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 A290890 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 (15, -95, 330, -685, 873, -685, 330, -95, 15, -1)
FORMULA
a(n) = 15*a(n-1) - 95*a(n-2) + 330*a(n-3) - 685*a(n-4) + 873*a(n-5) - 695*a(n-6) + 330*a(n-7) - 95*a(n-8) + 15*a(n-9) - a(n-10).
G.f.: (5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5. - Colin Barker, Aug 24 2017
MATHEMATICA
PROG
(PARI) Vec((5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5 + O(x^30)) \\ Colin Barker, Aug 24 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 18 2017
STATUS
approved