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A291415 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2.

%I A291415 #4 Sep 11 2017 12:06:07
%S A291415 3,11,37,126,427,1448,4909,16643,56424,191292,648529,2198680,7454090,
%T A291415 25271280,85676131,290464093,984747891,3338548317,11318536416,
%U A291415 38372746007,130093466328,441050269849,1495273713773,5069362002354,17186439428582,58266444593059
%N A291415 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2.
%C A291415 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 A291415 See A291382 for a guide to related sequences.
%H A291415 Clark Kimberling, <a href="/A291415/b291415.txt">Table of n, a(n) for n = 0..1000</a>
%H A291415 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 2, -2, -1)
%F A291415 G.f.: -(((1 + x) (-3 + x + x^2))/(1 - 3 x - 2 x^2 + 2 x^3 + x^4)).
%F A291415 a(n) = 3*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.
%t A291415 z = 60; s = x + x^2; p = 1 - 3 s + s^2;
%t A291415 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291415 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291415 *)
%Y A291415 Cf. A019590, A291382.
%K A291415 nonn,easy
%O A291415 0,1
%A A291415 _Clark Kimberling_, Sep 07 2017