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A292534 p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.

%I A292534 #4 Oct 05 2017 09:51:21
%S A292534 -1,-2,4,21,30,11,80,622,2055,4584,10711,34354,115480,341213,934750,
%T A292534 2640483,7874188,23564242,68738591,198108496,575654335,1688669686,
%U A292534 4951141372,14443935957,42064267934,122731975243,358682023576,1047906654118,3058580566407
%N A292534 p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.
%C A292534 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 A292534 See A292479 for a guide to related sequences.
%H A292534 Clark Kimberling, <a href="/A292534/b292534.txt">Table of n, a(n) for n = 0..1000</a>
%H A292534 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (5, -12, 22, -16, 7, -1)
%F A292534 G.f.: ((1 + x) (-1 + 4 x - 2 x^2 + x^3))/(1 - 5 x + 12 x^2 - 22 x^3 + 16 x^4 - 7 x^5 + x^6).
%F A292534 a(n) = 5*a(n-1) - 12*a(n-2) + 22*a(n-3) - 16*a(n-4) + 7*a(n-5) - a(n-6) for n >= 7.
%t A292534 z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - s^2;
%t A292534 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292534 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292534 *)
%Y A292534 Cf. A000290, A292479.
%K A292534 easy,sign
%O A292534 0,2
%A A292534 _Clark Kimberling_, Oct 04 2017