Number of partition of $[1,m]$ of cardinality $n$.

Question

Given an integer $m \geq 1$, let $\pi_{m, n}$ the set of partition $\pi$ of $[1,m] = \{1,\dots,m\}$ such that $|\pi|$ equal $n$.

I am interested in the cardinality of $\pi_{m, n} $.

For example let $m = 2$. Then $\pi_{2,2} = \{ [\{1\},\{2\}] \}$, $\pi_{2,1} = \{ [\{1,2\}]\}$

Thus $|\pi_{2,2}|=|\pi_{2,1}|=1$.

Let $m = 3$. Then

$\pi_{3,3} = \{ [\{1\},\{2\}, \{3\}] \}$

$\pi_{3,2} = \{ [\{1,2\}, \{3\}], [\{1,3\},\{2\}], [\{3,2\}, \{1\}]\}$

$\pi_{3,1} = \{ [\{1,2, 3\}]$

Of cardinality 1,3, and 1.

I have never encountered $\pi_{m, n} $. Is there a name for this problem? Are there any nice properties of $|\pi_{m, n}|$? Ideally I need non asymptotic lower and upper bound of $|\pi_{m, n}|$.

I apologize if the problem is not well formulated. I am not used to combinatorics.

Thanks,

Edit : context.

If it could help, I need to control :

$$\sum_{n=1}^m |\pi_{m, n}| (-1)^n (n-1)! q^n$$ for $0 \leq q \leq 1$.

Answer 1

Let us define functions $$F_m(q)=\sum_{n=1}^mS(m,n)(-1)^n(n-1)!q^n$$ where $S(m,n)=\pi_{m,n}$ the traditional notation for the Stirling numbers of the second kind (which I am not familiar with).

Then by using the identity $S(m,n)=S(m-1,n-1)+nS(m-1,n)$, we get $$\begin{align} F_m(q)&=\sum_{n=1}^mS(m-1,n-1)(-1)^n(n-1)!q^n+\sum_{n=1}^mS(m-1,n)(-1)^nn!q^n\\ &=\sum_{n=1}^{m-1}S(m-1,n)(-1)^{n+1}n!q^{n+1}+\sum_{n=1}^{m-1}S(m-1,n)(-1)^nn!q^n\\ &=-q^2F'_{m-1}(q)+qF'_{m-1}(q)\\ &=(q-q^2)F'_{m-1}(q). \end{align}$$ I found it here: Equation $(17)$. Let $G_m(q)=(-1)^m\text{Li}_{1-m}(1-\frac1q)$. Then, this function satisfies the same equation: $G_m(q)=(q-q^2)G'_{m-1}(q)$. So, $$F_m(q)=(-1)^m\text{Li}_{1-m}(1-\frac1q).$$

Answer 2

Fix $m\in\mathbb N$. By definition, $\left|\pi_{m,n}\right|$ counts the number of partitions $\pi$ of $\{1,\dots,m\}$ with $\left|\pi\right|=n$ ($1\leq n\leq m$). The sum of these numbers $$\left|\pi_{m,1}\right|+\left|\pi_{m,2}\right|+\cdots+\left|\pi_{m,m}\right|=\sum_{n=1}^m\left|\pi_{m,n}\right|$$ then counts the total number of partitions of $\{1,\dots,m\}$. Since this number is precisely the Bell number $B_m$, we have the following identity: $$B_m=\sum_{n=1}^m\left|\pi_{m,n}\right|$$ This gives us crude upper and lower bounds for $\sum_{n=1}^m\left|\pi_{m,n}\right|(-1)^n(n-1)!q^n$: \begin{align*} \left|\sum_{n=1}^m\left|\pi_{m,n}\right|(-1)^n(n-1)!q^n\right| &\leq \sum_{n=1}^m\left|\left|\pi_{m,n}\right|(-1)^n(n-1)!q^n\right|\\ &= \sum_{n=1}^m\left|\pi_{m,n}\right|(n-1)!|q|^n\\ &\leq\sum_{n=1}^m\left|\pi_{m,n}\right|(n-1)!\text{ (because }0\leq q\leq 1\text{)}\\ &\leq\sum_{n=1}^m\left|\pi_{m,n}\right|(m-1)!\text{ (since }n\leq m\text{)}\\ &=(m-1)!B_m \end{align*} $$\implies -(m-1)!B_m\leq \sum_{n=1}^m\left|\pi_{m,n}\right|(-1)^n(n-1)!q^n\leq (m-1)B_m$$