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%I #4 Sep 11 2017 20:04:36
%S 2,6,18,56,168,510,1544,4680,14176,42952,130128,394252,1194456,
%T 3618840,10963960,33217424,100638528,304903688,923764032,2798719872,
%U 8479257216,25689531840,77831351040,235804967056,714416256800,2164460716896,6557647800096
%N p-INVERT of (1,0,1,0,0,0,0,...), 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="/A291730/b291730.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, 4, 0, 2)
%F G.f.: -((2 (1 + x^2) (1 + x + x^3))/(-1 + 2 x + 2 x^2 + 2 x^3 + 4 x^4 + 2 x^6)).
%F a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) + 4*a(n-4) + 2*a(n-6) for n >= 7.
%t z = 60; s = x + x^3; p = 1 - 2 s - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291730 *)
%t u / 2 (*A291731)
%Y Cf. A154272, A291728, A291731.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 11 2017
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