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%I #24 Mar 16 2023 04:02:53
%S 1,43,169,379,673,1051,1513,2059,2689,3403,4201,5083,6049,7099,8233,
%T 9451,10753,12139,13609,15163,16801,18523,20329,22219,24193,26251,
%U 28393,30619,32929,35323,37801,40363,43009,45739,48553,51451,54433,57499,60649,63883,67201
%N a(n) = 42*n^2 + 1.
%C The identity (42*n^2 + 1)^2 - (441*n^2 + 21)*(2*n)^2 = 1 can be written as a(n)^2 - A158603(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158604/b158604.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 + 40*x + 43*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 16 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(42))*Pi/sqrt(42) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(42))*Pi/sqrt(42) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 43, 169}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (Magma) I:=[1, 43, 169]; [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 16 2012
%o (PARI) for(n=0, 40, print1(42*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158603.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 22 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009
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