If $G = (V, E)$ is a simple, undirected graph and $T \subseteq V$, let $$N(T) = \{v \in V: \{v, t\}\in E \text{ for some }t\in T\}.$$


Given $v\in V$ we let $N\_0(v) = \{v\}$ and $N\_{k+1}(v) = N\_k(v) \cup N(N\_k(v))$ for all $k\geq 1$. The *iterated degree sequence* of $v$, denoted by $(\text{deg}\_k(v))\_{k\in\omega}$, is defined by $$\text{deg}\_k(v) = |N\_k(v)|\text{ for every }k\in \omega.$$


To every finite graph $G = (V,E)$ we associate the *iterated degree matrix* $\mathbb{D}(G) \in \mathbb{N}^{n\times n}$ (where $n=|V|$) in the following way: for every $v\in V$, take the first $n$ elements of its iterated degree sequence; order these $n$-element integer vectors [lexicographically](https://en.wikipedia.org/wiki/Lexicographic_order), and put these lexicographically ordered vectors in the matrix.


**Question.** Are there finite $G\_i = (V\_i, E\_i)$ for $i = 1,2$ with $|V\_1| = |V\_2|$, $G\_1\not\cong G\_2$, but $\mathbb{D}(G\_1) = \mathbb{D}(G\_2)$?



Non-isomorphic graphs with identical iterated degree matrix
Yes. Consider all graphs with $V=\{1,\ldots,n\}$ for which the vertex $n$ has degree $n-1$. There are $2^{n^2/2+o(n^2)}$ isomorphism classes of such graphs. But $\deg\_k(v)=n$ for all $v$ and all $k\geqslant 2$, thus there exist only at most $n^{n-1}$ distinct matrices.


