A292324 p-INVERT of (1,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

2, 3, 4, 7, 12, 19, 28, 42, 64, 97, 144, 212, 312, 459, 672, 979, 1422, 2062, 2984, 4308, 6206, 8925, 12816, 18376, 26310, 37620, 53728, 76648, 109230, 155507, 221184, 314325, 446320, 633249, 897804, 1271993, 1800942, 2548242, 3603468, 5092747, 7193604

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, 2, -2, 0, 0, -1)

Formula

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

a(n) = 2*a(n-1) - a(n-2) + 2*a(n-4) - 2*a(n-5) - a(n-8) for n >= 9.

a(n) = a(n-1)+a(n-4)+A003269(n+2). - R. J. Mathar, Mar 19 2024

Mathematica

z = 60; s = x + x^4; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292324 *)

Crossrefs

Cf. A079978, A292322.

Sequence in context: A376987 A217786 A228494 * A289919 A293411 A227047

Adjacent sequences: A292321 A292322 A292323 * A292325 A292326 A292327

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2017

Status

approved