%I #22 Mar 21 2023 04:26:10
%S 61,247,557,991,1549,2231,3037,3967,5021,6199,7501,8927,10477,12151,
%T 13949,15871,17917,20087,22381,24799,27341,30007,32797,35711,38749,
%U 41911,45197,48607,52141,55799,59581,63487,67517,71671,75949,80351,84877,89527,94301,99199
%N a(n) = 62*n^2 - 1.
%C The identity (62*n^2 - 1)^2 - (961*n^2 - 31)*(2*n)^2 = 1 can be written as a(n)^2 - A158679(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158680/b158680.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.: x*(-61 - 64*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 21 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(62))*Pi/sqrt(62))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(62))*Pi/sqrt(62) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {61, 247, 557}, 50] (* _Vincenzo Librandi_, Feb 19 2012 *)
%o (Magma) I:=[61, 247, 557]; [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 19 2012
%o (PARI) for(n=1, 40, print1(62*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 19 2012
%Y Cf. A005843, A158679.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 24 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009