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A158587 a(n) = 289*n^2 - 17.

%I A158587 #28 Jan 16 2025 09:38:56
%S A158587 272,1139,2584,4607,7208,10387,14144,18479,23392,28883,34952,41599,
%T A158587 48824,56627,65008,73967,83504,93619,104312,115583,127432,139859,
%U A158587 152864,166447,180608,195347,210664,226559,243032,260083,277712,295919,314704,334067,354008,374527
%N A158587 a(n) = 289*n^2 - 17.
%C A158587 The identity (34*n^2 - 1)^2 - (289*n^2 - 17)*(2*n)^2 = 1 can be written as A158588(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158587 Vincenzo Librandi, <a href="/A158587/b158587.txt">Table of n, a(n) for n = 1..10000</a>
%H A158587 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 A158587 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158587 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158587 G.f.: 17*x*(-16 - 19*x + x^2)/(x-1)^3.
%F A158587 From _Amiram Eldar_, Mar 14 2023: (Start)
%F A158587 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(17))*Pi/sqrt(17))/34.
%F A158587 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(17))*Pi/sqrt(17) - 1)/34. (End)
%F A158587 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A158587 E.g.f.: 17*(exp(x)*(17*x^2 + 17*x - 1) + 1).
%F A158587 a(n) = 17*A321180(n). (End)
%t A158587 LinearRecurrence[{3, -3, 1}, {272, 1139, 2584}, 50] (* _Vincenzo Librandi_, Feb 15 2012 *)
%t A158587 289*Range[40]^2-17 (* _Harvey P. Dale_, Jan 30 2019 *)
%o A158587 (Magma) I:=[272, 1139, 2584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 15 2012
%o A158587 (PARI) for(n=1, 50, print1(289*n^2-17", ")); \\ _Vincenzo Librandi_, Feb 15 2012
%Y A158587 Cf. A005843, A158588, A321180.
%K A158587 nonn,easy
%O A158587 1,1
%A A158587 _Vincenzo Librandi_, Mar 22 2009
%E A158587 Comment rewritten by _R. J. Mathar_, Oct 16 2009