OFFSET
0,3
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 (0, 9, 3, -36, -18, 81, 45, -117, -59, 117, 45, -81, -18, 36, 3, -9, 0, 1)
FORMULA
G.f.: -((x^2 (3 - 18 x^2 - 3 x^3 + 45 x^4 + 9 x^5 - 59 x^6 - 9 x^7 + 45 x^8 + 3 x^9 - 18 x^10 + 3 x^12))/((-1 + x + x^2)^3 (1 + x - x^2 - x^3 + x^4)^3)).
a(n) = 9*a(n-2) + 3*a(n-3) - 36*a(n-4) - 18*a(n-5) + 81*a(n-6) + 45*a(n-7) - 117*a(n-8) - 59*a(n-9) + 117*a(n-10) + 45*a(n-11) - 81*a(n-12) - 18*a(n-13) + 36*a(n-14) + 3*a(n-15) - 9*a(n-16) + a(n-18) for n >= 19.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 29 2017
STATUS
approved