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

%I

%S 6,27,114,459,1788,6804,25440,93825,342258,1237329,4439778,15829992,

%T 56135274,198125703,696387570,2438803863,8513220696,29631246012,

%U 102865720452,356257472589,1231184095602,4246476696765,14620160955390,50252266808784,172462429888782

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

%C 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 See A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291232/b291232.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -7, -6, -1)

%F G.f.: -((3 (-2 + 3 x + 2 x^2))/(-1 + 3 x + x^2)^2).

%F a(n) = 6*a(n-1) - 7*a(n-2) -6*a(n-3) - a(n-4) for n >= 5.

%F 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 z = 60; s = x/(1 - x^2); p = (1 - 3 s)^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291232 *)

%t u/3 (* A291265 *)

%t LinearRecurrence[{6, -7, -6, -1}, {6, 27, 114, 459}, 25] (* _Vincenzo Librandi_, Aug 28 2017 *)

%o (PARI) Vec(3*(2 + x)*(1 - 2*x) / (1 - 3*x - x^2)^2 + O(x^30)) \\ _Colin Barker_, Aug 26 2017

%o (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 Cf. A000035, A291265, A291219.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 26 2017

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Last modified July 17 18:47 EDT 2019. Contains 325109 sequences. (Running on oeis4.)