%I #14 Aug 13 2015 04:00:23
%S 5,8,13,5,29,40,53,17,85,104,5,37,173,8,229,65,293,328,365,101,445,
%T 488,533,145,629,680,733,197,5,904,965,257,1093,1160,1229,13,1373,
%U 1448,61,401,1685,1768,1853,485,2029,2120,2213,577,2405,2504,2605,677,2813
%N Discriminant of the field of the number having constant continued fraction [n,n,n,...].
%C Central numbers of the triangle at A259911.
%C It appears that a(n) = 5 for n in A002878 = (1,4,11,29,...), a bisection of the Lucas sequence.
%H Clark Kimberling, <a href="/A259912/b259912.txt">Table of n, a(n) for n = 1..10000</a>
%e [3,3,3,...] = (1/2)(3 + sqrt(13)), so that a(3) = 13.
%t t = Table[FromContinuedFraction[{n, {n}}], {n, 1, 100}];
%t Flatten[NumberFieldDiscriminant[t]]
%Y Cf. A259911, A259913.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jul 20 2015