A290918 p-INVERT of the positive integers, where p(S) = (1 - S)^3.

3, 12, 43, 147, 486, 1566, 4944, 15351, 47009, 142278, 426315, 1266300, 3732705, 10928910, 31806583, 92069229, 265215756, 760621914, 2172669846, 6183333681, 17538237677, 49590486888, 139817553417, 393157465848, 1102792703055, 3086146454592, 8617872504643

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 (9, -30, 45, -30, 9, -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]  (* A290918 *)

Formula

G.f.: (3 - 15 x + 25 x^2 - 15 x^3 + 3 x^4)/(1 - 3 x + x^2)^3.

a(n) = 9*a(n-1) - 30*a(n-2) + 45*a(n-3) - 30*a(n-4) + 9*a(n-5) - a(n-6).

(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^3.