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A291035 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 2 S^2.

%I A291035 #7 Mar 14 2025 04:26:34
%S A291035 1,3,5,12,27,58,130,285,629,1389,3060,6753,14892,32844,72445,159775,
%T A291035 352401,777244,1714259,3780930,8339090,18392449,40565829,89470733,
%U A291035 197333940,435233685,959938112,2117210180,4669654005,10299246171,22715702489,50101058976
%N A291035 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 2 S^2.
%C A291035 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 A291035 See A291728 for a guide to related sequences.
%H A291035 Clark Kimberling, <a href="/A291035/b291035.txt">Table of n, a(n) for n = 0..1000</a>
%H A291035 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,2,-1,0,-1).
%F A291035 G.f.: -(1 + x)*(-1 - x + x^2)/((-1 - x + x^3)*(-1 + 2*x + x^3)).
%F A291035 a(n) = a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4) - a(n-6) for n >= 7.
%t A291035 z = 60; s = x/(x - x^3); p = 1 - s - 2 s^2;
%t A291035 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978 *)
%t A291035 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291035 *)
%Y A291035 Cf. A079978, A289918, A290616.
%K A291035 nonn,easy
%O A291035 0,2
%A A291035 _Clark Kimberling_, Sep 14 2017