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A291391 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 6 S)^2.

%I A291391 #4 Sep 06 2017 21:15:15
%S A291391 12,120,1080,9180,75168,599616,4691520,36164880,275503680,2078711424,
%T A291391 15559682688,115688917440,855249269760,6291326453760,46080184338432,
%U A291391 336227628720384,2445042642140160,17726787591690240,128173151784867840,924487654349822976
%N A291391 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 6 S)^2.
%C A291391 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 A291391 See A291382 for a guide to related sequences.
%H A291391 Clark Kimberling, <a href="/A291391/b291391.txt">Table of n, a(n) for n = 0..1000</a>
%H A291391 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12, -24, -72, -36)
%F A291391 G.f.: -((12 (1 + x) (-1 + 3 x + 3 x^2))/(-1 + 6 x + 6 x^2)^2).
%F A291391 a(n) = 12*a(n-1) - 24*a(n-2) - 72*a(n-3) - 36*a(n-4) for n >= 5.
%t A291391 z = 60; s = x + x^2; p = (1 - 6 s)^2;
%t A291391 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291391 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291391 *)
%t A291391 u / 12  (* A291392 *)
%Y A291391 Cf. A019590, A291382, A291392.
%K A291391 nonn,easy
%O A291391 0,1
%A A291391 _Clark Kimberling_, Sep 06 2017