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%I #4 Sep 14 2017 18:14:45
%S 2,6,16,46,132,376,1074,3066,8752,24986,71328,203624,581298,1659462,
%T 4737360,13524006,38607732,110215648,314638754,898216794,2564189568,
%U 7320134930,20897197344,59656394448,170304435554,486177568038,1387918211824,3962167507006
%N p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - 2 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).
%C See A291728 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291036/b291036.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 2, 2, -2, 0, -1)
%F G.f.: -((2 (-1 - x + x^3))/(1 - 2 x - 2 x^2 - 2 x^3 + 2 x^4 + x^6)).
%F a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.
%t z = 60; s = x/(x - x^3); p = 1 - 2 s - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A079978 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291036 *)
%t u/2 (* A291037 *)
%Y Cf. A079978, A290616, A291037.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 14 2017
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