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A365700 G.f. satisfies A(x) = 1 + x^5*A(x)^3 / (1 - x*A(x)).

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%I A365700 #14 May 29 2025 11:29:11
%S A365700 1,0,0,0,0,1,1,1,1,1,4,8,13,19,26,46,88,163,284,466,781,1369,2468,
%T A365700 4449,7856,13724,24084,42788,76759,137785,246418,439757,786132,
%U A365700 1411148,2541368,4581906,8259500,14889781,26871106,48573823,87934175,159333544,288857216
%N A365700 G.f. satisfies A(x) = 1 + x^5*A(x)^3 / (1 - x*A(x)).
%F A365700 a(n) = Sum_{k=0..floor(n/5)} binomial(n-4*k-1,n-5*k) * binomial(n-2*k+1,k) / (n-2*k+1).
%F A365700 a(n) ~ s*sqrt((1 - r*s)*(5 - 4*r*s)/(Pi*(3 - r*s*(3 - r*s)))) / (2*n^(3/2)*r^n), where r = 0.53247307479161512230023149440436598140650951738583 and s = 1.2504652351088857309836364363044636883260447207988... are roots of the system of equations r^5*s^3 = (s-1)*(1 - r*s), (s-1)*(3 - 2*r*s) = s*(1 - r*s). - _Vaclav Kotesovec_, May 29 2025
%t A365700 terms = 43; A[_] = 0; Do[A[x_] = 1 + x^5*A[x]^3 / (1 - x*A[x])+ O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* _Stefano Spezia_, May 29 2025 *)
%o A365700 (PARI) a(n) = sum(k=0, n\5, binomial(n-4*k-1, n-5*k)*binomial(n-2*k+1, k)/(n-2*k+1));
%Y A365700 Cf. A212364, A365698, A365699, A365701, A365702.
%K A365700 nonn
%O A365700 0,11
%A A365700 _Seiichi Manyama_, Sep 16 2023