A standard graph invariant is the degree sequence, but it is well-known, that the degree sequence is not a complete graph invariant, i.e. a graph cannot be reconstructed uniquely from its degree sequence.

That means: the degree sequence contains too little information about the graph. But what about generalizations of the degree sequence, containing more information but still relying on degrees only, i.e. counting?

[B⁰] Consider the set $D^0 = [n]^{[1]}$ of all functions $d^0:[1] \rightarrow [n]$ and assign to each $d^0 \in D^0$ the number of vertices with degree $d^0_i = d^0$ (precisely $d^0_i = d^0(1)$).

[C⁰] This yields a function $D_G^1: D^0 \rightarrow [n]$, which bears the same information as the degree sequence. Let's call it degree spectrum. Note, that it takes into account only the 1-neighbourhood of each node.

[A¹] Consider for each node $v_i$ the function $d^1_i: D^0 \rightarrow [n]$ assigning to each $d^0 \in D^0$ the number of its neighbours with degree $d^0_j = d^0$.

[B¹] Consider the set $D^1 = [n]^{D^0}$ of all functions $d^1: D^0 \rightarrow [n]$ and assign to each $d^1 \in D^1$ the number of vertices with $d^1_i = d^1$.

[C¹] This yields a function $D_G^2: D^1 \rightarrow [n]$, which is another graph invariant, taking into account the 2-neighbourhood of each node.

This process can be continued:

[A^k+1] Consider for each node $v_i$ the function $d^{k+1}_i: D^k \rightarrow [n]$ assigning to each $d^k \in D^k$ the number of its neighbours with $d^k_j = d^k$

[B^k+1] Consider the set $D^{k+1} = [n]^{D^k}$ of all functions $d^{k+1}: D^k \rightarrow [n]$ and assign to each $d^{k+1} \in D^{k+1}$ the number of vertices with $d^{k+1}_i = d^{k+1}$.

[C^k+1] This yields a function $D_G^{k+2}: D^{k+1} \rightarrow [n]$, which is another graph invariant, taking into account the k+2-neighbourhood of each node.

Question: Has this kind of generalized degree spectrum already been investigated? Under which name?

If it not has been investigated already I will feel free to continue this post, otherwise I will stop here.

A standard graph invariant is the degree sequence, but it is well-known, that the degree sequence is not a complete graph invariant, i.e. a graph cannot be reconstructed uniquely from its degree sequence.

That means: the degree sequence contains too little information about the graph. But what about generalizations of the degree sequence, containing more information but still relying on degrees only, i.e. counting?

Let $G$ be a undirected graph with $n$ nodes.

[B⁰] Consider the set $D^0 = [n]^{[1]}$ of all functions $d^0:[1] \rightarrow [n]$ and assign to each $d^0 \in D^0$ the number of vertices with degree $d^0_i = d^0$ (precisely $d^0_i = d^0(1)$).

[C⁰] This yields a function $D_G^1: D^0 \rightarrow [n]$, which bears the same information as the degree sequence. Let's call it degree spectrum. Note, that it takes into account only the 1-neighbourhood of each node.

[A¹] Consider for each node $v_i$ the function $d^1_i: D^0 \rightarrow [n]$ assigning to each $d^0 \in D^0$ the number of its neighbours with degree $d^0_j = d^0$.

[B¹] Consider the set $D^1 = [n]^{D^0}$ of all functions $d^1: D^0 \rightarrow [n]$ and assign to each $d^1 \in D^1$ the number of vertices with $d^1_i = d^1$.

[C¹] This yields a function $D_G^2: D^1 \rightarrow [n]$, which is another graph invariant, taking into account the 2-neighbourhood of each node.

This process can be continued:

[A^k+1] Consider for each node $v_i$ the function $d^{k+1}_i: D^k \rightarrow [n]$ assigning to each $d^k \in D^k$ the number of its neighbours with $d^k_j = d^k$

[B^k+1] Consider the set $D^{k+1} = [n]^{D^k}$ of all functions $d^{k+1}: D^k \rightarrow [n]$ and assign to each $d^{k+1} \in D^{k+1}$ the number of vertices with $d^{k+1}_i = d^{k+1}$.

[C^k+1] This yields a function $D_G^{k+2}: D^{k+1} \rightarrow [n]$, which is another graph invariant, taking into account the k+2-neighbourhood of each node.

Question: Has this kind of generalized degree spectrum already been investigated? Under which name?

If it not has been investigated already I will feel free to continue this post, otherwise I will stop here.