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A158543 a(n) = 144*n^2 - 12.

%I A158543 #34 Jan 16 2025 08:46:40
%S A158543 132,564,1284,2292,3588,5172,7044,9204,11652,14388,17412,20724,24324,
%T A158543 28212,32388,36852,41604,46644,51972,57588,63492,69684,76164,82932,
%U A158543 89988,97332,104964,112884,121092,129588,138372,147444,156804,166452,176388,186612,197124,207924
%N A158543 a(n) = 144*n^2 - 12.
%C A158543 The identity (24*n^2 - 1)^2 - (144*n^2 - 12)*(2*n)^2 = 1 can be written as A158544(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158543 Vincenzo Librandi, <a href="/A158543/b158543.txt">Table of n, a(n) for n = 1..10000</a>
%H A158543 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158543 From _Vincenzo Librandi_, Feb 14 2012: (Start)
%F A158543 G.f.: -x*(132 + 168*x - 12*x^2)/(x-1)^3.
%F A158543 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F A158543 From _Amiram Eldar_, Mar 06 2023: (Start)
%F A158543 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)))/24.
%F A158543 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)) - 1)/24. (End)
%F A158543 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A158543 E.g.f.: 12*(exp(x)*(12*x^2 + 12*x - 1) + 1).
%F A158543 a(n) = 12*A158463(n). (End)
%t A158543 LinearRecurrence[{3, -3, 1}, {132, 564, 1284}, 50] (* _Vincenzo Librandi_, Feb 14 2012 *)
%t A158543 144*Range[40]^2-12 (* _Harvey P. Dale_, Oct 20 2012 *)
%o A158543 (Magma) I:=[132, 564, 1284]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 14 2012
%o A158543 (PARI) for(n=1, 40, print1(144*n^2 - 12", ")); \\ _Vincenzo Librandi_, Feb 14 2012
%Y A158543 Cf. A005843, A158463, A158544.
%K A158543 nonn,easy
%O A158543 1,1
%A A158543 _Vincenzo Librandi_, Mar 21 2009
%E A158543 Comment rewritten by _R. J. Mathar_, Oct 16 2009