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 A291000 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,-16).
FORMULA
G.f.: 3*(2 - 5*x)/(1 - 4*x)^2.
a(n) = 8*a(n-1) - 16*a(n-2) for n >= 3.
a(n) = 3*A006234(n+3) for n >= 0.
a(n) = 3 * 4^(n-1) * (3*n+8). - Colin Barker, Aug 23 2017
MATHEMATICA
z = 60; s = x/(1-x); p = (1 - 3 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291004 *)
LinearRecurrence[{8, -16}, {6, 33}, 25] (* Vincenzo Librandi, Aug 27 2017 *)
PROG
(Magma) [3*(4^(n-1)*(3*n+8)): n in [0..30]]; // Vincenzo Librandi, Aug 27 2017
(SageMath) [3*4^n*(3*n+8)//4 for n in range(41)] # G. C. Greubel, Jun 01 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 23 2017
STATUS
approved