Let $\mathcal{E}$ be the poset of all idempotent matrices over $\mathbb{F}_q$ having only finitely many nonzero entries with the ordering $A \leq B$ iff $AB=BA=A$. Then $\mathcal{E}$ is a binomial poset with factorial function $B(n) = \frac{\gamma_n}{(q-1)^n} $ where $\gamma_n = |GL_n(\mathbb{F}_q)|$. Here are some examples of generating functions of the form $\sum_{n \geq 0}a_n\frac{x^n}{B(n)}$ that stem directly from the binomial poset $\mathcal{E}$ and the isomorphism from the reduced incidence algebra to the ring of formal power series. Let $E_{\mathcal{E}}(u) = \sum_{n \geq 0}\frac{u^n}{B(n)} $

Let $a_n$ be the number of idempotent matrices in $\text{Mat}_n(\mathbb{F}_q)$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^2(u)$

Let $a_{n,k}$ be the number of idempotent matrices in $\text{Mat}_n(\mathbb{F}_q)$ having rank $k$.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{n} a_{n,k} v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}(v u) E_\mathcal{E}(u)$

Let $a_n$ be the number of relations in the poset $\mathcal{E}_n$, i.e., the number of ordered pairs $(A,B)$ such that $A \leq B$ with $A,B \in \mathcal{E}_n $.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^3(u)$

Let $a_n$ be the number of covering relations in the poset $\mathcal{E}_n$, i.e., the number of ordered pairs $(A,B)$ such that $A$ is covered by $ B$ with $A,B \in \mathcal{E}_n $.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= u E_\mathcal{E}^2(u)$

Let $a_n$ be the number of diagonalizable matrices in $\text{Mat}_n(\mathbb{F}_q)$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^q(u)$

Let $a_n$ be the number of diagonalizable matrices in $\text{Mat}_n(\mathbb{F}_q)$ having rank $k$.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{n}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}(u) E_\mathcal{E}^{q-1}(v u)$

Let $a_{n,k}$ be the number of diagonalizable matrices in $GL_n(\mathbb{F}_q)$ with exactly $k$ distinct eigenvalues.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=(v E_\mathcal{E}( u) - v +1)^q$

Let $a_{n}$ be the number of direct sum decompositions of $\mathbb{F}_q^n$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= \exp(E_{\mathcal{E}}(u) -1)$

Let $a_{n,k}$ be the number of direct sum decompositions of $\mathbb{F}_q^n$ into exactly $k$ subspaces.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=\exp(v (E_{\mathcal{E}}(u) -1))$

Let $a_n$ be the number of \textit{periodic} matrices, i.e., elments that are contained in some (maximal) subroup of $\text{Mat}_n(\mathbb{F}_q)$. In other words, $ \displaystyle a_n = \sum_{e \in \mathcal{E}_n}|G_e|$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=E_{\mathcal{E}}(u) /(1- (q-1)u)$

Let $a_{n,k}$ be the number of ordered direct sum decompositions of $\mathbb{F}_q^n$ into exactly $k$ subspaces.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=1/(1- v (E_{\mathcal{E}}(u) -1))$

Substituting $v = -1$ in the generating function above gives $\frac{1}{E_{\mathcal{E}}(u)}$ (the image of the Moebius function $\mu$ under our isomorphism). So for the poset $\mathcal{E}_n$, we have that $\mu(\hat{0},\hat{1})$ is equal to the number of ordered direct sum decompositions of $\mathbb{F}_q^n$ into an even number of subspaces minus the number of such decompositions into an odd number of subspaces. This is an instance of Phillip Hall's Theorem.

Let $\mathcal{E}$ be the poset of all idempotent matrices over $\mathbb{F}_q$ having only finitely many nonzero entries with the ordering $A \leq B$ iff $AB=BA=A$. Then $\mathcal{E}$ is a binomial poset with factorial function $B(n) = \frac{\gamma_n}{(q-1)^n} $ where $\gamma_n = |GL_n(\mathbb{F}_q)|$. Here are some examples of generating functions of the form $\sum_{n \geq 0}a_n\frac{x^n}{B(n)}$ that stem directly from the binomial poset $\mathcal{E}$ and the isomorphism from the reduced incidence algebra to the ring of formal power series. Let $E_{\mathcal{E}}(u) = \sum_{n \geq 0}\frac{u^n}{B(n)} $

Let $a_n$ be the number of idempotent matrices in $\text{Mat}_n(\mathbb{F}_q)$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^2(u)$

Let $a_{n,k}$ be the number of idempotent matrices in $\text{Mat}_n(\mathbb{F}_q)$ having rank $k$.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{n} a_{n,k} v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}(v u) E_\mathcal{E}(u)$

Let $a_n$ be the number of relations in the poset $\mathcal{E}_n$, i.e., the number of ordered pairs $(A,B)$ such that $A \leq B$ with $A,B \in \mathcal{E}_n $.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^3(u)$

Let $a_n$ be the number of covering relations in the poset $\mathcal{E}_n$, i.e., the number of ordered pairs $(A,B)$ such that $A$ is covered by $ B$ with $A,B \in \mathcal{E}_n $.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= u E_\mathcal{E}^2(u)$

Let $a_n$ be the number of diagonalizable matrices in $\text{Mat}_n(\mathbb{F}_q)$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}^q(u)$

Let $a_n$ be the number of diagonalizable matrices in $\text{Mat}_n(\mathbb{F}_q)$ having rank $k$.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{n}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= E_\mathcal{E}(u) E_\mathcal{E}^{q-1}(v u)$

Let $a_{n,k}$ be the number of diagonalizable matrices in $GL_n(\mathbb{F}_q)$ with exactly $k$ distinct eigenvalues.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=(v E_\mathcal{E}( u) - v +1)^q$

Let $a_{n}$ be the number of direct sum decompositions of $\mathbb{F}_q^n$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}= \exp(E_{\mathcal{E}}(u) -1)$

Let $a_{n,k}$ be the number of direct sum decompositions of $\mathbb{F}_q^n$ into exactly $k$ subspaces.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=\exp(v (E_{\mathcal{E}}(u) -1))$

Let $a_n$ be the number of periodic matrices, i.e., elments that are contained in some (maximal) subroup of $\text{Mat}_n(\mathbb{F}_q)$. In other words, $ \displaystyle a_n = \sum_{e \in \mathcal{E}_n}|G_e|$.

$\displaystyle \sum_{n \geq 0}a_n \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=E_{\mathcal{E}}(u) /(1- (q-1)u)$

Let $a_{n,k}$ be the number of ordered direct sum decompositions of $\mathbb{F}_q^n$ into exactly $k$ subspaces.

$\displaystyle \sum_{n \geq 0}\sum_{k=0}^{q}a_{n,k}v^k \frac{u^n}{\frac{\gamma_n}{(q-1)^n}}=1/(1- v (E_{\mathcal{E}}(u) -1))$

Substituting $v = -1$ in the generating function above gives $\frac{1}{E_{\mathcal{E}}(u)}$ (the image of the Moebius function $\mu$ under our isomorphism). So for the poset $\mathcal{E}_n$, we have that $\mu(\hat{0},\hat{1})$ is equal to the number of ordered direct sum decompositions of $\mathbb{F}_q^n$ into an even number of subspaces minus the number of such decompositions into an odd number of subspaces. This is an instance of Phillip Hall's Theorem.