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%I #26 Feb 27 2024 11:59:27
%S 255,1022,2301,4092,6395,9210,12537,16376,20727,25590,30965,36852,
%T 43251,50162,57585,65520,73967,82926,92397,102380,112875,123882,
%U 135401,147432,159975,173030,186597,200676,215267,230370,245985,262112,278751,295902
%N a(n) = 256*n^2 - n.
%C The identity (512*n - 1)^2 - (256*n^2 - n)*32^2 = 1 can be written as A158011(n)^2 - a(n)*32^2 = 1. - _Vincenzo Librandi_, Feb 10 2012
%H Vincenzo Librandi, <a href="/A158010/b158010.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 14 in the first table at p. 85, case d(t) = t*(16^2*t-1)).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: x*(-255 - 257*x)/(x-1)^3. - _Vincenzo Librandi_, Feb 10 2012
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 10 2012
%t Table[256n^2-n,{n,50}] (* _Harvey P. Dale_, Mar 30 2011 *)
%t LinearRecurrence[{3, -3, 1}, {255, 1022, 2301}, 50] (* _Vincenzo Librandi_, Feb 10 2012 *)
%o (Magma) I:=[255, 1022, 2301]; [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 10 2012
%o (PARI) for(n=1, 50, print1(256*n^2 - n", ")); \\ _Vincenzo Librandi_, Feb 10 2012
%Y Cf. A158011.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 11 2009
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