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 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 (9, -27, 27)
FORMULA
G.f.: -(((1 + x) (3 - 15 x + 22 x^2 - 7 x^3 + x^4))/(-1 + 3 x)^3).
a(n) = 9*a(n-1) - 27*a(n-2) + 27*a(n-3) for n >= 6.
a(n) = 16*3^(n-5)*(51 + 22*n + 2*n^2) for n>2. - Colin Barker, Oct 03 2017
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 02 2017
STATUS
approved