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A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

%I A289976 #4 Aug 21 2017 19:43:29
%S A289976 0,0,1,1,2,5,9,18,36,70,137,268,522,1017,1980,3852,7492,14568,28321,
%T A289976 55051,106999,207952,404134,785366,1526186,2965752,5763103,11198858,
%U A289976 21761463,42286357,82169547,159668921,310262351,602888757,1171506956,2276419286
%N A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.
%C A289976 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 A289976 See A290890 for a guide to related sequences.
%H A289976 Clark Kimberling, <a href="/A289976/b289976.txt">Table of n, a(n) for n = 0..1000</a>
%H A289976 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, -1, -2, -1, 1)
%F A289976 G.f.: ((1 - x)^2 x^2 (1 + x))/(1 - 2 x - x^2 + x^3 + 2 x^4 + x^5 - x^6).
%F A289976 a(n) = 2*a(n-1) + a(n-2) - a(n-3) - 2*a(n-4) - a(n-5) + a(n-6).
%t A289976 z = 60; s = x^3/(1 - x - x^2); p = 1 - s - s^2;
%t A289976 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,5,... *)
%t A289976 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289976 *)
%Y A289976 Cf. A000045, A289975, A289780.
%K A289976 nonn,easy
%O A289976 0,5
%A A289976 _Clark Kimberling_, Aug 21 2017