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 A291219 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 (2, 4, -2, -1)
FORMULA
G.f.: -((2 (-1 - x + x^2))/(1 - 2 x - 4 x^2 + 2 x^3 + x^4)).
a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.
a(n) = 2*A291257(n) for n >= 0.
MATHEMATICA
z = 60; s = x/(1 - x^2); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291228 *)
u/2 (* A291257 *)
LinearRecurrence[{2, 4, -2, -1}, {2, 6, 18, 56}, 30] (* Harvey P. Dale, Aug 08 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 25 2017
STATUS
approved