TITLE: Sum of degree differences for simple graphs
QUESTION [8 upvotes]: For a simple graph $G$ on $n$ vertices, let us define 
$$\mathcal{I}_{n}(G)=\sum_{i,j=1}^{n}|\deg\ x_{i}-\deg\ x_{j}|^{3}.$$
I know that there are many different topological indices defined and studied for graphs. Have You ever seen such that was defined similar as above? Can You provide any references? 
I am highly interested in finding $\sup \mathcal{I}_{n}$ over all graphs with $n$ vertices (or at least some tight upper bound). What I have tried myself, was noticing that $\mathcal{I}_{n}$ must be maximized by a threshold graph - these graphs produce degree sequences that are extreme points of he convex hull of all degree sequences. But this didn't lead me too far. I will be glad for any insight.

REPLY [5 votes]: I will guess that the optimum occurs for $k$ isolated vertices and a complete graph on the other $n-k$ where $k=\lfloor\frac{n+1}5\rfloor.$ The same count occurs for $k$ vertices of degree $n-1$ and no other edges so the other $n-k$ have degree $k.$
Past that I have these observations:

A graph $G$ and the complement $\bar G$ give the same value to the sum.
If the maximum degree in an optimal $G$ is $\Delta$ then any degree $\Delta$ vertex is connected to any other. This is because connecting two such increases some of the $|\deg(x_i)-\deg(x_j)|$ but decreases none.
Similarly two vertices with the minimum degree are non-adjacent.
For the type of graph I defined above, the count is $k(n-k)(n-k-1)^3.$ The maximum over  the reals occurs at $$k=\frac{3\,n-\sqrt {4\,{n}^{2}-n+1}-1}5\approx \frac{n}{5}-\frac3{20}.$$

As commented, the exponent of $3$ is relevant. Take the conjectured optimal case of a $K_{4t}$ and $t$ isolated vertices. Deleting one edge reduces $2t$ degree difference from $4t-1$ to $4t-2$ and increases $2(4t-2)$ other differences from $0$ to $1.$ If one is summing the square or cubes of the differences that is worse. But with exponent $1$ that is an improvement.
NOTE Based on limited calculations, The same things seem maximal if we replace the exponent of 3 by 2