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A292402 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.

%I A292402 #10 Nov 07 2017 10:20:36
%S A292402 0,1,0,1,2,1,4,2,6,7,8,16,14,29,32,47,70,82,136,162,244,331,440,650,
%T A292402 834,1220,1632,2262,3176,4261,6056,8175,11414,15747,21568,30121,41094,
%U A292402 57210,78644,108521,150300,206456,286288,393865,544424,751675,1035980
%N A292402 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.
%C A292402 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).
%H A292402 Clark Kimberling, <a href="/A292402/b292402.txt">Table of n, a(n) for n = 0..1000</a>
%H A292402 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 0, 0, 2, 0, 0, 1)
%F A292402 G.f.: -((x (1 + x)^2 (1 - x + x^2)^2)/((-1 + x + x^4) (1 + x + x^4))).
%F A292402 a(n) = a(n-2) + 2*a(n-5) + a(n-8) for n >= 9.
%t A292402 z = 60; s = x + x^4; p = 1 - s^2;
%t A292402 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
%t A292402 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292402 *)
%Y A292402 Cf. A292324, A292403, A292404.
%K A292402 nonn,easy
%O A292402 0,5
%A A292402 _Clark Kimberling_, Sep 30 2017