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%I #5 Sep 21 2013 12:58:29
%S 1,1,1,1,3,1,1,5,2,15,2,13,6,3,1,5,3,1,5,13,12,55,6,97,6,9,2,1,5,39,
%T 28,7,14,31,11,83,21,51,2,31,30,51,4,7,3,1,40,37,21,27,2,95,15,9,14,5,
%U 2,1,11,107,6,51,18,31,9,13,13,13,9,81,7,90,13,19
%N Least k such that the numerator of the continued fraction [2,..,2,k] (n 2s) is prime.
%e The numerators of the continued fraction [2,2,2,2,2,k] for k=1,2,3 are 99,169,239; the first two are not prime and 239 is, so a(5) = 3.
%t z = 160; c[n_, k_] := Join[ContinuedFraction[1 + Sqrt[2], n], {k}]; x[n_, k_] := Numerator[FromContinuedFraction[c[n, k]]]; t[n_] := Table[x[n, k], {k, 1, z}]; u = Table[First[Select[t[n], PrimeQ]], {n, 1, z}]; Flatten[Table[Position[t[n], u[[n]]], {n, 1, z}]]
%Y Cf. A099088, A086395, A229287.
%K nonn,cofr
%O 1,5
%A _Clark Kimberling_, Sep 19 2013
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