A sequential optimizing task Find distinct positive real numbers $x_1$ , $x_2$ , ...  of least supremum such that, for each positive integer $n$, any two of 0, $x_1$ , $x_2$ ,..., $x_n$ differ by $1/n$ or more.
Note that the hurdle term $1/n$ is optimal in the sense that any replacement for it would need to stay below a constant multiple of it to allow a finite supremum. By a nonconstructive proof, there is a unique solution minimal with respect to the lexicographic ordering of real sequences; so a constructed solution (while eluding me) doesn't seem impossible.
    Although I haven't seen this problem anywhere, it looks too simple not to have been posed before. Any pointers would be welcome. 
 A: I have enough musings to post them as an answer, rather than fill 
up comment space.
First : Use a simple recursive construction to get
an upper bound on the supremum.  This places x_1 at 1,
x_2n at x_n - 1/2n, and x_(2n+1) at x_n + 1/(2n+1).  This
gives an upper bound of sum{i positive integer} 1/(2^i - 1)
which is some number less than 169/105.  Of course, you need
to prove this construction works.
Second: viewed as a tree with node n branching to children
x_(2n) and x_(2n+1), note that you can prune and graft the
tree, reshaping it as needed.  Specifically, start by
exchanging branches at nodes 11 and 7.  (This works because
1/2 + 1/5 + 2/7 < 2* 1/2 = 1.)  You may find that repruning
smaller branches leads more quickly to a near optimal bound.
Even with the one graft made, the upper bound is reduced to
less than 1147/759.
Third: start determining optimal placements for the first
n terms for small n, which meet the conditions and stay
below the bounds established above.  A computer
simulation should quickly run through placements for n
up to 12 which stay below the lower bound.  For example,
by hand one sees that x_1 < x_2 < x_n for n < 80 already
leads to non optimal placements, so that combined with
some analysis should prove that x_2 < x_1 in an optimal
placement. 
This approach should lead you quickly to the first four
decimal digits of the supremum.
UPDATE 07.11 :  I have what I think are two tools
to tackle the problem.  The first tool is the bounded width
branch:  Given n, form the branch suggested above starting
with x_n "representing" 1/n, placing x_2n at x_n - 1/2n and 
x_2n+1 at x_n + 1/(2n+1), and continuing recursively.  The
actual tool is the lemma that this branch meets the criteria
for extending the sequence and does so using up at most 2/n
space, and actually at most 1/n + 1/(2n+1) + 1/(4n+3) + ... .
Formally the lemma should read: Let  for j in S be
the subsequence described above, where n in S is given
and for k in S one has both 2k and 2k+1 in S, and no other
integers or objects are in S otherwise.  This subsequence
can be part of a sequence that satisfies the spacing
criterion given in the problem, and max(x_i - x_j) for
i,j coming from S is less than 2/n.
The second tool is that, given any starting sequence,
there is a way to extend it using bounded width branches
to get a solution.  Formally: Let  for m <= M be
a finite subsequence which satisfies the spacing criterion
given.  Then there are M+1 bounded width branches that can
be grafted on to the sequence, given a complete sequence
that also satisfies the spacing requirements.
Proof sketch: start with x_M, and place x_2M and x_2M+1
adjacent to it.  Then go backwards up to x_M+1, placing
bounded width branches in the space next to the smallest
undecorated leaf.  The spacing requirements guarantee that
the branches will fit without needing to move any of the
first M x_i .  Also, show that the branches aren't close
enough to each other to conflict with the spacing requirement.
So for any suitable sequence of length M, one can extend
it to a complete suitable sequence at a cost of at most
2/(M+1).  Now with this estimate, one can go through the
first few finite sequences and weed out those that are
provably nonoptimal.
