OFFSET
0,2
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 A291000 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, -13, 14, -7)
FORMULA
G.f.: -((-1 + 3 x - 4 x^2 + 2 x^3 + x^4)/(1 - 6 x + 13 x^2 - 14 x^3 + 7 x^4) ).
a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 7*a(n-4) for n >= 6.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 23 2017
STATUS
approved