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G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 3*x)) / (1 + 3*x).
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%I #6 Feb 08 2022 05:18:03

%S 1,1,1,-2,4,-11,55,-359,2359,-15230,100840,-716555,5580145,-47230091,

%T 425472229,-4013326982,39379161136,-402010392971,4279164575167,

%U -47533936734179,550239127112107,-6618018093867506,82447377648018700,-1061324336149876667,14095604842846277617

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

%C Shifts 2 places left under 3rd-order inverse binomial transform.

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

%t nmax = 24; A[_] = 0; Do[A[x_] = 1 + x + x^2 A[x/(1 + 3 x)]/(1 + 3 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = a[1] = 1; a[n_] := a[n] = Sum[Binomial[n - 2, k] (-3)^k a[n - k - 2], {k, 0, n - 2}]; Table[a[n], {n, 0, 24}]

%Y Cf. A010739, A051139, A317996, A350456, A351049, A351185, A351186, A351187.

%K sign

%O 0,4

%A _Ilya Gutkovskiy_, Feb 04 2022