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 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 (1, 0, 2, 0, 3, 0, 3, 0, 1)
FORMULA
G.f.: -(((1 + x^2) (1 + x^2 + 2 x^4 + x^6))/(-1 + x + 2 x^3 + 3 x^5 + 3 x^7 + x^9)).
a(n) = a(n-1) + 2*a(n-3) + 3*a(n-5) + 3*a(n-7) + a(n-9) for n >= 10.
MATHEMATICA
z = 60; s = x + x^3; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291735 *)
LinearRecurrence[{1, 0, 2, 0, 3, 0, 3, 0, 1}, {1, 1, 3, 5, 10, 19, 35, 67, 124}, 40] (* Harvey P. Dale, Aug 25 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 11 2017
STATUS
approved