%I #19 Jun 11 2023 16:58:04
%S 402,1604,3606,6408,10010,14412,19614,25616,32418,40020,48422,57624,
%T 67626,78428,90030,102432,115634,129636,144438,160040,176442,193644,
%U 211646,230448,250050,270452,291654,313656,336458,360060,384462,409664
%N 400n^2 + 2n.
%C The identity (400*n+1)^2-(400*n^2+2*n)*(20)^2=1 can be written as A158313(n)^2-a(n)*(20)^2=1.
%H Vincenzo Librandi, <a href="/A158312/b158312.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&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 15 in the first table at p. 85, case d(t) = t*(20^2*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).
%F G.f.: x*(402+398*x)/(1-x)^3.
%t LinearRecurrence[{3,-3,1},{402,1604,3606},50]
%t Table[400n^2+2n,{n,40}] (* _Harvey P. Dale_, Jun 11 2023 *)
%o (Magma) I:=[402, 1604, 3606]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
%o (PARI) a(n) = 400*n^2 + 2*n.
%Y Cf. A158313.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 16 2009