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Generating function satisfies A(x) = exp(2*A(x)*x + 2*A(x^3)*x^3/3 + 2*A(x^5)*x^5/5 + 2*A(x^7)*x^7/7 +...).
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%I #19 Oct 29 2019 13:47:37

%S 1,2,6,22,86,358,1554,6950,31822,148434,702802,3369046,16319050,

%T 79749294,392711090,1946732854,9706813790,48651303118,244972282734,

%U 1238621756174,6286144819506,32011282859598,163517409895602,837631563577814,4301996341244810

%N Generating function satisfies A(x) = exp(2*A(x)*x + 2*A(x^3)*x^3/3 + 2*A(x^5)*x^5/5 + 2*A(x^7)*x^7/7 +...).

%C Which kind of trees is counted by this sequence (see formulas)? - _Joerg Arndt_, Mar 04 2015

%F G.f.: A(x) = exp(sum_{n>=0} 2*A(x^(2n+1))*x^(2n+1)/(2n+1)), A(0)=1, where A(x) = 1 + 2x + 6x^2 + 22x^3 + 86x^4 + 358x^5 +...

%F Let b(n) = a(n-1) for n>=1, then sum(n>=1, b(n)*x^n ) = x * prod(n>=1, ((1+x^n)/(1-x^n))^b(n) ); compare to A000081, A004111, and A115593. - _Joerg Arndt_, Mar 04 2015

%p spec := [S,{B=Set(S),C=PowerSet(S),S=Prod(Z,B,C)},unlabeled]: seq(combstruct[count](spec,size=n), n=1..20); # _Vladeta Jovovic_, Feb 10 2005

%t m = 23; A[_] = 0;

%t Do[A[x_] = Exp[Sum[2 A[x^k] x^k/k, {k, 1, m, 2}]] + O[x]^m // Normal, {m}];

%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Oct 29 2019 *)

%K easy,nonn

%O 0,2

%A _Paul D. Hanna_, Aug 17 2002