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 A291728 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 (4, -6, 8, -13, 12, -10, 12, -6, 4, -4, 0, -1)
FORMULA
G.f.: -(((-1 + x) (1 + x^2) (2 + x + x^2) (2 - 2 x + x^2 - 2 x^3 + 2 x^4 + x^6))/(-1 + x + x^3)^4).
a(n) = 4*a(n-1) - 6*a(n-2) + 8*a(n-3) - 13*a(n-4) + 12*a(n-5) - 10*a(n-6) + 12*a(n-7) - 6*a(n-8) +4*a(n-9) - 4*a(n-10) - a(n-12) for n >= 13.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
_Clark Kimberling_, Sep 08 2017
STATUS
approved