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%I #11 Aug 17 2017 16:02:11
%S 1,3,9,29,95,307,976,3073,9645,30283,95207,299625,943363,2970320,
%T 9351621,29439359,92671625,291715157,918275995,2890621063,9099375792,
%U 28643956245,90168412937,283841284899,893503898503,2812659866565,8853968158791,27871395427616
%N p-INVERT of the positive integers, where p(S) = 1 - S - S^3.
%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 A290890 for a guide to related sequences.
%H Clark Kimberling, <a href="/A290897/b290897.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 (7, -19, 27, -19, 7, -1)
%F a(n) = 7*a(n-1) - 19*a(n-2) + 27*a(n-3) - 19*a(n-4) + 7*a(n-5) - a(n-6).
%F G.f.: (1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)). - _Colin Barker_, Aug 16 2017
%t z = 60; s = x/(1 - x)^2; p = 1 - s - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290897 *)
%o (PARI) Vec((1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)) + O(x^30)) \\ _Colin Barker_, Aug 16 2017
%Y Cf. A000027, A290890.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 15 2017
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