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 A291219 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 (6, -8, -6, 8, 6, 1)
FORMULA
G.f.: (-6 + 11 x + 6 x^2 - 11 x^3 - 6 x^4)/(-1 + 6 x - 8 x^2 - 6 x^3 + 8 x^4 + 6 x^5 + x^6).
a(n) = 6*a(n-1) - 8*a(n-2) - 6*a(n-3) + 8*a(n-4) + 6*a(n-5) + a(n-6) for n >= 7.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 25 2017
STATUS
approved