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%I #15 Aug 18 2015 15:39:32
%S 5,12,21,8,5,60,77,24,13,140,165,12,221,28,285,5,357,44,437,120,21,
%T 572,69,168,29,780,93,56,957,1020,1085,8,1221,1292,1365,40,1517,1596,
%U 1677,440,205,1932,2021,33,5,92,2397,156,53,12,2805,728,3021,348,3245
%N Discriminant of the number field containing the number with periodic continued fraction [1,n,1,n,1,n,...].
%C a(n) is the first term in row n of the triangle at A259911.
%C It appears that a(n) = 5 if n is a nonzero term of A004146.
%H Clark Kimberling, <a href="/A259913/b259913.txt">Table of n, a(n) for n = 1..10000</a>
%e [1,3,1,3,1,3,...] = (1/6)(3 + sqrt(21)), so that a(3) = 21.
%t v = Table[FromContinuedFraction[{1, {n, 1}}], {n, 1, 60}];
%t Flatten[NumberFieldDiscriminant[v]]
%Y Cf. A259911.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jul 20 2015
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