OFFSET
0,3
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 A292480 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 (3, -4, 7, -1)
FORMULA
G.f.: ((1 + x) (-1 + 3 x))/(1 - 3 x + 4 x^2 - 7 x^3 + x^4).
a(n) = 3*a(n-1) - 4*a(n-2) + 7*a(n-3) - a(n-4) for n >= 5.
MATHEMATICA
z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292484 *)
LinearRecurrence[{3, -4, 7, -1}, {-1, -1, 4, 9}, 40] (* Harvey P. Dale, Sep 22 2024 *)
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Clark Kimberling, Oct 02 2017
STATUS
approved