A290926 p-INVERT of the positive integers, where p(S) = (1 - S^2)^2.

0, 2, 8, 23, 64, 182, 520, 1475, 4152, 11624, 32408, 90028, 249272, 688140, 1894600, 5203665, 14260968, 39004962, 106486512, 290226621, 789776888, 2146082610, 5823823120, 15784464728, 42731452816, 115556460982, 312175750152, 842537682283, 2271900155120

Offset: 0

Author

Clark Kimberling, Aug 19 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 (8, -26, 48, -59, 48, -26, 8, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = (1 - s^2)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290926 *)

Formula

G.f.: (2 x - 8 x^2 + 11 x^3 - 8 x^4 + 2 x^5)/(1 - 4 x + 5 x^2 - 4 x^3 + x^4)^2.

a(n) = 8*a(n-1) - 26*a(n-2) + 48*a(n-3) - 59*a(n-4) + 48*a(n-5) - 26*a(n-6) + 8*a(n-7) - a(n-8).