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A291232 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - 3 S)^2.

%I A291232 #11 Sep 08 2022 08:46:19
%S A291232 6,27,114,459,1788,6804,25440,93825,342258,1237329,4439778,15829992,
%T A291232 56135274,198125703,696387570,2438803863,8513220696,29631246012,
%U A291232 102865720452,356257472589,1231184095602,4246476696765,14620160955390,50252266808784,172462429888782
%N A291232 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - 3 S)^2.
%C A291232 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 A291232 See A291219 for a guide to related sequences.
%H A291232 Clark Kimberling, <a href="/A291232/b291232.txt">Table of n, a(n) for n = 0..1000</a>
%H A291232 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -7, -6, -1)
%F A291232 G.f.: -((3 (-2 + 3 x + 2 x^2))/(-1 + 3 x + x^2)^2).
%F A291232 a(n) = 6*a(n-1) - 7*a(n-2) -6*a(n-3) - a(n-4) for n >= 5.
%F A291232 a(n) = 3 * (((3-sqrt(13))/2)^n*(-3+sqrt(13))*(-39+17*sqrt(13)-39*n) + 2^(-n)*(3+sqrt(13))^(1+n)*(39+17*sqrt(13)+39*n)) / 338. - _Colin Barker_, Aug 26 2017
%t A291232 z = 60; s = x/(1 - x^2); p = (1 - 3 s)^2;
%t A291232 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291232 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291232 *)
%t A291232 u/3   (* A291265 *)
%t A291232 LinearRecurrence[{6, -7, -6, -1}, {6, 27, 114, 459}, 25] (* _Vincenzo Librandi_, Aug 28 2017 *)
%o A291232 (PARI) Vec(3*(2 + x)*(1 - 2*x) / (1 - 3*x - x^2)^2 + O(x^30)) \\ _Colin Barker_, Aug 26 2017
%o A291232 (Magma) I:=[6,27,114,459]; [n le 4 select I[n] else 6*Self(n-1)-7*Self(n-2)-6*Self(n-3)-Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Aug 28 2017
%Y A291232 Cf. A000035, A291265, A291219.
%K A291232 nonn,easy
%O A291232 0,1
%A A291232 _Clark Kimberling_, Aug 26 2017