A351186 G.f. A(x) satisfies: A(x) = 1 + x + x^2 * A(x/(1 + 5*x)) / (1 + 5*x).

1, 1, 1, -4, 16, -69, 371, -2719, 24691, -243804, 2479276, -25931249, 284075601, -3320433179, 41744590941, -561939568544, 8008026088996, -119496752915869, 1854697111334891, -29870689367146379, 499291484226079551, -8668202648905259624, 156301404533216141576

Offset: 0

Author

Ilya Gutkovskiy, Feb 04 2022

Keywords

• sign

Comments

Shifts 2 places left under 5th-order inverse binomial transform.

Crossrefs

Cf. A010739, A051139, A318180, A350456, A351056, A351184, A351185, A351187.

Programs

Mathematica

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]
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}]

Formula

a(0) = a(1) = 1; a(n) = Sum_{k=0..n-2} binomial(n-2,k) * (-5)^k * a(n-k-2).