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G.f. A(x) satisfies: A(x) = 1 / (1 - x + x^3 * A(x)).
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%I #11 Nov 07 2021 11:17:57

%S 1,1,1,0,-2,-5,-7,-4,10,38,70,68,-40,-329,-767,-1012,-214,2842,8642,

%T 14332,10136,-21622,-96578,-196412,-213080,96264,1037344,2608788,

%U 3698996,1121127,-10234567,-33425980,-58537486,-45735382,83471346,408899204,871127040

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

%H Seiichi Manyama, <a href="/A349047/b349047.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: (-1 + x + sqrt((1 - x)^2 + 4*x^3)) / (2*x^3).

%F a(0) = 1; a(n) = a(n-1) - Sum_{k=0..n-3} a(k) * a(n-k-3).

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

%F a(n) = F([(1-n)/3, (2-n)/3, -n/3], [2, -n], -27), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 07 2021

%t nmax = 36; A[_] = 0; Do[A[x_] = 1/(1 - x + x^3 A[x]) + O[x]^(nmax + 1) // Normal,nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 1; a[n_] := a[n] = a[n - 1] - Sum[a[k] a[n - k - 3], {k, 0, n - 3}]; Table[a[n], {n, 0, 36}]

%t Table[Sum[(-1)^k Binomial[n - k, 2 k] CatalanNumber[k], {k, 0, Floor[n/3]}], {n, 0, 36}]

%Y Cf. A000108, A007440, A023431, A100223, A214649, A343773, A349048.

%K sign

%O 0,5

%A _Ilya Gutkovskiy_, Nov 06 2021