Chromatic polynomial of harmonious coloring has anyone encounter a chromatic polynomial of a harmonious coloring? The assumption I made is that since the harmonious graph looks like a tree, then the chromatic polynomial of a tree can be used to determine the chromatic poly of a harmonious coloring.
 A: Just about any variant of graph coloring has its own analogue of the chromatic polynomial. It is the function $f_G(k)$ counting the number of $k$-colorings of $G$ of the type we're interested in.
Here is why this function is a polynomial in the case of harmonious coloring. (You will see that this argument is very general.) For each $i$, let $p_G(i)$ be the number of $i$-part "harmonious partitions" of $G$. A harmonious partition is a partition of $V(G)$ into $i$ nonempty parts, such that if we color each part with its own separate color, we get a harmonious coloring of $G$.
Each harmonious coloring corresponds to a harmonious partition: the color classes of the coloring. On the other hand, given an $i$-part harmonious partition and a supply of $k$ colors, there are $k(k-1)(k-2)\cdots(k-i+1)$ ways to fill in the color of each part. So the function $f_G(k)$ has the formula
$$
   f_G(k) = \sum_{i=1}^n p_G(i) k(k-1)(k-2)\cdots(k-i+1)
$$
which is a degree-$n$ polynomial in $k$.
However, I do not expect that this polynomial has properties as nice as the properties of an ordinary chromatic polynomial. For one, it does not behave nicely with combining graphs: even taking the disjoint union of two graphs is tricky!
We can find a formula for the harmonic chromatic polynomial in some special cases. For example, if $G$ is any $n$-vertex graph with diameter $2$, then its harmonic chromatic polynomial is $f_G(k) = n(n-1)(n-2)\cdots(n-k+1)$, because any two vertices at distance $2$ must have different colors in any chromatic coloring.
