I am not a combinatorialist by training, but I need insight on the following question for a current project. I cannot find this as a duplicate here.
Let $[n]$ be the set $\{1, 2, \ldots, n\}$. I want to count certain partitions of $[n]$. I wish for no ordering of my cells or within my cells. I think I am correct in that the Bell number $B_n$ counts this for me. So, for $[5]$, the Bell number $B_5=52$ takes into account that the two partitions $$ \{1, 2, 4\}, \{3\}, \{5\} $$ versus $$ \{3\}, \{1, 2, 4\}, \{5\} $$ are counted as the same, and there are then 52 distinct partitions. Correct?
This is close to what I need, but not quite. My problem is: Count the number of partitions of $[n]$ in which no cell is a singleton. Call this count $B^*_n$ for the lack of any good notation.
By nothing clever, just brute force, I have example calculations
\begin{array}{c c c} n & B_n & B^*_n\\\hline 3 & 5 & 1\\ 4 & 15 & 4\\ 5 & 52 & 11\\ \end{array}
I've tried getting $B^*_n$ from $B_n$ by in/exclusion, but this is not going well for me. I've also tried to find some recurrence, but I am just as stuck there. Does anyone have any insight or a reference for this count? I would take just about anything, even some asymptotics.
Edit: a paper in the OEIS link below proves via generating functions that $$ B^*_{n+1} = B_n - B^*_n $$ and this is not clear to me at all. Is this deep?
Here is an easy combinatorial proof for $B_{n+1}^* = B_n - B_{n}^*$. Note that $B_n - B_{n}^*$ is the number of partitions of $[n]$ that have at least one singleton. Take all elements that are in singletons and add $n+1$ to that to form a part in a partition of $[n+1]$ that has no singletons. This is clearly a bijection.