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A291004 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 3*S)^2.

%I A291004 #22 Jun 02 2023 21:52:37
%S A291004 6,33,168,816,3840,17664,79872,356352,1572864,6881280,29884416,
%T A291004 128974848,553648128,2365587456,10066329600,42681237504,180388626432,
%U A291004 760209211392,3195455668224,13400297963520,56075093016576,234195976716288,976366325465088,4063794976260096
%N A291004 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 3*S)^2.
%C A291004 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 A291004 See A291000 for a guide to related sequences.
%H A291004 Clark Kimberling, <a href="/A291004/b291004.txt">Table of n, a(n) for n = 0..1000</a>
%H A291004 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-16).
%F A291004 G.f.: 3*(2 - 5*x)/(1 - 4*x)^2.
%F A291004 a(n) = 8*a(n-1) - 16*a(n-2) for n >= 3.
%F A291004 a(n) = 3*A006234(n+3) for n >= 0.
%F A291004 a(n) = 3 * 4^(n-1) * (3*n+8). - _Colin Barker_, Aug 23 2017
%t A291004 z = 60; s = x/(1-x); p = (1 - 3 s)^2;
%t A291004 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291004 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291004 *)
%t A291004 LinearRecurrence[{8, -16}, {6, 33}, 25] (* _Vincenzo Librandi_, Aug 27 2017 *)
%o A291004 (Magma) [3*(4^(n-1)*(3*n+8)): n in [0..30]]; // _Vincenzo Librandi_, Aug 27 2017
%o A291004 (SageMath) [3*4^n*(3*n+8)//4 for n in range(41)] # _G. C. Greubel_, Jun 01 2023
%Y A291004 Cf. A000012, A289780, A291000.
%K A291004 nonn,easy
%O A291004 0,1
%A A291004 _Clark Kimberling_, Aug 23 2017