This is a question more from combinatorics although its background is in linear algebra.
Let $A$ be a real matrix of dimension $n \times n$. We know that in general the sequence of matrices $A, A^2, A^3, \dots $ can have decreasing rank - in extreme case if for some $k$ we have $A^k=0$ then the matrix is (called) nilpotent and the patterns of transforming the rank from an initial value to $0$ can be very different, in general we could write such rank pattern $ m_1\rightarrow m_2\rightarrow \dots \rightarrow m_k=0 $.
Of course if a matrix is of full rank the exponentiation preserves rank: we have always $ n \rightarrow n$ (single possible pattern).
When the rank is less than $n$ it is also possible that exponentiation doesn't change the rank.
But the situation when the rank is decreasing from some value of $m_1$ to $m_k$ is possible in many different ways.
I'm interested in the number of different ways how it could be done.
It obviously depends on the possible Jordan Normal Forms for $n \times n$ matrix.
- What is the number formula for these forms if the only thing which is used in classification here is the way the rank decreases?
(what corresponds to unique possible sequences of changes in rank for n-dimensional matrix $ m_1\rightarrow m_2\rightarrow \dots \rightarrow m_k$)
It suffices to consider Jordan matrices. Further, it suffices to look at each Jordan block separately, since the decrease in rank of the entire Jordan matrix is the sum of the decreases in each block, upon doing the exponentiation operation.
By thinking about what happens when you raise a Jordan block to some power, you see that the rank remains the same if the eigenvalue of the Jordan block is nonzero, and otherwise decreases by $1$ each time you raise the power by one.
As an example, suppose we have a Jordan matrix, with Jordan blocks of size $5, 3, 4, 2$ corresponding to eigenvalues $0, 7, -1, 0$.
The first Jordan block's rank decreases as $4 \to 3 \to 2 \to 1 \to 0$. The last Jordan block's rank decreases as $1 \to 0$. The other Jordan blocks do not decrease rank. Thus overall the matrix rank decreases as $12 \to 10 \to 9 \to 8 \to 7 \to 7 \to \cdots$.
From the above, you can see that one restriction on the sequence of ranks is that the decreases at each step are themselves nonincreasing: the rank decrease at a particular step is the number of zero-eigenvalue Jordan blocks that have not yet become the zero matrix from the exponentiation thus far, and this number of "active" zero-eigenvalue Jordan blocks is nonincreasing. This recovers the claim in Arthur's comment.
Another restriction is that the largest decrease (the first decrease) is $\le n - r$ where $r$ is the rank of the full matrix (and the first number in the sequence), since the number of zero-eigenvalue Jordan blocks is $\le n - r$.
[Have I missed anything else?]
Actually now, I think it would make more sense to count the number of Jordan block configurations, rather than the sequence of ranks. The size configurations of the zero-eigenvalue Jordan blocks are in bijection with the sequences of ranks(?), so I believe the number of sequences is $p(0) + \cdots + p(n)$ where $p(k)$ is the number of partitions of $k$.
[Please correct me if I have made a mistake anywhere.]