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A291398 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.

%I A291398 #6 Dec 13 2017 19:13:34
%S A291398 0,1,3,5,9,19,39,76,150,301,600,1191,2370,4721,9396,18696,37212,74069,
%T A291398 147417,293398,583956,1162257,2313237,4604037,9163443,18238042,
%U A291398 36299229,72246487,143792475,286190708,569606421,1133689810,2256387135,4490895817,8938246848
%N A291398 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.
%C A291398 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 A291398 See A291382 for a guide to related sequences.
%H A291398 Clark Kimberling, <a href="/A291398/b291398.txt">Table of n, a(n) for n = 0..1000</a>
%H A291398 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 3, 4, 3, 1)
%F A291398 G.f.: -((x (1 + x)^2 (1 + x + x^2))/(-1 + x^2 + 3 x^3 + 4 x^4 + 3 x^5 + x^6)).
%F A291398 a(n) = a(n-2) + 3*a(n-3) + 4*a(n-4) + 3*a(n-5) + a(n-6) for n >= 7.
%t A291398 z = 60; s = x + x^2; p = 1 - s^2 - s^3;
%t A291398 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291398 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291398 *)
%t A291398 LinearRecurrence[{0,1,3,4,3,1},{0,1,3,5,9,19},40] (* _Harvey P. Dale_, Dec 13 2017 *)
%Y A291398 Cf. A019590, A291382.
%K A291398 nonn,easy
%O A291398 0,3
%A A291398 _Clark Kimberling_, Sep 06 2017