END UPDATE 07.11
Gerhard "Ask Me About System Design" Paseman, 2010.07.08
A: @Gerhard: Thank you for your interest. (Because my MO identity was lost, I have to reply as a new user.) What makes the problem hard is that, even if you have a minimal-supremum sequence of n points, adding the (n + 1)th point may bump up the supremum by 1/(n + 1); whereas, if you had chosen the first n points just a whisker less optimally, you might have been able to fit in an extra point or two before you are forced to bump up: The best (n + k)th sequence is not always the best k-th extension of the best n-th sequence. What makes the problem tantalizing is that approaching stepwise with egyptian fractions, as you have outlined, can probably get very near indeed to the optimum. However, while the best sequence for each finitized problem is rational, my guess is that the best infinite sequence isn't rational. ...........  John Bentin 
A: I'll now put forward my candidate solution to the problem. It clearly satisfies the hurdle condition, but I can't prove its optimality. To get a handle on the algorithm, let's represent $x_1$ , $x_2$ , … as hotel guests numbered accordingly. Recall Hilbert's hotel, where the unfortunate guests were ever being shunted from their room to a higher-numbered room. The hotel in this case, rather than having a countably infinite number of discrete rooms, is a continuum of “rooms”, represented by the points of a closed bounded real interval; [0 , 1.4] is big enough, as it turns out. The “guests”, numbered 1, 2, … , are movable tags, each assigned to a rational point in the interval. Unlike Hotel Hilbert, which is always full, this hotel starts empty apart from the proprietor who resides permanently at 0, and the guests arrive one by one in the order 1, 2, … . The proprietor operates the strict rule that, when there are a total of $m$ guests in the hotel, there must be a space of at least $1/m$ between the residents (including himself), for $m$ = 1, 2, … .
     Guest 1 is assigned to the point 1. Guest 2 is placed at 1/2. When guest 3 arrives, she is put at 1/3, while guests 1 and 2 are moved up by a distance 2/3 – 2/(3 + 1) = 1/6. When guest 4 comes, he is allotted to 1/2 + 1/6 + 1/4 = 11/12. The general rule is as follows.
For $k$ = 1, 2, … , after $2^k$  guests have been accommodated, the next $2^k$  are put consecutively into the $2^k$ spaces between the proprietor and the first $2^k$  guests, in left-to-right order: An odd-numbered arrival, say guest $2n$ – $1$, is assigned to the point below her right-hand neighbour (guest $n$) that is 1/($2n$ – $1$) above her left-hand neighbour, while all the guests to her right are moved up by a distance 2/($2n$ – 1) – $1/n$; when the next, even-numbered, guest arrives (guest $2n$), he can just go to the midpoint of the next space up, at a distance 1/$2n$ above his left-hand neighbour (guest $n$), and mercifully no resident has to move until the next (odd-numbered) guest arrives.
     The result is that the guests, now identified with their limiting room positions, are each the sum of two parts: The first part is a sum of a finite number of distinct fractions of the form $1/n$, while the second is an infinite series whose terms are of the form 1/$n$($2n$ – 1), where $n$ is a positive integer. Generally an infinite number of terms of the latter type are absent; only in the case of $x_1$ is the series free of gaps, and then the first “sum” has only one term. Thus $x_1$ = 1 + ∑{1/$j$($2j$ – 1) : $j$ = 2, 3,  …} =  ln 4, and this is also the supremum of the sequence.
An alternative characterization, suggested by Gerhard Paseman, is as follows:   For j from 2^(k-1)+1 to 2^k, you will arrange to place guest (2j-1) to the left of guest j and guest 2j to the right of guest j. Since space to the left of guest j has previously been guaranteed to be 1/j from his/her lefthand neighbor, guest (2j-1) needs more since it needs 1/(2j-1) space to his/her left and right. So add the difference delta_j = (2/(2j-1) – 1/j) to the left of guest j and shift guest j and every guest on the right of guest j by this difference delta_j. Since guest 2j does not need more than 1/j = (2/2j) space, no such adjustment is needed for guest 2j. This gives guest 1 infinitely many adjustments; 1 ends up at place 1 + sum(j > 1) delta_j = 1 + sum(j > 1) {2 [ 1/(2j-1) - 1/2j ] } = 2 ln(2).
A: Belated thanks to Kevin O'Bryant for his pointer to discrepancy theory. This led me eventually to a source where the problem is solved: See Theorem 6.7 in Harald Niederreiter's book Random Number Generation and Quasi Monte Carlo Methods (SIAM 1992). The logarithmic sequence described by Tracy Hall is due to Rusza and is indeed optimal. 
