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

%I #15 Aug 20 2017 11:07:02

%S 0,0,2,12,42,115,288,738,2020,5751,16362,45697,125538,342318,933050,

%T 2547630,6960042,18990309,51699042,140439411,380871538,1031705466,

%U 2792009100,7548723827,20389716864,55020917232,148334534420,399562167420,1075432476492

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

%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="/A290928/b290928.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,222,-507,822,-965,822,-507,222,-66,12,-1)

%F a(n) = 12*a(n-1) - 66*a(n-2) + 222*a(n-3) - 507*a(n-4) + 822*a(n-5) - 965*a(n-6) + 822*a(n-7) - 507*a(n-8) + 222*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).

%F G.f.: x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2). - _Colin Barker_, Aug 19 2017

%t z = 60; s = x/(1 - x)^2; p = (1 - s^3)^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290928 *)

%t LinearRecurrence[{12, -66, 222, -507, 822, -965, 822, -507, 222, -66, 12, -1}, {0, 0, 2, 12, 42, 115, 288, 738, 2020, 5751, 16362, 45697}, 40] (* _Vincenzo Librandi_, Aug 20 2017 *)

%o (PARI) concat(vector(2), Vec(x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2) + O(x^30))) \\ _Colin Barker_, Aug 19 2017

%Y Cf. A000027, A033453, A290890.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Aug 19 2017