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%I #25 Mar 23 2023 03:30:07
%S 75,303,683,1215,1899,2735,3723,4863,6155,7599,9195,10943,12843,14895,
%T 17099,19455,21963,24623,27435,30399,33515,36783,40203,43775,47499,
%U 51375,55403,59583,63915,68399,73035,77823,82763,87855,93099,98495,104043,109743,115595
%N a(n) = 76*n^2 - 1.
%C The identity (76*n^2 - 1)^2 - (1444*n^2 - 38)*(2*n)^2 = 1 can be written as a(n)^2 - A158764(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158765/b158765.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*(-75 - 78*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 23 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(19)))*Pi/(2*sqrt(19)))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(19)))*Pi/(2*sqrt(19)) - 1)/2. (End)
%t 76*Range[40]^2-1 (* or *) LinearRecurrence[{3,-3,1},{75,303,683},40] (* _Harvey P. Dale_, Jan 18 2012 *)
%o (Magma) I:=[75,303,683]; [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 21 2012
%o (PARI) for(n=1, 40, print1(76*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158764.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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