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A291387 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 4 S)^2.

%I A291387 #7 Sep 09 2017 21:43:01
%S A291387 8,56,352,2096,12032,67328,369664,2000128,10696704,56666112,297836544,
%T A291387 1555066880,8073379840,41709076480,214558048256,1099562549248,
%U A291387 5616171483136,28599668703232,145249047412736,735884541427712,3720035809886208,18767645931208704
%N A291387 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 4 S)^2.
%C A291387 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 A291387 See A291382 for a guide to related sequences.
%C A291387 All terms = 0 mod 8. - _Muniru A Asiru_, Sep 07 2017
%H A291387 Clark Kimberling, <a href="/A291387/b291387.txt">Table of n, a(n) for n = 0..1000</a>
%H A291387 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8, -8, -32, -16)
%F A291387 G.f.: -((8 (1 + x) (-1 + 2 x + 2 x^2))/(-1 + 4 x + 4 x^2)^2).
%F A291387 a(n) = 8*a(n-1) - 8*a(n-2) - 32*a(n-3) - 16*a(n-4) for n >= 5.
%t A291387 z = 60; s = x + x^2; p = (1 - 4 s)^2;
%t A291387 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291387 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291387 *)
%t A291387 u / 8  (* A291388 *)
%o A291387 (GAP)
%o A291387 a:=8*[1,7,44,262];; for n in [5..10^2] do a[n]:=8*a[n-1]-8*a[n-2]-32*a[n-3]-16*a[n-4]; od; a;  # _Muniru A Asiru_, Sep 07 2017
%Y A291387 Cf. A019590, A291382, A291388.
%K A291387 nonn,easy
%O A291387 0,1
%A A291387 _Clark Kimberling_, Sep 04 2017