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Discriminant of the field of the number having constant continued fraction [n,n,n,...].
4

%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