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A158730 a(n) = 68*n^2 - 1.

%I A158730 #25 Jan 17 2025 07:47:55
%S A158730 67,271,611,1087,1699,2447,3331,4351,5507,6799,8227,9791,11491,13327,
%T A158730 15299,17407,19651,22031,24547,27199,29987,32911,35971,39167,42499,
%U A158730 45967,49571,53311,57187,61199,65347,69631,74051,78607,83299,88127,93091,98191,103427,108799
%N A158730 a(n) = 68*n^2 - 1.
%C A158730 The identity (68*n^2 - 1)^2 - (1156*n^2 - 34)*(2*n)^2 = 1 can be written as a(n)^2 - A158729(n)*A005843(n)^2 = 1.
%H A158730 Vincenzo Librandi, <a href="/A158730/b158730.txt">Table of n, a(n) for n = 1..10000</a>
%H A158730 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 A158730 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158730 G.f.: x*(-67 - 70*x + x^2)/(x-1)^3.
%F A158730 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158730 From _Amiram Eldar_, Mar 22 2023: (Start)
%F A158730 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(17)))*Pi/(2*sqrt(17)))/2.
%F A158730 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(17)))*Pi/(2*sqrt(17)) - 1)/2. (End)
%F A158730 From _Elmo R. Oliveira_, Jan 17 2025: (Start)
%F A158730 E.g.f.: exp(x)*(68*x^2 + 68*x - 1) + 1.
%F A158730 a(n) = A141759(2*n). (End)
%t A158730 LinearRecurrence[{3, -3, 1}, {67, 271, 611}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%o A158730 (Magma) I:=[67, 271, 611]; [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 20 2012
%o A158730 (PARI) for(n=1, 40, print1(68*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 20 2012
%Y A158730 Cf. A005843, A141759, A158729.
%K A158730 nonn,easy
%O A158730 1,1
%A A158730 _Vincenzo Librandi_, Mar 25 2009
%E A158730 Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009