In the past year, I found a short elementary proof of Conjecture 6 on this website on bohemian matrices which is currently still marked as “open”.

Bohemian matrices are just matrices with all entries being integers.

The conjecture states that the number $a(n)$ of nilpotent matrices with entries in $\{0,1,2\}$ and of dimension $n\times n$ is equal to the definition of the sequence A188457 of the OEIS.

A nilpotent matrix is a matrix $A$ such that there exists a natural number $k$ such that $A^k=0$.

According to a recent comment on the OEIS (from February 2021), the sequence A188457 is equal to the number of labeled acyclic $2$-multidigraphs with $n$ vertices.

A digraph is a graph with directed edges and a $2$-multidigraph allows for up to $2$ directed edges with same source and target, i.e. it allows for up to $2$ parallel edges.

To prove the conjecture, we prove a bijective correspondence between nilpotent matrices with entries in $\{0,1,\dots ,p\}$ and of dimension $n\times n$ on the one hand and labeled acyclic $p$-multidigraphs with $n$ vertices on the other hand. Conjecture $6$ is then covered by the special case $p=2$.

The correspondence which we want to prove is given by adjacency matrices and we can see that as follows: Let $A$ be an adjacency matrix of a $p$-multidigraph $D$ whose vertices are labeled with the numbers $1,\dots ,n$.

A $p$-multidigraph corresponds to an adjacency matrix with entries in $\{0,1,\dots ,p\}$.

Now, consider a natural power $A^k$. The entry with indices $(i,j)$ of $A^k$ counts the number of walks from $i$ to $j$ with exactly $k$ steps (proof).

So $A$ is nilpotent if and only if the corresponding multidigraph, $D$, doesn’t contain any walks for some fixed length $K\in \mathbb{N}$. But this is equivalent to the multidigraph, $D$, being acyclic which finishes the proof.

Note that the special case $p=1$ of our proof also covers Conjecture 1 on the same website.