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).
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 (9, -27, 36, -27, 9, -1)
FORMULA
G.f.: (3 - 12 x + 16 x^2 - 12 x^3 + 3 x^4)/(1 - 9 x + 27 x^2 - 36 x^3 + 27 x^4 - 9 x^5 + x^6).
a(n) = 9*a(n-1) - 27*a(n-2) + 36*a(n-3) - 27*a(n-4) + 90*a(n-5) - a(n-6).
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 19 2017
STATUS
approved