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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 15 2017
STATUS
approved