OFFSET
0,4
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 A292480 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 (6, -15, 21, -12, 9)
FORMULA
G.f.: -((x^2 (1 + x)^3)/((-1 + 3 x) (1 - 3 x + 6 x^2 - 3 x^3 + 3 x^4))).
a(n) = 6*a(n-1) - 25*a(n-2) + 21*a(n-3) - 12*a(n-4) + 9*a(n-5) for n >= 6.
MATHEMATICA
z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292481 *)
LinearRecurrence[{6, -15, 21, -12, 9}, {0, 0, 1, 9, 42, 139}, 30] (* Harvey P. Dale, Jun 06 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 02 2017
STATUS
approved