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A291725 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.

%I A291725 #6 Sep 11 2017 12:06:48
%S A291725 2,3,6,11,18,30,50,81,130,208,330,520,816,1275,1984,3077,4758,7337,
%T A291725 11286,17322,26532,40563,61908,94336,143540,218112,331008,501749,
%U A291725 759726,1149159,1736534,2621751,3954826,5960902,8977686,13511461,20320854,30542064,45875998
%N A291725 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.
%C A291725 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 A291725 See A291728 for a guide to related sequences.
%H A291725 Clark Kimberling, <a href="/A291725/b291725.txt">Table of n, a(n) for n = 0..1000</a>
%H A291725 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 2, -2, 0, -1)
%F A291725 G.f.: -(((-1 + x) (1 + x^2) (2 + x + x^2))/(-1 + x + x^3)^2).
%F A291725 a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.
%t A291725 z = 60; s = x + x^3; p = (1 - s)^2;
%t A291725 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291725 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291725 *)
%t A291725 LinearRecurrence[{2, -1, 2, -2, 0, -1}, {2, 3, 6, 11, 18, 30}, 40] (* _Vincenzo Librandi_, Sep 10 2017 *)
%Y A291725 Cf. A154272, A291728.
%K A291725 nonn,easy
%O A291725 0,1
%A A291725 _Clark Kimberling_, Sep 08 2017