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A158673 a(n) = 60*n^2 + 1.

%I A158673 #24 Mar 20 2023 03:01:59
%S A158673 1,61,241,541,961,1501,2161,2941,3841,4861,6001,7261,8641,10141,11761,
%T A158673 13501,15361,17341,19441,21661,24001,26461,29041,31741,34561,37501,
%U A158673 40561,43741,47041,50461,54001,57661,61441,65341,69361,73501,77761,82141,86641,91261,96001
%N A158673 a(n) = 60*n^2 + 1.
%C A158673 The identity (60*n^2+1)^2 - (900*n^2+30)*(2*n)^2 = 1 can be written as a(n)^2 - A158672(n)*A005843(n)^2 = 1.
%H A158673 Vincenzo Librandi, <a href="/A158673/b158673.txt">Table of n, a(n) for n = 0..10000</a>
%H A158673 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A158673 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158673 G.f.: (1+58*x+61*x^2)/(1-x)^3.
%F A158673 a(n)= 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158673 From _Amiram Eldar_, Mar 20 2023: (Start)
%F A158673 Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)) + 1)/2.
%F A158673 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)) + 1)/2. (End)
%t A158673 LinearRecurrence[{3, -3, 1}, {1, 61, 241}, 50] (* _Vincenzo Librandi_, Feb 19 2012 *)
%t A158673 60*Range[0,40]^2+1 (* _Harvey P. Dale_, Jun 18 2021 *)
%o A158673 (Magma) I:=[1, 61, 241]; [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 19 2012
%o A158673 (PARI) for(n=0, 40, print1(60*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 19 2012
%Y A158673 Cf. A005843, A158672.
%K A158673 nonn,easy
%O A158673 0,2
%A A158673 _Vincenzo Librandi_, Mar 24 2009
%E A158673 Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009