OFFSET
0,5
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 (0, 0, 1, 0, 3, 1, 3, 6, 1, 15, 0, 20, 0, 15, 0, 6, 0, 1)
FORMULA
G.f.: -((x^2 (1 + x^2)^3 (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x^3 + 3 x^5 + x^6 + 3 x^7 + 6 x^8 + x^9 + 15 x^10 + 20 x^12 + 15 x^14 + 6 x^16 + x^18)).
a(n) = a(n-3) + 3*a(n-5) + a(n-6) + 3*a(n-7) + 6*a(n-8) + a(n-9) + 15*a(n-10) + 20 *a(n-12) + 15*a(n-14) + 6*a(n-16) + a(n-18) for n >= 19.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 11 2017
STATUS
approved