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%I #4 Aug 28 2017 20:15:17
%S 4,11,30,79,202,508,1262,3109,7614,18569,45152,109560,265448,642463,
%T 1553790,3755843,9075302,21923060,52949458,127869209,308767326,
%U 745537309,1800065788,4346043888,10492781068,25332654899,61159842270,147655261111,356475233986
%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^2 (1 - 2 S).
%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 A291219 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291238/b291238.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 (4, -2, -6, 2, 4, 1)
%F G.f.: (-4 + 5 x + 6 x^2 - 5 x^3 - 4 x^4)/((-1 + x + x^2)^2 (-1 + 2 x + x^2)).
%F a(n) = 4*a(n-1) - 2*a(n-2) - 6*a(n-3) + 2*a(n-4) + 4*a(n-5) + a(n-6) for n >= 7.
%t z = 60; s = x/(1 - x^2); p = (1 - s)^2(1 - 2s);
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291238 *)
%Y Cf. A000035, A291219.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Aug 28 2017
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