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%I #7 Feb 08 2022 05:17:17
%S 1,1,1,-4,16,-69,371,-2719,24691,-243804,2479276,-25931249,284075601,
%T -3320433179,41744590941,-561939568544,8008026088996,-119496752915869,
%U 1854697111334891,-29870689367146379,499291484226079551,-8668202648905259624,156301404533216141576
%N G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 5*x)) / (1 + 5*x).
%C Shifts 2 places left under 5th-order inverse binomial transform.
%F a(0) = a(1) = 1; a(n) = Sum_{k=0..n-2} binomial(n-2,k) * (-5)^k * a(n-k-2).
%t nmax = 22; A[_] = 0; Do[A[x_] = 1 + x + x^2 A[x/(1 + 5 x)]/(1 + 5 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] (-5)^k a[n - k - 2], {k, 0, n - 2}]; Table[a[n], {n, 0, 22}]
%Y Cf. A010739, A051139, A318180, A350456, A351056, A351184, A351185, A351187.
%K sign
%O 0,4
%A _Ilya Gutkovskiy_, Feb 04 2022