A351185 G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 4*x)) / (1 + 4*x).

1, 1, 1, -3, 9, -31, 153, -1075, 8689, -72031, 605201, -5282051, 49239225, -497094079, 5410919273, -62597718643, 759331611489, -9586004915007, 125701843190689, -1713676634245251, 24313707650733289, -358906747784541151, 5502327502961296825, -87382907614533531443

Offset: 0

Author

Ilya Gutkovskiy, Feb 04 2022

Keywords

• sign

Comments

Shifts 2 places left under 4th-order inverse binomial transform.

Crossrefs

Cf. A010739, A051139, A318179, A350456, A351050, A351184, A351186, A351187.

Programs

Mathematica

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]
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}]

Formula

a(0) = a(1) = 1; a(n) = Sum_{k=0..n-2} binomial(n-2,k) * (-4)^k * a(n-k-2).