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A291393 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

%I A291393 #4 Sep 06 2017 21:15:30
%S A291393 3,10,29,83,232,643,1771,4862,13321,36455,99696,272535,744839,2035358,
%T A291393 5561381,15195075,41515496,113425323,309888403,846638398,2313071313,
%U A291393 6319448079,17265085152,47169141487,128868574671,352075628734,961888724621,2627929220939
%N A291393 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).
%C A291393 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 A291393 See A291382 for a guide to related sequences.
%H A291393 Clark Kimberling, <a href="/A291393/b291393.txt">Table of n, a(n) for n = 0..1000</a>
%H A291393 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 1, -4, -2)
%F A291393 G.f.: -(((1 + x) (-3 + 2 x + 2 x^2))/((-1 + x + x^2) (-1 + 2 x + 2 x^2))).
%F A291393 a(n) = 3*a(n-1) + a(n-2) - 4*a(n-3) - 2*a(n-4) for n >= 5.
%t A291393 z = 60; s = x + x^2; p = (1 - s)(1 - 2s);
%t A291393 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291393 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291393 *)
%Y A291393 Cf. A019590, A291382.
%K A291393 nonn,easy
%O A291393 0,1
%A A291393 _Clark Kimberling_, Sep 06 2017