|
%I #4 Sep 14 2017 18:14:24
%S 1,2,3,6,12,22,42,80,151,287,544,1031,1956,3708,7031,13333,25280,
%T 47936,90895,172350,326806,619677,1175008,2228011,4224672,8010672,
%U 15189552,28801880,54613096,103555397,196358029,372327066,705993241,1338679088,2538355336
%N p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 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="/A289920/b289920.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 (1, 1, 2, -1, 0, -1)
%F G.f.: -((-1 - x + x^3)/(1 - x - x^2 - 2 x^3 + x^4 + x^6)).
%F a(n) = a(n-1) + a(n-2) + 2*a(n-3) - a(n-4) - a(n-6) for n >= 7.
%t z = 60; s = x/(x - x^3); p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A079978 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289920 *)
%Y Cf. A079978, A289918.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Sep 14 2017
|
