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%I #29 Sep 08 2022 08:45:50
%S 50,198,444,788,1230,1770,2408,3144,3978,4910,5940,7068,8294,9618,
%T 11040,12560,14178,15894,17708,19620,21630,23738,25944,28248,30650,
%U 33150,35748,38444,41238,44130,47120,50208,53394,56678,60060,63540,67118,70794,74568,78440,82410,86478,90644,94908,99270,103730,108288
%N a(n) = 49*n^2 + n.
%C The identity (98*n+1)^2-(49*n^2+n)*(14)^2 = 1 can be written as A157947(n)^2-a(n)*14^2 = 1. - _Vincenzo Librandi_, Feb 10 2012
%H Vincenzo Librandi, <a href="/A173141/b173141.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*(7^2*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*(-50-48*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 LinearRecurrence[{3, -3, 1}, {50, 198, 444}, 50] (* _Vincenzo Librandi_, Feb 10 2012 *)
%t Table[49n^2+n,{n,50}] (* _Harvey P. Dale_, Aug 22 2015 *)
%o (Magma) [ 49*n^2+n: n in [1..50] ];
%o (PARI) for(n=1, 50, print1(49*n^2 + n", ")); \\ _Vincenzo Librandi_, Feb 10 2012
%Y Cf. A157947.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Nov 22 2010
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