A290891 p-INVERT of the positive integers, where p(S) = 1 - S^3.

0, 0, 1, 6, 21, 57, 138, 330, 827, 2175, 5826, 15519, 40836, 106584, 277696, 724968, 1897380, 4972113, 13029534, 34125561, 89336141, 233831262, 612074526, 1602358863, 4195173507, 10983645498, 28756340047, 75285234408, 197097337248, 516002648064

Offset: 0

Author

Clark Kimberling, Aug 15 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 A280890 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, -15, 21, -15, 6, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290891 *)

Formula

G.f.: x^2/(1 - 6 x + 15 x^2 - 21 x^3 + 15 x^4 - 6 x^5 + x^6).

a(n) = 6*a(n-1) - 15*a(n-2) + 21*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).