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A290911 p-INVERT of the positive integers, where p(S) = 1 - 6*S^2.

%I A290911 #8 Aug 19 2017 13:20:57
%S A290911 0,6,24,96,408,1722,7248,30528,128592,541638,2281416,9609504,40475976,
%T A290911 170487930,718108320,3024727680,12740386464,53663491206,226034767224,
%U A290911 952075887072,4010217126648,16891344084282,71147645118192,299679373092288,1262272651579632
%N A290911 p-INVERT of the positive integers, where p(S) = 1 - 6*S^2.
%C A290911 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 A290911 See A290890 for a guide to related sequences.
%H A290911 Clark Kimberling, <a href="/A290911/b290911.txt">Table of n, a(n) for n = 0..1000</a>
%H A290911 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 0, 4, -1)
%F A290911 G.f.: (6 x)/(1 - 4 x - 4 x^3 + x^4).
%F A290911 a(n) = 4*a(n-1) + 4*a(n-3) - a(n-4).
%F A290911 a(n) = 6*A290912(n) for n >= 0.
%t A290911 z = 60; s = x/(1 - x)^2; p = 1 - 6 s^2;
%t A290911 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290911 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290911 *)
%t A290911 u/6 (* A290912 *)
%Y A290911 Cf. A000027, A290890, A290912.
%K A290911 nonn,easy
%O A290911 0,2
%A A290911 _Clark Kimberling_, Aug 18 2017