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%I #13 Aug 18 2017 19:08:46
%S 0,1,5,17,51,149,439,1308,3916,11728,35093,104943,313773,938199,
%T 2805439,8389163,25086356,75016104,224321012,670787533,2005857561,
%U 5998122649,17936209267,53634716681,160384099011,479597177352,1434141243492,4288517958652
%N p-INVERT of the positive integers, where p(S) = 1 - S^2 - 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="/A290900/b290900.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 (6, -14, 19, -14, 6, -1)
%F a(n) = 6*a(n-1) - 14*a(n-2) + 19*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).
%F G.f.: x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6). - _Colin Barker_, Aug 18 2017
%t z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290900 *)
%o (PARI) concat(0, Vec(x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6) + O(x^40))) \\ _Colin Barker_, Aug 18 2017
%Y Cf. A000027, A290890.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_, Aug 17 2017
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