%I #21 Mar 16 2023 04:04:10
%S 45,183,413,735,1149,1655,2253,2943,3725,4599,5565,6623,7773,9015,
%T 10349,11775,13293,14903,16605,18399,20285,22263,24333,26495,28749,
%U 31095,33533,36063,38685,41399,44205,47103,50093,53175,56349,59615,62973,66423,69965,73599
%N a(n) = 46*n^2 - 1.
%C The identity (46*n^2 - 1)^2 - (529*n^2 - 23)*(2*n)^2 = 1 can be written as a(n)^2 - A158633(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158634/b158634.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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: x*(-45 - 48*x + x^2)/(x-1)^3.
%F From _Amiram Eldar_, Mar 16 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(46))*Pi/sqrt(46))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(46))*Pi/sqrt(46) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {45, 183, 413}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%o (Magma) I:=[45, 183, 413]; [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 17 2012
%o (PARI) for(n=1, 40, print1(46*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158633.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009