A229288 Least k such that the numerator of the continued fraction [2,..,2,k] (n 2s) is prime.
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, 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, 2, 1, 11, 107, 6, 51, 18, 31, 9, 13, 13, 13, 9, 81, 7, 90, 13, 19
Offset: 1
Keywords
Examples
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.
Programs
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Mathematica
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}]]