Take the set $A\_n=\{a\_1,...,a\_n\}$. Let $S\_n$ be the set of subset-sums of $A\_n$. (The subset-sum of the empty set is assumed to be zero.) Assume that there are $2^n$ unique members of $S\_n$. How many possible sortings are there of set $S\_n$?


For instance, if $n=2$, we have $S\_2=\{0,a\_1,a\_2,a\_1+a\_2\}$. The number of possible sortings of $S\_2$ is 8: 
$$\left\{ \begin{matrix} 0<a\_1<a\_2<a\_1+a\_2,\\ 0<a\_2<a\_1<a\_1+a\_2,\\ a\_2<0<a\_1+a\_2<a\_1, \\ a\_1<0<a\_1+a\_2<a\_2, \\ a\_2<a\_1+a\_2<0<a\_1, \\ a\_1<a\_1+a\_2<0<a\_2, \\ a\_1+a\_2<a\_1<a\_2<0, \\ a\_1+a\_2<a\_2<a\_1<0 \end{matrix} \right\}.$$



Number of unique sortings of subset-sums
As David Speyer points out in a comment, this is equivalent to the number of regions resulting when $\mathbb{R}^n$ is divided by hyperplanes of the form $\sum\_{i \in I}a\_i = \sum\_{i \in J}a\_i$ for all disjoint pairs of subsets $I,J \subseteq [n]$. Dual to this description, it is the number of ways that the $3^n-1$ nonzero vectors in $\{-1,0,1\}^n$ can be divided into "positive" and "negative" by a hyperplane (in general position) passing through the origin, which makes it easy to see that the answer is indeed $8$ for $n=2$.


As I mentioned in a comment, the total is always $2^n$ times what you get when making the assumption $a\_i > 0$ for all $i$. There is another symmetry that can also be exploited: making the assumption $a\_1 < a\_2 < \cdots < a\_n$ reduces the total by a further factor of $n!$, which gives a known sequence:


<http://www.oeis.org/A009997>


<http://arxiv.org/abs/math.CO/9809134>


Various key words are "coherent boolean term order", "coherent generalized term order", and "additive antisymmetric comparative probability order". It doesn't look like anyone knows the values beyond $n=7$. You'll want to check Maclagan's reference to Fine and Gill 1976 to see if they give any asymptotics.


Including the $2^nn!$ symmetries gives these values:


1. $2$
2. $8$
3. $96$
4. $5$ $376$
5. $1$ $981$ $440$
6. $5$ $722$ $536$ $960$
7. $138$ $430$ $238$ $607$ $360$


