How is this combinatorial behavior called? If you sum all possible combinatios from a set of numbers, you will get a pattern on the deltas between each element if you order from lowest to highest
How is this called? Is there more information on this behavior?
For example, $\{2,5,8,12\}$ gives you the sorted sums $$2,5,7,8,10,12,13,14,15,17,19,20,22,25,27\,,$$ whose deltas  $$2,3,2,1,2,2,1,1,1,2,2,1,2,3,2$$ show a pattern.
Thanks.
 A: The deltas will always form a palindrome (read the same from left to right as from right to left). 
I think that's not too hard to prove. Here's the idea: (in your example) the first number is the smallest, the last is the sum of everything. The second number is the second smallest, and the second from the end is the sum of everything but the smallest. So the first and last deltas are the smallest number.
A: The name of this in combinatorics is partitioning, which is the number of ways of separating a number into sets of smaller numbers which add up to it.  Your initial set is just one way of partitioning the number $27$ and you are describing the number of different ways of further partitioning that partition of $27$.
There is a massive pattern of ways to partition any number $n$, which has $n$ at one end, and $n$ 1's at the other end, and partitions into a certain number of sets in-between.  The number of partitions of any number is given by an incredible equation derived by Ramanujan and an extensive theory of partitioning which you can easily find in Wikipedia here: https://en.wikipedia.org/wiki/Partition_(number_theory)
If you add the empty set to your list $2,5,7,\ldots$ then that list becomes the power set of your starting set.  The reason the "deltas" form a palindrome is derived from the fact that for every integer $n$, the set of all partitionings of $n$ contains every set of the partitionings of all integers $m< n$.
A: I hope I understand you correctly.  Please check if this is exactly what you ask.

Let $n$ be a positive integer.  Given $n$ real numbers $a_1,a_2,\ldots,a_n$ (not necessarily positive nor distinct), $M$ denotes the multiset of all sums of the form $\sum\limits_{i\in S}\,a_i$, where $S$ is an arbitrary subset of $\{1,2,\ldots,n\}$ (I will include $S=\emptyset$ for the sake of completeness).
Then, you sort the elements of $M$, say, $M=\left\{m_1,m_2,\ldots,m_{2^n}\right\}$, where $m_1\leq m_2\leq \ldots \leq m_{2^n}$.  Note that $$m_1=\sum\limits_{\substack{{i\in[n]}\\{a_i<0}}}\,a_i\text{ and }m_{2^n}=\sum\limits_{\substack{{i\in[n]}\\{a_i>0}}}\,a_i$$ always.
Your claim is that, if the $i$-th delta is given by $d_{i}:=m_{i+1}-m_i$ for $i=1,2,\ldots,2^n$, then the sequence $\left(d_1,d_2,\ldots,d_{2^n-1}\right)$ is palindromic, namely, $$d_i=d_{2^n-i}$$ for each $i=1,2,\ldots,2^{n}-1$.

This follows immediately from the observation that $m_{j}+m_{2^n+1-j}$ is precisely $\sigma:=\sum\limits_{i=1}^n\,a_i$ for all $j=1,2,\ldots,2^{n}$.  Therefore,
$$d_{j}=m_{j+1}-m_j=\left(\sigma-m_{2^n-j}\right)-\left(\sigma-m_{2^n+1-j}\right)=m_{2^n+1-j}-m_{2^{n}-j}=d_{2^n-j}$$
for every $j=1,2,\ldots,2^{n}-1$.
In fact, the same is true for the underlying set $\tilde{M}$ of $M$.  Write $\tilde{M}:=\left\{\tilde{m}_1,\tilde{m}_2,\ldots,\tilde{m}_k\right\}$ with $\tilde{m}_1<\tilde{m}_2<\ldots<\tilde{m}_k$.  Then, it is still true that
$$\tilde{m}_j+\tilde{m}_{k+1-j}=\sigma$$
for all $j=1,2,\ldots,k$.  Consequently, if $\tilde{d}_i:=\tilde{m}_{i+1}-\tilde{m}_i$ for $i=1,2,\ldots,k-1$, then the sequence $\left(\tilde{d}_1,\tilde{d}_2,\ldots,\tilde{d}_{k-1}\right)$ is also palindromic for the same reason.  (It is easier to prove the claim first for $\tilde{M}$, and then for $M$.)
