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%I #23 Mar 22 2023 08:14:12
%S 36,1332,5220,11700,20772,32436,46692,63540,82980,105012,129636,
%T 156852,186660,219060,254052,291636,331812,374580,419940,467892,
%U 518436,571572,627300,685620,746532,810036,876132,944820,1016100,1089972,1166436,1245492,1327140,1411380
%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 A158740(n)^2-a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158739/b158739.txt">Table of n, a(n) for n = 0..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*(1+34*x+37*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>=0} 1/a(n) = (coth(Pi/6)*Pi/6 + 1)/72.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/6)*Pi/6 + 1)/72. (End)
%p A158739:=n->1296*n^2+36: seq(A158739(n), n=0..40); # _Wesley Ivan Hurt_, Nov 20 2014
%t LinearRecurrence[{3, -3, 1}, {36, 1332, 5220}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%o (Magma) I:=[36, 1332, 5220]; [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 21 2012
%o (PARI) for(n=0, 40, print1(1296*n^2 + 36", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158740.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 25 2009
%E Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009
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