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289n^2 - 2n.
2

%I #17 Sep 08 2022 08:45:42

%S 287,1152,2595,4616,7215,10392,14147,18480,23391,28880,34947,41592,

%T 48815,56616,64995,73952,83487,93600,104291,115560,127407,139832,

%U 152835,166416,180575,195312,210627,226520,242991,260040,277667,295872,314655

%N 289n^2 - 2n.

%C The identity (289*n-1)^2-(289*n^2-2*n)*(17)^2=1 can be written as A158253(n)^2-a(n)*(17)^2=1.

%H Vincenzo Librandi, <a href="/A158252/b158252.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</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 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

%F G.f.: x*(-287-291*x)/(x-1)^3.

%t LinearRecurrence[{3,-3,1},{287,1152,2595},50]

%o (Magma) I:=[287, 1152, 2595]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];

%o (PARI) a(n) = 289*n^2 - 2*n.

%Y Cf. A158253.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 15 2009