A122700 Numbers k such that the length of the continued fraction for (k/m) is maximized for a unique m (1 < m < k).
2, 3, 4, 5, 8, 12, 13, 15, 21, 30, 34, 39, 40, 42, 48, 55, 56, 65, 72, 74, 80, 89, 102, 110, 130, 144, 168, 170, 176, 180, 185, 193, 194, 208, 233, 272, 275, 288, 297, 299, 312, 336, 340, 377, 400, 445, 456, 468, 517, 546, 550, 610, 638, 699, 715, 720, 754, 777
Offset: 1
Keywords
Examples
5/2 = [2, 2], 5/3 = [1, 1, 2], 5/4 = [1, 4]; thus the continued fraction for 5/m is maximized at the unique value m=3, and 5 is in the sequence.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..500
Crossrefs
Cf. A089642. - R. J. Mathar, Aug 02 2008
Programs
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Mathematica
cfQ[v_] := Count[v, Max[v]] == 1; q[n_] := cfQ[Length[ContinuedFraction[#/n]] & /@ Range[2, n - 1]]; Join[{2}, Select[Range[3, 1000], q]](* Amiram Eldar, Jun 25 2022 *)
Comments