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%I #27 Mar 07 2023 02:26:47
%S 90,333,738,1305,2034,2925,3978,5193,6570,8109,9810,11673,13698,15885,
%T 18234,20745,23418,26253,29250,32409,35730,39213,42858,46665,50634,
%U 54765,59058,63513,68130,72909,77850,82953,88218,93645,99234,104985,110898,116973,123210
%N a(n) = 81*n^2 + 9.
%C The identity (18n^2 + 1)^2 - (81n^2 + 9)*(2n)^2 = 1 can be written as A157889(n)^2 - a(n)*A005843(n+1)^2 = 1. - _Vincenzo Librandi_, Feb 05 2012
%H Vincenzo Librandi, <a href="/A157888/b157888.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 From _Vincenzo Librandi_, Feb 05 2012: (Start)
%F G.f: x*(90 + 63*x + 9*x^2)/(1-x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F From _Amiram Eldar_, Mar 07 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (coth(Pi/3)*Pi/3 - 1)/18.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/3)*Pi/3)/18. (End)
%t LinearRecurrence[{3, -3, 1}, {90, 333, 738}, 40] (* _Vincenzo Librandi_, Feb 05 2012 *)
%t 81*Range[40]^2+9 (* _Harvey P. Dale_, Aug 05 2015 *)
%o (Magma) I:=[90, 333, 738]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 05 2012
%o (PARI) for(n=1, 40, print1(81*n^2 + 9", ")); \\ _Vincenzo Librandi_, Feb 05 2012
%Y Cf. A005843, A157889.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 08 2009
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