Sum_{n>=0} x^n * C(x)^(n^2), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
%F a(n) = Sum_{k=0..n-1} binomial((n-k)^2+2k, k) * (n-k)^2/((n-k)^2 + 2k) for n>0 with a(0)=1.
%F G.f.: A(x) = Sum_{n>=0} x^n*C(x)^n*Product_{k=1..n} (1-x*C(x)^(4*k-3))/(1-x*C(x)^(4*k-1)) where C(x) = 1 + x*C(x)^2.
%F Let q = C(x) = 1 + x*C(x)^2, then g.f. A(x) equals the continued fraction:
%F A(x) = 1/(1- q*x/(1- q*(q^2-1)*x/(1- q^5*x/(1- q^3*(q^4-1)*x/(1- q^9*x/(1- q^5*(q^6-1)*x/(1- q^13*x/(1- q^7*(q^8-1)*x/(1- ...)))))))))
%F due to an identity of a partial elliptic theta function.
%F G.f.: A(x) = 1 + x*C(x)* G( x*C(x)^2 ), where G(x) = Sum_{k>=0} x^k*(1+x)^(k^2) is the g.f. of A121689.
%e G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 29*x^4 + 133*x^5 + 658*x^6 +...
%t Flatten[{1,Table[Sum[Binomial[(n-k)^2+2*k, k] * (n-k)^2/((n-k)^2 + 2*k),{k,0,n-1}],{n,1,20}]}] (* _Vaclav Kotesovec_, Mar 06 2014 *)
%o (PARI) {a(n)=if(n<0,0,0^n+sum(k=0, n-1, binomial((n-k)^2+2*k, k)*(n-k)^2/((n-k)^2+2*k)))}
%Y Cf. A121689, A000108.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 15 2011
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