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A291005 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 2 S - 2 S^3.

%I A291005 #19 Jun 02 2023 21:51:36
%S A291005 2,6,20,68,230,774,2600,8732,29330,98526,330980,1111868,3735110,
%T A291005 12547374,42150440,141596132,475664450,1597901526,5367837140,
%U A291005 18032197268,60575633990,203491974774,683591422280,2296391457932,7714277207570,25914602943726,87055031555300
%N A291005 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 2 S - 2 S^3.
%C A291005 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 A291005 See A291000 for a guide to related sequences.
%H A291005 Clark Kimberling, <a href="/A291005/b291005.txt">Table of n, a(n) for n = 0..1000</a>
%H A291005 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-7,5).
%F A291005 G.f.: 2*(1 - 2*x + 2*x^2)/(1 - 5*x + 7*x^2 - 5*x^3).
%F A291005 a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) for n >= 4.
%F A291005 a(n) = 2*A291337(n) for n >= 0.
%t A291005 z = 60; s = 1 - 2 s - 2 s^3;
%t A291005 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291005 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291005 *)
%t A291005 u / 2  (* A291337 *)
%t A291005 LinearRecurrence[{5, -7, 5}, {2, 6, 20}, 30] (* _Vincenzo Librandi_, Aug 27 2017 *)
%o A291005 (Magma) I:=[2,6,20]; [n le 3 select I[n] else 5*Self(n-1)-7*Self(n-2)+5*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Aug 27 2017
%o A291005 (SageMath)
%o A291005 def A291005_list(prec):
%o A291005     P.<x> = PowerSeriesRing(ZZ, prec)
%o A291005     return P( 2*(1-2*x+2*x^2)/(1-5*x+7*x^2-5*x^3) ).list()
%o A291005 A291005_list(30) # _G. C. Greubel_, Jun 01 2023
%Y A291005 Cf. A000012, A289780, A291000, A291337.
%K A291005 nonn,easy
%O A291005 0,1
%A A291005 _Clark Kimberling_, Aug 23 2017