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%I #6 Jun 16 2016 23:27:42
%S 1,1,2,5,15,47,153,515,1782,6293,22576,82043,301417,1117693,4177687,
%T 15723545,59538258,226656336,866983080,3330496250,12843380569,
%U 49700905874,192942347560,751191150163,2932439491143,11475503589091
%N G.f. satisfies: A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = Sum_{n>=0} x^(n(n+1)/2)*(1+x)^n.
%F G.f.: A(x) = Sum_{n>=0} [x*A(x)]^(n(n+1)/2) * (1 + x*A(x))^n.
%F Let q = x*A(x), then g.f. A(x) satisfies:
%F (1) A(x) = Sum_{n>=0} (1+q)^n*q^n*Product_{k=1..n} (1 - (1+q)*q^(2k-1))/(1 - (1+q)*q^(2k)),
%F (2) A(x) = 1/(1- q*(1+q)/(1+ q*(1-q)*(1+q)/(1- q^3*(1+q)/(1+ q^2*(1-q^2)*(1+q)/(1- q^5*(1+q)/(1+ q^3*(1-q^3)*(1+q)/(1- q^7*(1+q)/(1+ q^4*(1-q^4)*(1+q)/(1-...))))))))) (continued fraction).
%F These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression.
%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 47*x^5 + 153*x^6 +...
%e where G(x) = A(x/G(x)) = Sum_{n>=0} x^(n(n+1)/2)*(1+x)^n is illustrated by:
%e G(x) = 1 + x*(1+x) + x^3*(1+x)^2 + x^6*(1+x)^3 + x^10*(1+x)^4 +...
%e which has the binomial coefficients of the flattened Pascal's triangle:
%e G(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) +...
%e G.f. A(x) = G(x*A(x)) satisfies:
%e A(x) = 1 + x*A(x)*(1 + x*A(x)) + x^3*A(x)^3*(1 + x*A(x))^2 + x^6*A(x)^6*(1 + x*A(x))^3 + x^10*A(x)^10*(1 + x*A(x))^4 +...
%o (PARI) {a(n)=local(G=sum(m=0,(sqrtint(8*n+1)+1)\2,x^(m*(m+1)/2)*(1+x)^m)+x*O(x^n));polcoeff(G^(n+1)/(n+1),n)}
%o (PARI) /* Using a q-series expansion: */
%o {a(n)=local(A=1+x*O(x^n));for(i=1,n,q=x*A;A=sum(m=0,n,(1+q)^m*q^m*prod(k=1,m,(1-(1+q)*q^(2*k-1))/(1-(1+q)*(q+x*O(x^n))^(2*k)))));polcoeff(A,n)}
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 06 2010
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