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A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

%I A289977 #11 Jul 14 2018 15:50:03
%S A289977 0,0,0,1,1,2,3,7,12,23,41,77,140,258,470,861,1570,2867,5225,9526,
%T A289977 17352,31607,57547,104766,190684,347029,631476,1148985,2090427,
%U A289977 3803044,6918379,12585209,22892932,41641932,75744383,137772396,250592150,455792833,829016539
%N A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.
%C A289977 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 A289977 See A290890 for a guide to related sequences.
%H A289977 Clark Kimberling, <a href="/A289977/b289977.txt">Table of n, a(n) for n = 0..1000</a>
%H A289977 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, -2, 0, -1, -1, 0, 1)
%F A289977 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-5) - a(n-6) + a(n-8).
%F A289977 G.f.: x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8). - _Colin Barker_, Aug 24 2017
%t A289977 z = 60; s = x^4/(1 - x - x^2); p = 1 - s - s^2;
%t A289977 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,5,... *)
%t A289977 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289977 *)
%t A289977 LinearRecurrence[{2,1,-2,0,-1,-1,0,1},{0,0,0,1,1,2,3,7},40] (* _Harvey P. Dale_, Jul 14 2018 *)
%o A289977 (PARI) concat(vector(3), Vec(x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8) + O(x^50))) \\ _Colin Barker_, Aug 24 2017
%Y A289977 Cf. A000045, A289975, A289780.
%K A289977 nonn,easy
%O A289977 0,6
%A A289977 _Clark Kimberling_, Aug 21 2017