TITLE: Repeatedly taking mean values of non-empty subsets of a set: $2,\,3,\,5,\,15,\,875,\,...$
QUESTION [38 upvotes]: Consider the following iterative process. We start with a 2-element set $S_0=\{0,1\}$. At $n^{\text{th}}$ step $(n\ge1)$ we take all non-empty subsets of $S_{n-1}$, then for each subset compute the arithmetic mean of its elements, and collect the results to a new set $S_n$. Let $a_n$ be the size of $S_n$. Note that, because some subsets of $S_{n-1}$ may have identical mean values, $a_n$ may be less than the number of non-empty subsets of $S_{n-1}$ (that is, $2^{a_{n-1}}-1$).
For example, at the $1^{\text{st}}$ step we get the subsets $\{\{0\},\,\{1\},\,\{0,1\}\}$, and their means are $\{0,\,1,\,1/2\}.$ So $S_1=\{0,\,1,\,1/2\}$ and $a_1=|S_1|=3.$
At the $2^{\text{nd}}$ step we get the subsets $\{\{0\},\,\{1\},\,\{1/2\},\,\{0,\,1\},\,\{0,\,1/2\},\,\{1,\,1/2\},\,\{0,\,1,\,1/2\}\},$ and their means are $\{0,\,1,\,1/2,\,1/2,\,1/4,\,3/4,\,1/2\}.$ So, after removing duplicate values, we get $S_2=\{0,\,1,\,1/2,\,1/4,\,3/4\}$ and $a_2=|S_2|=5.$ And so on.
The sequence $\{a_n\}_{n=0}^\infty$ begins: $2,\,3,\,5,\,15,\,875,\,...$ 
I submitted it as A273525 in OEIS.
A brute-force algorithmic approach allows to easily find its elements up to $a_4=875$, but becomes computationally unfeasible after that. My question is:

What is the value of $a_5$?

It's easy to see that $5\times10^5<a_5<2^{875}<10^{264}$ (the lower bound $5\times10^5$ is found by direct enumeration of some subsets of $S_4$ on computer). Greg Martin in his answer below proves stricter bounds $2\times10^6<a_5<7\times10^{12}$. Can we find the exact value of $a_5$?

REPLY [2 votes]: First I should want to confirm the Japhet's results for $|S_{5,125}|$ and $|S_{5,200}|$. I cannot use the large amount of disk ( 17 Tb ? ) on a sole system and NFS units are not fast enough for intensive computation. But it's not the purpose of this answer and if I find a fast solution between brute force and soft computation, I'll answer twice.

The values of A(n) are very big and we cannot expect to compute them by brute force, $A(5)$ being a feat. I should want to share some observations, perhaps someone can take this further ( before me ).
First, come back to the question : How to compute A(n) ?. It is difficult because many elements of S(n) are found many times ( multiples occurences ) and we cannot merely add them.
Let $D(n)$ = all the partitions found at step $n$ , in surjection with $S(n)$. Let $F(n)$ the frequencies of each result.
Then  we can write : 
cardinal $D(n) = \sum_{i \in S(n)} F(i)$
while
$A(n) = \sum_{i \in S(n)} 1$
$D(n)$ is known and if we find all the $S(i)$ coupled with their $F(i)$, we have a way to check the result against the $D(n)$ value.
It may seem harder but obviously we are about to search effective optimization operators for inductions between consecutive $n$ to make the results computationable. It's not an useless complication.
Let's compute the couples $C(n) = \{ (s,f) \in S(n)xF(n)  \} $  for $n=4$.

Don't ask symmetries-lovers if they are happy ...
 
Frequencies of the means on the segment

 
Zoom : Frequencies of the means on the segment
Note : the repetition width is constant as it seems. For $n=4$ , the $width = 1/48$

 
another Zoom : Frequencies of the means on the segment
yes, in the cyclic width, you see an image of all the curve at the previous step. There are duplicates.

Note : many high ABC quality numbers and primaries numbers are in the remarquable values involved in this question. I'm very troubled by a confuse view giving a semi cyclic function pointing on H-Q ABC numbers and primaries. Anyway, it's an excellent summer end's puzzle.
I like too much this question because it has concerns with simulations I am following around thermodynamics , QM and information propagation. I hope that another answers will help and that this one will evolve.
Feel free to ask for my scripts. The tools are ( the very efficient ) linux sort , C++ and Grapher. The last version mixes brute force and tricks coming from the fractal approach. Next version in preparation.