show/hide this revision's text 2 suggested a new approach

UPDATE 07.11 : I have what I think are two toolsto tackle the problem. The first tool is the bounded widthbranch: Given n, form the branch suggested above startingwith 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. Theactual tool is the lemma that this branch meets the criteriafor extending the sequence and does so using up at most 2/nspace, and actually at most 1/n + 1/(2n+1) + 1/(4n+3) + ... .

Formally the lemma should read: Let for j in S bethe subsequence described above, where n in S is givenand for k in S one has both 2k and 2k+1 in S, and no otherintegers or objects are in S otherwise. This subsequencecan be part of a sequence that satisfies the spacingcriterion given in the problem, and max(x_i - x_j) fori,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 branchesto get a solution. Formally: Let for m <= M bea finite subsequence which satisfies the spacing criteriongiven. Then there are M+1 bounded width branches that canbe grafted on to the sequence, given a complete sequencethat also satisfies the spacing requirements.

Proof sketch: start with x_M, and place x_2M and x_2M+1adjacent to it. Then go backwards up to x_M+1, placingbounded width branches in the space next to the smallestundecorated leaf. The spacing requirements guarantee thatthe branches will fit without needing to move any of thefirst M x_i . Also, show that the branches aren't closeenough to each other to conflict with the spacing requirement.

So for any suitable sequence of length M, one can extendit to a complete suitable sequence at a cost of at most2/(M+1). Now with this estimate, one can go through thefirst few finite sequences and weed out those that areprovably nonoptimal.

END UPDATE 07.11
show/hide this revision's text 1

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.

Gerhard "Ask Me About System Design" Paseman, 2010.07.08