# Finding the cardinality of classes that divide all possible directed graphs into those that share k-subgraph cardinalities?

Let us have a set of nodes $$V$$, such that $$|V|=N$$. Let $$G= (V,E)$$ be an arbitrary directed graph on $$V$$. Let $$U$$ be the set of all possible directed graphs on $$V$$.Hence, $$|U| = 2^{|V|^2}$$. Now, for a given directed graph $$G$$, we define $$G[k]$$ as the set of all $$k$$-subgraphs on $$G$$. Hence, for any $$G$$, $$|G[k]| = \binom{|V|}{k}$$. Now, let $$O_{k}$$ be the set of all possible $$k$$-sized directed graphs, then all the k-subgraphs in $$G[k]$$ must be isomorphic to one of the elements of $$O_k$$, which has a size $$2^{k^2}$$.

For any graph $$G$$ let $$G(k) = $$ be a vector, such that $$a_i$$ is the number of k-subgraph in $$G$$ which are isomorphic to the $$i^{th}$$ element in $$O_{k}$$. What is the most efficient way of finding the number of $$G's$$ that share the same $$G(k)$$ ?

PS: The number of $$G's$$ counts isomorphic graphs distinctly.