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G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 6*x)) / (1 + 6*x).
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%I #6 Feb 08 2022 05:16:45

%S 1,1,1,-5,25,-131,793,-6137,60049,-670919,7930321,-96775853,

%T 1225237609,-16333089227,232150489129,-3531321746465,57178717416097,

%U -975918663642767,17400776511175201,-322309002081819221,6188520430773389881,-123166171374344928275,2542231599282355411897

%N G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 6*x)) / (1 + 6*x).

%C Shifts 2 places left under 6th-order inverse binomial transform.

%F a(0) = a(1) = 1; a(n) = Sum_{k=0..n-2} binomial(n-2,k) * (-6)^k * a(n-k-2).

%t nmax = 22; A[_] = 0; Do[A[x_] = 1 + x + x^2 A[x/(1 + 6 x)]/(1 + 6 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] (-6)^k a[n - k - 2], {k, 0, n - 2}]; Table[a[n], {n, 0, 22}]

%Y Cf. A010739, A051139, A318181, A350456, A351057, A351184, A351185, A351186.

%K sign

%O 0,4

%A _Ilya Gutkovskiy_, Feb 04 2022