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