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A291394 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 3 S).

%I A291394 #4 Sep 06 2017 21:15:36
%S A291394 4,17,66,254,968,3679,13962,52957,200812,761396,2886768,10944725,
%T A291394 41494856,157319353,596443614,2261290498,8573204920,32503490435,
%U A291394 123230092830,467200760741,1771292578424,6715480046152,25460317920096,96527393973769,365963135802988
%N A291394 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 3 S).
%C A291394 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 A291394 See A291382 for a guide to related sequences.
%H A291394 Clark Kimberling, <a href="/A291394/b291394.txt">Table of n, a(n) for n = 0..1000</a>
%H A291394 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 1, -6, -3)
%F A291394 G.f.: -(((1 + x) (-4 + 3 x + 3 x^2))/((-1 + x + x^2) (-1 + 3 x + 3 x^2))).
%F A291394 a(n) = 4*a(n-1) + a(n-2) - 6*a(n-3) - 3*a(n-4) for n >= 5.
%t A291394 z = 60; s = x + x^2; p = (1 - s)(1 - 3s);
%t A291394 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291394 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291394 *)
%Y A291394 Cf. A019590, A291382.
%K A291394 nonn,easy
%O A291394 0,1
%A A291394 _Clark Kimberling_, Sep 06 2017