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%I #21 Mar 20 2023 03:02:05
%S 57,231,521,927,1449,2087,2841,3711,4697,5799,7017,8351,9801,11367,
%T 13049,14847,16761,18791,20937,23199,25577,28071,30681,33407,36249,
%U 39207,42281,45471,48777,52199,55737,59391,63161,67047,71049,75167,79401,83751,88217,92799
%N a(n) = 58*n^2 - 1.
%C The identity (58*n^2 - 1)^2 - (841*n^2 - 29)*(2*n)^2 = 1 can be written as a(n)^2 - A158667(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158668/b158668.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*(-57 - 60*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(58))*Pi/sqrt(58))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(58))*Pi/sqrt(58) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {57, 231, 521}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)
%o (Magma) I:=[57, 231, 521]; [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(58*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 18 2012
%Y Cf. A005843, A158667.
%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