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

%I #27 Sep 08 2022 08:45:42

%S 0,66,260,582,1032,1610,2316,3150,4112,5202,6420,7766,9240,10842,

%T 12572,14430,16416,18530,20772,23142,25640,28266,31020,33902,36912,

%U 40050,43316,46710,50232,53882,57660,61566,65600,69762,74052,78470,83016

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

%C The identity (64*n + 1)^2 - (64*n^2 + 2*n)*8^2 = 1 can be written as A158071(n)^2 - a(n)*8^2 = 1. - _Vincenzo Librandi_, Feb 11 2012

%H Ray Chandler, <a href="/A158070/b158070.txt">Table of n, a(n) for n = 0..10000</a> (corrected by Ray Chandler, Jan 19 2019)

%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*(8^2*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), a(1)=66, a(2)=260, a(3)=582. - _Harvey P. Dale_, Jul 25 2011 [corrected by _M. F. Hasler_, Oct 09 2014]

%F G.f.: x*(66 + 62*x)/(1-x)^3. - _Vincenzo Librandi_, Feb 11 2012

%t Table[64n^2+2n,{n,40}] (* or *) LinearRecurrence[{3,-3,1},{66,260,582},40] (* _Harvey P. Dale_, Jul 25 2011 *)

%o (Magma) I:=[66, 260, 582]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 11 2012

%o (PARI) for(n=1, 50, print1(64*n^2 + 2*n", ")); \\ _Vincenzo Librandi_, Feb 11 2012

%Y Cf. A158071.

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Mar 12 2009

%E Extended to a(0)=0 by _M. F. Hasler_, Oct 09 2014