%I #7 Feb 08 2022 05:17:39
%S 1,1,1,-3,9,-31,153,-1075,8689,-72031,605201,-5282051,49239225,
%T -497094079,5410919273,-62597718643,759331611489,-9586004915007,
%U 125701843190689,-1713676634245251,24313707650733289,-358906747784541151,5502327502961296825,-87382907614533531443
%N G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 4*x)) / (1 + 4*x).
%C Shifts 2 places left under 4th-order inverse binomial transform.
%F a(0) = a(1) = 1; a(n) = Sum_{k=0..n-2} binomial(n-2,k) * (-4)^k * a(n-k-2).
%t nmax = 23; A[_] = 0; Do[A[x_] = 1 + x + x^2 A[x/(1 + 4 x)]/(1 + 4 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] (-4)^k a[n - k - 2], {k, 0, n - 2}]; Table[a[n], {n, 0, 23}]
%Y Cf. A010739, A051139, A318179, A350456, A351050, A351184, A351186, A351187.
%K sign
%O 0,4
%A _Ilya Gutkovskiy_, Feb 04 2022
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