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%I #4 Aug 23 2017 23:37:43
%S 1,3,9,25,67,178,472,1249,3297,8685,22843,60014,157540,413289,1083693,
%T 2840521,7443331,19500394,51079696,133782385,350354841,917456901,
%U 2402365387,6290338310,16470047644,43122600825,112903347237,295598625697,773914899475
%N p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.
%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 A291000 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291021/b291021.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (6, -13, 14, -9, 2)
%F G.f.: (-1 + 3 x - 4 x^2 + 4 x^3 - x^4)/(-1 + 6 x - 13 x^2 + 14 x^3 - 9 x^4 + 2 x^5).
%F a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 9*a(n-4) + 2*a(n-5) for n >= 5.
%t z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 + s^4 + s^5;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291021 *)
%Y Cf. A000012, A289780, A291000.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 23 2017
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