OFFSET
0,5
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 A291219 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 (0, 4, 1, -5, -1, 4, 0, -1)
FORMULA
a(n) = 4*a(n-2) + a(n-3) - 5*a(n-4) - a(n-5) + 4*a(n-6) - a(n-8) for n >= 9.
G.f.: x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8). - Colin Barker, Aug 25 2017
MATHEMATICA
PROG
(PARI) concat(vector(2), Vec(x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8) + O(x^50))) \\ Colin Barker, Aug 25 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 24 2017
STATUS
approved