Was any progress ever made on Max's problem?? Thanks in anticipation of a response. R. ---------- Forwarded message ---------- Date: Thu, 23 Nov 2006 17:04:09 -0800 From: Max A. <maxale@gmail.com> To: David Wilson <davidwwilson@comcast.net> Cc: Sequence Fans <seqfan@ext.jussieu.fr> Subject: Re: Chestnut On 10/21/06, David Wilson <davidwwilson@comcast.net> wrote:

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.