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A158670 a(n) = 60*n^2 - 1.

%I A158670 #22 Mar 20 2023 03:02:02
%S A158670 59,239,539,959,1499,2159,2939,3839,4859,5999,7259,8639,10139,11759,
%T A158670 13499,15359,17339,19439,21659,23999,26459,29039,31739,34559,37499,
%U A158670 40559,43739,47039,50459,53999,57659,61439,65339,69359,73499,77759,82139,86639,91259,95999
%N A158670 a(n) = 60*n^2 - 1.
%C A158670 The identity (60*n^2 - 1)^2 - (900*n^2 - 30)*(2*n)^2 = 1 can be written as a(n)^2 - A158669(n)*A005843(n)^2 = 1.
%H A158670 Vincenzo Librandi, <a href="/A158670/b158670.txt">Table of n, a(n) for n = 1..10000</a>
%H A158670 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A158670 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158670 G.f.: x*(-59 - 62*x + x^2)/(x-1)^3.
%F A158670 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158670 From _Amiram Eldar_, Mar 20 2023: (Start)
%F A158670 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)))/2.
%F A158670 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)) - 1)/2. (End)
%t A158670 LinearRecurrence[{3, -3, 1}, {59, 239, 539}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)
%o A158670 (Magma) I:=[59, 239, 539]; [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 A158670 (PARI) for(n=1, 40, print1(60*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 18 2012
%Y A158670 Cf. A005843, A158669.
%K A158670 nonn,easy
%O A158670 1,1
%A A158670 _Vincenzo Librandi_, Mar 24 2009
%E A158670 Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009