A290903 p-INVERT of the positive integers, where p(S) = 1 - 5*S.

5, 35, 240, 1645, 11275, 77280, 529685, 3630515, 24883920, 170556925, 1169014555, 8012544960, 54918800165, 376419056195, 2580014593200, 17683683096205, 121205767080235, 830756686465440, 5694091038177845, 39027880580779475, 267501073027278480

Offset: 0

Author

Clark Kimberling, Aug 17 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.

Examples

			(See the example at A290902.)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (7, -1)

Crossrefs

Cf. A000027, A004187, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 5 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
u/5 (* A004187 shifted *)

Formula

G.f.: 5/(1 - 7 x + x^2).

a(n) = 7*a(n-1) - a(n-2).

a(n) = 5*A004187(n+1) for n >= 0.