Number of connected components of a graph with "3-partitions" as vertices and "doublings" as edges Let $n\in\mathbb{N}$ $(=\{0,1,2,3,\dots\})$. Let $G_n$ be a (directed) graph with vertices
$$V_n = \{ \{a,b,c\} \subset\mathbb{N} : \space a+b+c = n \}$$
and edges
$$E_n = \{  (A, B) \in V^2 : \space B \text{ can be formed from } A \text{ by 'doubling'}\},$$
where 'doubling' means that we double one number of $A$ and take that same amount away from some other number of $A$ (so the sum stays at $n$).
Example of doubling (take $2$ from $3$ and add it to $2$):
$$ \{ 3, 2, 2 \} \rightarrow \{1, 4, 2\}.$$
Here's an example graph for $n=7$:

(Notes: I have ordered the numbers biggest first in the node sets. Therefore the numbers can change places after forming the doubling. I haven't included loops that are formed when the doubling leads to the same set of number)
Here are some more cases :  jsfiddle having pictures of the graphs (use the arrow keys to change $n$).
Question: How many (weakly) connected components does $G_n$ have? Denote that number by $c_n$. I have hypothesized that $c_n$ is the number of odd divisors of $n$ (OEISA001227). I have checked this upto $n=200$ and for $n=225$ and $n=315.$
My thoughts:
Let's use ordered tuples as the nodes for ease of notation.
For odd $n$ the node $(n, 0, 0)$ is by itself since it can't be formed by doubling and doubling it can only lead to itself. So this could correspond to the odd factor $n$.
Now I think I got it: A component is formed when each number of each node is divisible by an odd factor $d$ of $n$. The doubling can't lead to other components where some number isn't divisible by $d$ since $ d | (x+2dk) \iff d|x$. In other words the components are formed by considering the greatest common divisor of each node. That divisor must also be a divisor of $n$ since the numbers sum to $n$. On the other hand, we have a (undirected) path between each node that has the same $\gcd$ by ... (solving linear diophantine equation and factors of $2$ don't matter ?) How could one finish this proof?
 A: We can finish the proof as follows. We can do two transformations $A\to B$ with the triple $A$. The first one is the doubling and $(A,B)$ is the edge, whereas the second corresponds to the case when $(B,A)$ is the edge and in this case we divide by two one even number of $A$ and add that same amount to some other number of $A$ (so the sum stays at $n$). It remains to show that any triple $\{a,b,c\}$ can be transformed to the form $\{d, n-d, 0\}$, where $d$ is the greatest common odd divisor of the numbers $a$, $b$, and $c$. For this purpose it suffices to show $\{a,b,c\}$ can be transformed to the form $\{D, n-D, 0\}$, where $D=GCD(a,b,c)$ is the greatest common divisor of the numbers $a$, $b$, and $c$, because the triple $\{D, n-D, 0\}$ can be transformed to the triple $\{d, n-d, 0\}$ as follows:
$$\{D, n-D, 0\}\to \{D/2, n-D/2, 0\}\to\dots\to  \{D/2^k, n-D/2^k, 0\}\to\dots\to  \{d, n-d, 0\}.$$
If $E=GCD(a,b)$ then $D=GCD(E,c)=GCD(E,n-E)$. If $a\le b$ then by our transformations we can emulate Euclidean algorithm for $E$ as follows: 
$$\{a, b, c\}\to \{2a, b-a, c\}\to \{a, b-a, c+a\}\to\dots\to \{E, 0, n-E\}.$$
Similarly, by our transformations we can emulate Euclidean algorithm from the triple $\{E, 0, n-E\}$ for $D$.
