Chestnut

[Max A.]

On 10/21/06, David Wilson <davidwwilson at comcast.net> wrote:
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> This is a chestnut that I am convinced it is true, but I cannot prove it. I
> have posed it on math-fun occasionally, and never got an answer. I would
> like to see it posed to some better minds that some of you might know, e.g,
> JHC or other number theorist.
>
> Let S and T be sets of real numbers. Call T a divider of S if some element
> of T lies strictly between any two elements of S.
>
> For integer n >= 1, let Fence(n) be the set of all rationals with
> denominator n, that is, { k/n : k in Z }.
>
> For real set S, let f(S) be the least n such that Fence(n) is a divider of
> S, if such an n exists.
>
> Let Recip(n) be the set of all integer reciprocals on [0,1] with denominator
> <= n, that is, { 1/b : 1 <= b <= n }
>
> Let Farey(n) be the set of all rationals on [0,1] with denominator <= n,
> that is, { a/b : 0 <= a <= b, 1 <= b <= n }
>
> Is f(Farey(n)) = f(Recip(n)) for every n?
>
> The apparently common sequences a(n) = f(Recip(n)) =? f(Farey(n)) is in the
> OEIS, I cannot find it in the short time I have.