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%I #25 Mar 16 2023 12:23:49
%S 290,579,868,1157,1446,1735,2024,2313,2602,2891,3180,3469,3758,4047,
%T 4336,4625,4914,5203,5492,5781,6070,6359,6648,6937,7226,7515,7804,
%U 8093,8382,8671,8960,9249,9538,9827,10116,10405,10694,10983,11272,11561
%N a(n) = 289n + 1.
%C The identity (289*n+1)^2-(289*n^2+2*n)*(17)^2=1 can be written as a(n)^2-A158254(n)*(17)^2=1.
%H Vincenzo Librandi, <a href="/A158255/b158255.txt">Table of n, a(n) for n = 1..10000</a>
%H E. J. Barbeau, <a href="http://www.math.toronto.edu/barbeau/home.html">Polynomial Excursions</a>, Chapter 10: <a href="http://www.math.toronto.edu/barbeau/hxpol10.pdf">Diophantine equations</a> (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(17^2*t+2)).
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F G.f.: x*(290-x)/(1-x)^2. - _Bruno Berselli_, Mar 21 2011
%F a(n) = 2*a(n-1)-a(n-2).
%t 289Range[50]+1 (* _Harvey P. Dale_, Mar 21 2011 *)
%t LinearRecurrence[{2,-1},{290,579},50]
%o (Magma) I:=[290, 579]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];
%o (PARI) a(n)=289*n+1 \\ _Charles R Greathouse IV_, Mar 22, 2011
%Y Cf. A158254.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 15 2009
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