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A216494 G.f. satisfies A(x) = 1 + x*A(x)^3/(1 - x^4*A(x)^10).

%I A216494 #12 Aug 28 2023 10:51:32
%S A216494 1,1,3,12,55,274,1444,7923,44803,259326,1529034,9151733,55459124,
%T A216494 339595673,2097962269,13060078469,81842038111,515867610612,
%U A216494 3268440469234,20803681980270,132963257157430,852981624781996,5490522454007139,35450567948693263,229537971398979212,1490074420399924169,9696064337840077823
%N A216494 G.f. satisfies A(x) = 1 + x*A(x)^3/(1 - x^4*A(x)^10).
%F A216494 G.f. satisfies: A(x) = 1/A(-x*A(x)^5); note that the g.f. of A001764, G(x) = 1 + x*G(x)^3, also satisfies this condition.
%F A216494 a(n) = Sum_{k=0..floor(n/4)} binomial(n-3*k-1,k) * binomial(3*n-2*k+1,n-4*k)/(3*n-2*k+1). - _Seiichi Manyama_, Aug 28 2023
%e A216494 G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 274*x^5 + 1444*x^6 + 7923*x^7 +...
%e A216494 such that
%e A216494 A(x) = 1 + x*A(x)^3 + x^5*A(x)^13 + x^9*A(x)^23 + x^13*A(x)^33 + x^17*A(x)^43 +...
%e A216494 Given (1) A(x) = 1 + x*A(x)^3 / (1 - x^4*A(x)^10),
%e A216494 suppose (2) A(x) = 1/A(-x*A(x)^5),
%e A216494 then substituting x in (1) with -x*A(x)^5 yields:
%e A216494 1/A(x) = 1 - x*A(x)^5/A(x)^3 / (1 - x^4*A(x)^20/A(x)^10),
%e A216494 which illustrates that (2) is consistent with (1).
%o A216494 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+x*A^3/(1 - x^4*A^10 +x*O(x^n))); polcoeff(A, n)}
%o A216494 for(n=0, 30, print1(a(n), ", "))
%Y A216494 Cf. A216493, A001764.
%K A216494 nonn
%O A216494 0,3
%A A216494 _Paul D. Hanna_, Sep 07 2012