A290917 p-INVERT of the positive integers, where p(S) = (1 - S)^2.

2, 7, 22, 67, 200, 588, 1708, 4913, 14018, 39725, 111922, 313752, 875702, 2434747, 6746350, 18636343, 51340988, 141089508, 386857888, 1058572325, 2891193242, 7882921697, 21458980582, 58330331952, 158339542250, 429274563823, 1162435429318, 3144299295403

Offset: 0

Author

Clark Kimberling, Aug 18 2017

Keywords

• nonn
• easy

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A290890 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (6, -11, 6, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290917 *)
LinearRecurrence[{6,-11,6,-1},{2,7,22,67},30] (* Harvey P. Dale, Jul 22 2024 *)

PARI

Vec((2 - x)*(1 - 2*x) / (1 - 3*x + x^2)^2 + O(x^30)) \\ Colin Barker, Aug 24 2017

Formula

G.f.: (2 - 5 x + 2 x^2)/(1 - 3 x + x^2)^2.

a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4).

(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^2.

a(n) = (((3-sqrt(5))/2)^n * (-3+sqrt(5)) * (-5+7*sqrt(5)-5*n) + 2^(-n) * (3+sqrt(5))^(n+1) * (5+7*sqrt(5)+5*n)) / 50. - Colin Barker, Aug 24 2017