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A289779 p-INVERT of the squares, where p(S) = 1 - S - S^2.

%I A289779 #20 Aug 13 2017 22:57:02
%S A289779 1,6,28,125,546,2371,10304,44838,195209,849896,3700025,16107530,
%T A289779 70121400,305262325,1328913506,5785228011,25185131956,109639724218,
%U A289779 477300202625,2077855302992,9045633454817,39378817534750,171429815189636,746294159430429,3248880433597858
%N A289779 p-INVERT of the squares, where p(S) = 1 - S - S^2.
%C A289779 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 A289779 See A289780 for a guide to related sequences.
%H A289779 Clark Kimberling, <a href="/A289779/b289779.txt">Table of n, a(n) for n = 0..1000</a>
%H A289779 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -16, 22, -12, 5, -1)
%F A289779 G.f.: (1 - x + 2 x^2 + 3 x^3 - x^4)/(1 - 7 x + 16 x^2 - 22 x^3 + 12 x^4 - 5 x^5 + x^6).
%F A289779 a(n) = 7*a(n-1) - 16*a(n-2) + 22*a(n-3) - 12*a(n-4) + 5*a(n-5) - a(n-6).
%t A289779 z = 60; s = (x + x^2)/(1 - x)^3; p = 1 - s - s^2;
%t A289779 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000290 *)
%t A289779 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289779 *)
%Y A289779 Cf. A000290, A033453, A289780.
%K A289779 nonn,easy
%O A289779 0,2
%A A289779 _Clark Kimberling_, Aug 10 2017