cp's OEIS Frontend

A289975 p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - S^2.

%I A289975 #6 Aug 21 2017 22:07:44
%S A289975 0,1,1,4,7,18,37,85,183,407,888,1956,4284,9409,20630,45270,99289,
%T A289975 217819,477776,1048053,2298912,5042783,11061455,24263687,53223023,
%U A289975 116746272,256086074,561731936,1232174181,2702807740,5928681960,13004724921,28526216361
%N A289975 p-INVERT of the Fibonacci numbers (A000045, including 0), where p(S) = 1 - S - S^2.
%C A289975 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).
%C A289975 See A290890 for a guide to related sequences.
%H A289975 Clark Kimberling, <a href="/A289975/b289975.txt">Table of n, a(n) for n = 0..1000</a>
%H A289975 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 2, -3, -1)
%F A289975 G.f.: (x - x^2)/(1 - 2 x - 2 x^2 + 3 x^3 + x^4).
%F A289975 a(n) = 2*a(n-1) + 2*a(n-2) - 3*a(n-3) - a(n-4).
%t A289975 z = 60; s = x^2/(1 - x - x^2); p = 1 - s - s^2;
%t A289975 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t A289975 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289975 *)
%Y A289975 Cf. A000045, A289781, A289976, A289780.
%K A289975 nonn,easy
%O A289975 0,4
%A A289975 _Clark Kimberling_, Aug 21 2017