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 (8,-15,8,-1).
FORMULA
G.f.: (4 - 9 x + 4 x^2)/(1 - 8 x + 15 x^2 - 8 x^3 + x^4).
a(n) = 8*a(n-1) - 15*a(n-2) + 8*a(n-3) - a(n-4).
MATHEMATICA
z = 60; s = x/(1 - x)^2; p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291026 *)
LinearRecurrence[{8, -15, 8, -1}, {4, 23, 128, 711}, 30] (* Harvey P. Dale, May 18 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 19 2017
STATUS
approved