Number of unique sortings of subset-sums 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: 
$$\left\{ \begin{matrix} 0<a_1<a_2<a_1+a_2,\\ 0<a_2<a_1<a_1+a_2,\\ a_2<0<a_1+a_2<a_1, \\ a_1<0<a_1+a_2<a_2, \\  a_2<a_1+a_2<0<a_1, \\ a_1<a_1+a_2<0<a_2, \\ a_1+a_2<a_1<a_2<0, \\ a_1+a_2<a_2<a_1<0 \end{matrix} \right\}.$$
 A: 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:


*

* $2$ 

* $8$ 

* $96$ 

* $5$ $376$ 

* $1$ $981$ $440$ 

* $5$ $722$ $536$ $960$ 

* $138$ $430$ $238$ $607$ $360$ 

