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A290927 p-INVERT of the positive integers, where p(S) = (1 - S^2)^3.

%I A290927 #11 Aug 20 2017 11:07:16
%S A290927 0,3,12,36,108,331,1008,3027,8992,26502,77592,225806,653544,1882224,
%T A290927 5396776,15411399,43847688,124331457,351448620,990586686,2784612380,
%U A290927 7808372811,21845061504,60983031772,169897677504,472435652577,1311365875700,3633925019190
%N A290927 p-INVERT of the positive integers, where p(S) = (1 - S^2)^3.
%C A290927 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 A290927 See A290890 for a guide to related sequences.
%H A290927 Clark Kimberling, <a href="/A290927/b290927.txt">Table of n, a(n) for n = 0..1000</a>
%H A290927 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12, -63, 196, -414, 636, -731, 636, -414, 196, -63, 12, -1)
%F A290927 a(n) = 12*a(n-1) - 63*a(n-2) + 196*a(n-3) - 414*a(n-4) + 636*a(n-5) - 731*a(n-6) + 636*a(n-7) - 414*a(n-8) + 196*a(n-9) - 63*a(n-10) + 12*a(n-11) - a(n-12).
%F A290927 G.f.: x*(3 - 24*x + 81*x^2 - 156*x^3 + 193*x^4 - 156*x^5 + 81*x^6 - 24*x^7 + 3*x^8) / ((1 - 3*x + x^2)^3*(1 - x + x^2)^3). - _Colin Barker_, Aug 19 2017
%t A290927 z = 60; s = x/(1 - x)^2; p = (1 - s^2)^3;
%t A290927 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290927 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290927 *)
%o A290927 (PARI) concat(0, Vec(x*(3 - 24*x + 81*x^2 - 156*x^3 + 193*x^4 - 156*x^5 + 81*x^6 - 24*x^7 + 3*x^8) / ((1 - 3*x + x^2)^3*(1 - x + x^2)^3) + O(x^30))) \\ _Colin Barker_, Aug 19 2017
%Y A290927 Cf. A000027, A033453, A290890.
%K A290927 nonn,easy
%O A290927 0,2
%A A290927 _Clark Kimberling_, Aug 19 2017