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p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.
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%I #10 Nov 07 2017 10:20:36

%S 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 834,1220,1632,2262,3176,4261,6056,8175,11414,15747,21568,30121,41094,

%U 57210,78644,108521,150300,206456,286288,393865,544424,751675,1035980

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

%C 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 Clark Kimberling, <a href="/A292402/b292402.txt">Table of n, a(n) for n = 0..1000</a>

%H <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 G.f.: -((x (1 + x)^2 (1 - x + x^2)^2)/((-1 + x + x^4) (1 + x + x^4))).

%F a(n) = a(n-2) + 2*a(n-5) + a(n-8) for n >= 9.

%t z = 60; s = x + x^4; p = 1 - s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292402 *)

%Y Cf. A292324, A292403, A292404.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Sep 30 2017