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%I #22 Mar 22 2023 08:14:24
%S 1260,5148,11628,20700,32364,46620,63468,82908,104940,129564,156780,
%T 186588,218988,253980,291564,331740,374508,419868,467820,518364,
%U 571500,627228,685548,746460,809964,876060,944748,1016028,1089900,1166364,1245420,1327068,1411308,1498140
%N a(n) = 1296*n^2 - 36.
%C The identity (72*n^2 - 1)^2 - (1296*n^2 - 36)*(2*n)^2 = 1 can be written as A158738(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158737/b158737.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: 36*x*(-35 - 38*x + x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 22 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/6)*Pi/6)/72 = (1 - Pi/(2*sqrt(3)))/72.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/6)*Pi/6 - 1)/72. (End)
%t LinearRecurrence[{3, -3, 1}, {1260, 5148, 11628}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%o (Magma) I:=[1260, 5148, 11628]; [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 20 2012
%o (PARI) for(n=1, 40, print1(1296*n^2 - 36", ")); \\ _Vincenzo Librandi_, Feb 20 2012
%Y Cf. A005843, A158738.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 25 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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