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A289845 p-INVERT of A079977, where p(S) = 1 - S - S^2.

%I A289845 #8 Jan 16 2019 15:10:59
%S A289845 1,2,4,9,19,43,91,202,433,952,2055,4494,9737,21236,46099,100403,
%T A289845 218164,474833,1032256,2245929,4883690,10623848,23103985,50255443,
%U A289845 109298635,237734446,517055409,1124617945,2446001258,5320100761,11571106298,25167245524,54738437517
%N A289845 p-INVERT of A079977, where p(S) = 1 - S - S^2.
%C A289845 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 A289845 See A289780 for a guide to related sequences.
%H A289845 Clark Kimberling, <a href="/A289845/b289845.txt">Table of n, a(n) for n = 0..999</a>
%H A289845 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, -1, 1, -1, -2, 0, -1)
%F A289845 G.f.: (1 + x - x^2 - x^4)/(1 - x - 3 x^2 + x^3 - x^4 + x^5 + 2 x^6 + x^8).
%F A289845 a(n) = a(n-1) + 3*a(n-2) - a(n-3) + a(n-4) - a(n-5) - 2*a(n-6) - a(n-8).
%t A289845 z = 60; s = -x/(x^4 + x^2 - 1); p = 1 - s - s^2;
%t A289845 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A079977 *)
%t A289845 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (*A289845*)
%t A289845 LinearRecurrence[{1,3,-1,1,-1,-2,0,-1},{1,2,4,9,19,43,91,202},40] (* _Harvey P. Dale_, Jan 16 2019 *)
%Y A289845 Cf. A079977, A289780.
%K A289845 nonn,easy
%O A289845 0,2
%A A289845 _Clark Kimberling_, Aug 14 2017