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G.f. A(x) satisfies: A(x) = 1 + x*A(x)*(1 + 2*x*A(x))/(1 - x*A(x) - x^2*A(x)^2).
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%I #6 Apr 08 2019 09:46:32

%S 1,1,4,14,60,267,1254,6071,30156,152714,785682,4094752,21573258,

%T 114709363,614777462,3317589966,18011350796,98307220409,539121535194,

%U 2969177051678,16415395615190,91070109305056,506843759000184,2828968117483929,15831944500607010,88818114923080102

%N G.f. A(x) satisfies: A(x) = 1 + x*A(x)*(1 + 2*x*A(x))/(1 - x*A(x) - x^2*A(x)^2).

%F G.f. A(x) satisfies: A(x) = 1 + Sum_{k>=1} Lucas(k)*x^k*A(x)^k, where Lucas = A000204.

%F G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - x - x^2)/(1 + x^2)).

%t terms = 25; A[_] = 0; Do[A[x_] = 1 + x A[x] (1 + 2 x A[x])/(1 - x A[x] - x^2 A[x]^2) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]

%t terms = 26; A[_] = 0; Do[A[x_] = 1 + Sum[LucasL[k] x^k A[x]^k, {k, 1, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]

%t terms = 26; CoefficientList[1/x InverseSeries[Series[x (1 - x - x^2)/(1 + x^2), {x, 0, terms}], x], x]

%Y Cf. A000204, A049124, A307412, A307413.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Apr 07 2019