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A292533 p-INVERT of the squares (A000290), where p(S) = (1 - S)^2.

%I A292533 #15 Feb 15 2018 14:11:24
%S A292533 2,11,46,187,748,2948,11480,44273,169374,643601,2431526,9140616,
%T A292533 34212350,127563959,474022478,1756118055,6488228880,23912815820,
%U A292533 87935847700,322713694333,1182114988606,4322734288413,15782353895178,57537481431056,209479529802682
%N A292533 p-INVERT of the squares (A000290), where p(S) = (1 - S)^2.
%C A292533 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 A292533 See A292479 for a guide to related sequences.
%H A292533 Clark Kimberling, <a href="/A292533/b292533.txt">Table of n, a(n) for n = 0..1000</a>
%H A292533 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (8, -20, 18, -12, 4, -1)
%F A292533 G.f.: -(((1 + x) (-2 + 7 x - 5 x^2 + 2 x^3))/(-1 + 4 x - 2 x^2 + x^3)^2).
%F A292533 a(n) = 8*a(n-1) - 20*a(n-2) + 18*a(n-3) - 12*a(n-4) + 4*a(n-5) - a(n-6) for n >= 7.
%t A292533 z = 60; s = x (x + 1)/(1 - x)^3; p = (1 - s)^2;
%t A292533 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292533 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292533 *)
%Y A292533 Cf. A000290, A292479.
%K A292533 nonn,easy
%O A292533 0,1
%A A292533 _Clark Kimberling_, Oct 04 2017