OFFSET
0,6
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 A291382 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, 0, 0, 1, 5, 10, 10, 5, 1)
FORMULA
G.f.: -((x^4 (1 + x)^5)/((-1 + x + x^2) (1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8))).
a(n) = a(n-5) + 5*a(n-6) + 10*a(n-7) + 10*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 04 2017
STATUS
approved