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A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3.

%I A290900 #18 May 14 2024 17:00:18
%S A290900 0,1,5,17,51,149,439,1308,3916,11728,35093,104943,313773,938199,
%T A290900 2805439,8389163,25086356,75016104,224321012,670787533,2005857561,
%U A290900 5998122649,17936209267,53634716681,160384099011,479597177352,1434141243492,4288517958652
%N A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3.
%C A290900 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 A290900 See A290890 for a guide to related sequences.
%H A290900 Clark Kimberling, <a href="/A290900/b290900.txt">Table of n, a(n) for n = 0..1000</a>
%H A290900 Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, <a href="https://arxiv.org/abs/2405.05357">Flattened Catalan Words</a>, arXiv:2405.05357 [math.CO], 2024. See p. 22.
%H A290900 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,19,-14,6,-1).
%F A290900 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 A290900 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 A290900 z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^3;
%t A290900 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290900 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290900 *)
%o A290900 (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 A290900 Cf. A000027, A033453, A290890.
%K A290900 nonn,easy
%O A290900 0,3
%A A290900 _Clark Kimberling_, Aug 17 2017