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 A280890 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 (8, -28, 56, -69, 56, -28, 8, -1)
FORMULA
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 69*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).
G.f.: x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)). - Colin Barker, Aug 16 2017
MATHEMATICA
PROG
(PARI) concat(vector(3), Vec(x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)) + O(x^50))) \\ Colin Barker, Aug 16 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 15 2017
STATUS
approved