A291033 p-INVERT of the positive integers, where p(S) = 1 - 6*S.

6, 48, 378, 2976, 23430, 184464, 1452282, 11433792, 90018054, 708710640, 5579667066, 43928625888, 345849340038, 2722866094416, 21437079415290, 168773769227904, 1328753074407942, 10461250826035632, 82361253533877114, 648428777444981280, 5105068966025973126

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, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 6 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291033 *)

Formula

G.f.: 6/(1 - 8 x + x^2).

a(n) = 8*a(n-1) - a(n-2).

a(n) = 6*A001090(n) for n >= 1.