%I #23 Nov 27 2024 15:17:01
%S 62,252,570,1016,1590,2292,3122,4080,5166,6380,7722,9192,10790,12516,
%T 14370,16352,18462,20700,23066,25560,28182,30932,33810,36816,39950,
%U 43212,46602,50120,53766,57540,61442,65472,69630,73916,78330,82872
%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 (A152691(n+1) - 1)^2 - a(n)*8^2 = 1. - _Vincenzo Librandi_, Feb 11 2012
%H Vincenzo Librandi, <a href="/A158067/b158067.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*(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 G.f.: x*(-62 - 66*x)/(x-1)^3. - _Vincenzo Librandi_, Feb 11 2012
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 11 2012
%t LinearRecurrence[{3, -3, 1}, {62, 252, 570}, 50] (* _Vincenzo Librandi_, Feb 11 2012 *)
%t Table[64n^2-2n,{n,40}] (* _Harvey P. Dale_, Nov 27 2024 *)
%o (Magma)[64*n^2 - 2*n: n in [1..50]]
%o (PARI) for(n=1, 50, print1(64*n^2 - 2*n ", ")); \\ _Vincenzo Librandi_, Feb 11 2012
%Y Cf. A152691.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 12 2009