%I #21 Mar 20 2023 03:02:40
%S 870,3570,8070,14370,22470,32370,44070,57570,72870,89970,108870,
%T 129570,152070,176370,202470,230370,260070,291570,324870,359970,
%U 396870,435570,476070,518370,562470,608370,656070,705570,756870,809970,864870,921570,980070,1040370,1102470
%N a(n) = 900*n^2 - 30.
%C The identity (60*n^2 - 1)^2 - (900*n^2 - 30)*(2*n)^2 = 1 can be written as A158670(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158669/b158669.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.: 30*x*(-29 - 32*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 20 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(30))*Pi/sqrt(30))/60.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(30))*Pi/sqrt(30) - 1)/60. (End)
%t LinearRecurrence[{3, -3, 1}, {870, 3570, 8070}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)
%o (Magma) I:=[870, 3570, 8070]; [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 18 2012
%o (PARI) for(n=1, 40, print1(900*n^2 - 30", ")); \\ _Vincenzo Librandi_, Feb 18 2012
%Y Cf. A005843, A158670.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 24 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009