In how many ways can the power set of the set $\{1,2,\cdots ,n\}$ be partitioned into two sets $S_1,S_2$ such that; $$ \bigcup_{X\in S_1} X=\{1,2,\cdots ,n\}$$ where the minimum cardinality of each element of $S_1$ is $k>1$
For example when $n=3, k=2$, one acceptable $S_1$ would be
$$S_1 = \Big\{\{1,2\},\{1,3\},\{1,2,3\} \Big\}$$
My question is:
Can we find a general formula for number of such partitions as a function of $n$ and $k$?
My attempt:
I recently found that it has something to do with Dedekind numbers. If we
consider all possible strings of length $n$ ,place $1$ in the place of $k$-th number if $k$-th number is present in a $S_1$ and $0$ otherwise. Like, for $S_1= \{\{1,2\},\{1,3\}\}$ we have 110 and 101. Now, consider a function which will give output $0$ or $1$ , when the string is accepted then output will be $1$ and if not then $0$. Observe that, if a string is accepted then also any string bigger than the accepted one is also a accepted string. So actually counting the number of functions $f:\{0,1\}^n\to \{0,1\}$ which are monotone are the number of possible such sets(haven't considered the condition that union of all subsets equals to $\{1,2,\cdots ,n\}$ ). Which is Dedekind number. Now, if consider the condition that the union covers the set $\{1,2\cdots ,n\}$ then we can use the method of counting number of onto functions and get the desired number of ways $$\sum_{i=0}^n(-1)^i\binom{n}{i}M(n-i)$$
where $M(n)$ is the $n$-th Dedekind number.
Note:$S_2$ is of no use, just stated.
There are $2^n-n-1$ sets that are candidates for $S_1$ so there are $2^{2^n-n-1}$ candidates for $S_1$. We are asking how many ways there is to select some set of them with union all of $[1,n]$. It looks easier to count the sets that do not include some element and subtract, using inclusion-exclusion. There are $2^{n-1}-(n-1)-1$ sets missing $n$ and $n2^{2^{n-1}-(n-1)-1}$ sets missing one number, but we have double counted the sets missing two numbers. Using inclusion-exclusion we get $$2^{2^n-n-1}+\sum_{m=1}^{n-2}(-1)^m{n \choose m}2^{2^{n-m}-(n-m)-1}$$