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 A290890 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, -27, 52, -63, 52, -27, 8, -1)
FORMULA
a(n) = 8*a(n-1) - 27*a(n-2) + 52*a(n-3) - 63*a(n-4) + 52*a(n-5) - 27*a(n-6) + 8*a(n-7) - a(n-8).
G.f.: x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8). - Colin Barker, Aug 18 2017
MATHEMATICA
PROG
(PARI) concat(0, Vec(x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8) + O(x^40))) \\ Colin Barker, Aug 18 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 17 2017
STATUS
approved