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%I #22 Mar 19 2023 02:46:12
%S 1,53,209,469,833,1301,1873,2549,3329,4213,5201,6293,7489,8789,10193,
%T 11701,13313,15029,16849,18773,20801,22933,25169,27509,29953,32501,
%U 35153,37909,40769,43733,46801,49973,53249,56629,60113,63701,67393,71189,75089,79093,83201
%N a(n) = 52*n^2 + 1.
%C The identity (52*n^2 + 1)^2 - (676*n^2 + 26)*(2*n)^2 = 1 can be written as a(n)^2 - A158643(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158644/b158644.txt">Table of n, a(n) for n = 0..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.: -(1 + 50*x + 53*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 19 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(13)))*Pi/(2*sqrt(13)) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(13)))*Pi/(2*sqrt(13)) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 53, 209}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%o (Magma) I:=[1, 53, 209]; [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=0, 40, print1(52*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158643.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009
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