A292325 p-INVERT of (1,0,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

2, 3, 4, 5, 8, 13, 20, 29, 40, 56, 80, 115, 164, 230, 320, 445, 620, 864, 1200, 1660, 2290, 3155, 4344, 5975, 8206, 11252, 15408, 21078, 28810, 39344, 53680, 73173, 99662, 135640, 184480, 250740, 340578, 462316, 627200, 850420, 1152480, 1561043, 2113420

Offset

0,1

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).

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, 2, -2, 0, 0, 0, -1)

Formula

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

a(n) = 2*a(n-1) - a(n-2) + 2*a(n-5) - 2*a(n-6) - a(n-10) for n >= 11.

Mathematica

z = 60; s = x + x^5; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292325 *)

Crossrefs

Cf. A079978, A292324.

Sequence in context: A222110 A249581 A051706 * A367692 A307971 A346073

Adjacent sequences: A292322 A292323 A292324 * A292326 A292327 A292328

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved