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A291228 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - 2 S^2.

%I A291228 #6 Aug 08 2019 10:19:00
%S A291228 2,6,18,56,170,522,1594,4880,14922,45654,139642,427176,1306690,
%T A291228 3997146,12227058,37402144,114411538,349980390,1070575586,3274847512,
%U A291228 10017625050,30643508586,93737246762,286738430256,877121205338,2683078129590,8207426973258
%N A291228 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - 2 S^2.
%C A291228 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 A291228 See A291219 for a guide to related sequences.
%H A291228 Clark Kimberling, <a href="/A291228/b291228.txt">Table of n, a(n) for n = 0..1000</a>
%H A291228 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 4, -2, -1)
%F A291228 G.f.: -((2 (-1 - x + x^2))/(1 - 2 x - 4 x^2 + 2 x^3 + x^4)).
%F A291228 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.
%F A291228 a(n) = 2*A291257(n) for n >= 0.
%t A291228 z = 60; s = x/(1 - x^2); p = 1 - 2 s - 2 s^2;
%t A291228 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291228 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291228 *)
%t A291228 u/2         (* A291257 *)
%t A291228 LinearRecurrence[{2,4,-2,-1},{2,6,18,56},30] (* _Harvey P. Dale_, Aug 08 2019 *)
%Y A291228 Cf. A000035, A291219, A291257.
%K A291228 nonn,easy
%O A291228 0,1
%A A291228 _Clark Kimberling_, Aug 25 2017