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a(n) = 64*n^2 - n.
2

%I #19 Sep 08 2022 08:45:42

%S 63,254,573,1020,1595,2298,3129,4088,5175,6390,7733,9204,10803,12530,

%T 14385,16368,18479,20718,23085,25580,28203,30954,33833,36840,39975,

%U 43238,46629,50148,53795,57570,61473,65504,69663,73950,78365,82908

%N a(n) = 64*n^2 - n.

%C The identity (128*n - 1)^2 - (64*n^2 - n)*16^2 = 1 can be written as A157949(n)^2 - a(n)*16^2 = 1. - _Vincenzo Librandi_, Jan 29 2012

%H Vincenzo Librandi, <a href="/A157948/b157948.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 14 in the first table at p. 85, case d(t) = t*(8^2*t-1)).

%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). - _Vincenzo Librandi_, Jan 29 2012

%F G.f.: x*(-63-65*x)/(x-1)^3. - _Vincenzo Librandi_, Jan 29 2012

%t LinearRecurrence[{3,-3,1},{63,254,573},50] (* _Vincenzo Librandi_, Jan 29 2012 *)

%t Table[64n^2-n,{n,40}] (* _Harvey P. Dale_, May 30 2017 *)

%o (Magma) I:=[63, 254, 573]; [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_, Jan 29 2012

%o (PARI) for(n=1, 40, print1(64*n^2 - n", ")); \\ _Vincenzo Librandi_, Jan 29 2012

%Y Cf. A157949.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 10 2009