Is this variant on set partition explored?

Question

Let $[n]=\{1,2,\dots,n\}$ and fix $r\in[n]$. Define the set $\mathcal{B}_{n,r}$ as the set of all set partitions of $[n]$ into disjoint non-empty blocks such that each block $B$ satisfies:

$\,\,\,\,\,\,\,\,\,\,$ either $\vert B\vert=1$, or $r\in B$, or there exist $i,j\in B$ such that $i<r<j$.

For example, there are $5$ set partitions of $\{1,2,3\}$, namely $$\{1,2,3\}; \{1\},\{2,3\}; \{2\},\{1,3\}; \{3\},\{1,2\}; \{1\}, \{2\},\{3\}.$$ If $r=1$, then we have $\vert\mathcal{B}_{3,1}\vert=4$ since $$\mathcal{B}_{3,1}=\{\{\{1,2,3\}\}, \{\{2\},\{1,3\}\}, \{\{3\},\{1,2\}\} \{\{1\},\{2\},\{3\}\}\}$$

Question. Is there any study of these sets $\mathcal{B}_{n,r}$ or their enumeration $\vert\mathcal{B}_{n,r}\vert$, directly or equivalently?

Answer 1

Here is an alternative approach to what is proposed by Brendan.

The key idea is to notice that the only forbidden parts in the set partitions are non-singleton subsets of $\{1,2,\dots,r-1\}$ and $\{r+1,r+2,\dots,n\}$, which enables one to employ the inclusion-exclusion principle to enumerate the set partitions without such forbidden parts.

First, we consider the associated Stirling numbers of second kind (A008299), which enumerate set partitions without singleton parts: $$S'(n,k) = \sum_{j=0}^k (-1)^j\cdot \binom{n}{j}\cdot S(n-j,k-j),$$ where $S(\cdot,\cdot)$ are the conventional Stirling numbers of second kind.

Now, we can claim that $$\begin{split}|\mathcal B_{n,r}| &= \sum_{p=0}^{r-1} \sum_{q=0}^{n-r} B_{n-p-q}\cdot \binom{r-1}{p}\cdot \binom{n-r}{q}\cdot \sum_{i=0}^{\lfloor p/2\rfloor} \sum_{j=0}^{\lfloor q/2\rfloor} (-1)^{i+j}\cdot S'(p,i)\cdot S'(q,j)\\ &=\sum_{p=0}^{r-1} \sum_{q=0}^{n-r} B_{n-p-q}\cdot \binom{r-1}{p}\cdot \binom{n-r}{q}\cdot {\bar B}_{p+1}\cdot {\bar B}_{q+1}, \end{split}$$ where $B_{m}$ are Bell numbers (A000110), ${\bar B}_{m}$ are complementary Bell numbers (A000587), $i$ and $j$ stand for the number of forbidden parts from $\{1,2,\dots,r-1\}$ and $\{r+1,r+2,\dots,n\}$ respectively, and $p$ and $q$ stand for the size of the unions of these forbidden parts.

Answer 2

The number can be expressed as a sum, though it isn't too enlightening. Let $m$ be the number of cells of the third type (there exists $i,j\in B$ such that $i\lt r\lt j$). Let $k_A$ be the number of elements $\lt r$ covered by such cells and $k_B$ be the number of elements $\gt r$ covered by such cells.

The number of possibilities for the cells of the third type with parameters $m,k_A,k_B$ is $$ \binom{r-1}{k_A} \binom{n-r}{k_B} S(k_A,m)S(k_B,m) m!,$$ where $S(N,R)$ is the Stirling number of the second kind. The idea is to take a partition with $m$ cells covering the $k_A$ points, a partition with $m$ cells covering the $k_B$ points, then combine them in $m!$ ways to make $m$ cells of the third type.

Now $k_A+k_B$ points have been covered. Choose the cell containing $r$ in $2^{n-k_A-k_B-1}$ ways. Finally, cover everything else with singeltons. So the total is $$ \;\sum_{0\le k_A\le r-1} \;\sum_{0\le k_B\le n-r} \; \sum_{0\le m\le \min(k_A,k_B)} \binom{r-1}{k_A} \binom{n-r}{k_B} 2^{n-k_A-k_B-1} S(k_A,m)S(k_B,m) m!\,.$$ Maybe it can be simplified a bit; I'm not sure.