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%I #6 Sep 04 2017 19:30:54
%S 2,8,28,98,344,1208,4240,14884,52248,183408,643824,2260040,7933504,
%T 27849280,97760384,343171984,1204649632,4228727296,14844261824,
%U 52108375328,182918006400,642104016000,2254002082560,7912309005376,27774878417792,97499209219328
%N p-INVERT of (1,1,0,0,0,0,...), 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 A291382 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291383/b291383.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 4, 4, 2)
%F G.f.: -((2 (1 + x) (1 + x + x^2))/(-1 + 2 x + 4 x^2 + 4 x^3 + 2 x^4)).
%F a(n) = 2*a(n-1) + 4*a(n-2) + 4*a(n-3) + 2*a(n-4) for n >= 5.
%t z = 60; s = x + x^2; p = 1 - 2 s - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291383 *)
%t u / 2 (* A291384 *)
%Y Cf. A019590, A291382, A291384.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 04 2017
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