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%I #4 Aug 14 2017 23:12:31
%S 1,2,4,9,18,39,80,170,353,744,1553,3262,6824,14313,29970,62823,131596,
%T 275782,577777,1210704,2536657,5315210,11136700,23334969,48893202,
%U 102446199,214654136,449764562,942387569,1974580920,4137324929,8668915558,18163921856
%N p-INVERT of (1,0,1,0,1,0,1,0,1,...) (A059841), 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 A289780 for a guide to related sequences.
%H Clark Kimberling, <a href="/A289846/b289846.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 (1, 3, -1, -1)
%F G.f.: (1 + x - x^2)/(1 - x - 3 x^2 + x^3 + x^4).
%F a(n) = a(n-1) + 3*a(n-2) - a(n-3) - a(n-4).
%t z = 60; s = x/(1 - x^2); p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A059841 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289846 *)
%Y Cf. A059841, A289780.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 14 2017
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