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G.f. satisfies A(x) = 1 + x^3*A(x)^2 / (1 - x*A(x)).
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%I #17 Sep 16 2023 10:42:31

%S 1,0,0,1,1,1,3,6,10,20,42,84,170,354,740,1549,3269,6945,14811,31711,

%T 68177,147091,318313,690837,1503351,3279445,7169907,15708485,34482475,

%U 75830981,167042763,368548926,814341362,1801867812,3992172298,8855912464,19668236110

%N G.f. satisfies A(x) = 1 + x^3*A(x)^2 / (1 - x*A(x)).

%F a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k-1,n-3*k) * binomial(n-k+1,k) / (n-k+1).

%F G.f.: A(x) = 2/(1 + x + sqrt(1 + x*(-2 + x - 4*x^2))). - _Vaclav Kotesovec_, Sep 16 2023

%t CoefficientList[Series[2/(1 + x + Sqrt[1 + x*(-2 + x - 4*x^2)]), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Sep 16 2023 *)

%o (PARI) a(n) = sum(k=0, n\3, binomial(n-2*k-1, n-3*k)*binomial(n-k+1, k)/(n-k+1));

%Y Cf. A023426, A023432, A054514, A114997, A365695.

%Y Cf. A365243.

%K nonn

%O 0,7

%A _Seiichi Manyama_, Sep 16 2023