This is the problem I am trying.
Define $S_k(n)$ to be the number of partitions of the set $[n]$ so that if $i$ and $j$ are in the same block, then $|i-j| > k$ holds. Show that $S_k(n) = B(n - k)$, for all $n \geq k$.
I first tried to find a bijection and also tried to find a recursive formula for the left hand side, but wasn't succesful. Can anyone help with this question?
In what follows, $[a,b]$ will always refer to the integer interval $\{a,a+1,\dots,b-1,b\}$, when $a\le b$.
I shall give a bijective proof of this equality. Suppose we are given a partition, $P$, of $[n-k]$. We shall output a partition, $P'$, of $[n]$, with the property that any two numbers in the same part are at distance more than $k$.
The $k$ numbers in the interval $[n-k+1,n]$ will be initially placed in $k$ new parts, each in a distinct part. Call a number $x\in [n-k]$ bad if there exists a $y$ in the same $P$-part as $x$ such that $x < y\le x + k$. The idea is that any bad numbers in $P$ will be moved to one of the $k$ new parts. We must do this carefully so that these two conditions are satisfied:
$P'$ is valid, in the sense that any two numbers in the same part are more than $k$ apart.
Our process is reversible, so we can remember where the numbers were moved from, and recover $P$ from $P'$.
For each part in $P$, we first group the numbers into "bad chains" as follows. We say that $x$ is linked to $y$ if they are in the same part and $|x-y|\le k$. The connected components of this linking relation are our bad chains. Each bad chain has a maximal element, which I will call the "leader." Note that the leader itself is not bad, because by definition, a number is bad if there is another number above it which is too close, but the leader is greatest in the chain, so has nothing above it.
The rules for moving the bad numbers is this:
For each bad number $x$ in $P$, let $\ell$ be the leader of $x$'s bad chain. If $\ell\equiv x\pmod{k+1}$, then $x$ will remain in its current part. Otherwise, $x$ is moved to the part containing the unique number $z$ in $[n-k+1,n]$ for which $$n+1-z\equiv \ell-x\pmod {k+1}$$
Finally, I shall prove the required points (1.) and (2.).
We have removed all bad numbers from the original parts, so we need only prove there are no numbers in the new parts that are too close. The only exceptions are the bad numbers $x$ with a leader $\ell$ for which $x\equiv \ell$, but these remaining bad numbers are all in the same equivalence class modulo $k+1$, so the remaining numbers are spaced far enough apart.
For any $x$ moved to a part containing $z\in [n-k+1,n]$, we have $|x-z|>k$. This is because $\ell \le n-k$, so $$z-x\ge z-(n-k)+\ell-x\equiv 0\pmod {k+1}.$$Since the quantity $z-(n-k)+\ell-x$ is stricly positive, and equivalent to zero mod $k+1$, this quantity must be at least $k+1$, we we have shown $|z-x|\ge k+1$.
Furthermore, for any $x_1,x_2$ in the same bad chain, we must have $x_1\equiv x_2\pmod {k+1}$, which implies $|x_1-x_2|\ge k+1$.
Looking at the outputted partition $P'$, you can determine all numbers that were originally bad by looking at the numbers in the parts containing the numbers in the interval $[n-k+1,n]$. This allows you to reconstruct the bad chains, by looking at the connected components of the relation of "at-most-$k$-apart" on these bad numbers. Letting $x$ be the largest number in such a chain, you can then recover the leader of the chain by solving the equation $n+1-z\equiv \ell-x$ for $\ell$, where $\ell$ must be chosen from the interval $[x+1,x+2,\dots,x+k]$. This allows you to return all of the numbers in the new parts to the original parts, so my claimed bijection is indeed reversible.
Claim. Each set partition of $[n]$ into $k$ parts can be bijectively correspond to set of $n-k$ ordered pairs of form $(i, j)$ $(1\leq i<j\leq k)$ such that following condition holds;
Each number in $[n]$ can appear in those ordered pair's left(right) position at most once.
For example, these set of ordered pairs cannot contain $(1, 2)$, $(1, 3)$ simultaneously. Now let's see why this is possible.
Proof. For given set partition of $[n]$, if there is a block $B=\{b_1, b_2, ..., b_m\}$(it's increasing order), we make pairs $(b_1, b_2), (b_2, b_3), ..., (b_{m-1}, b_m)$ from it. If size of $B$ was $1$, don't make any pair. In this way, we can make $|B|-1$ pairs for each, so total number of pairs will be $n-k$, which is sum of value $|B|-1$ for all blocks. And each number in $[n]$ must appear at most once among all left(right) position of pairs.
Now let's show this process is reversible. That is, we can make original set partition by given set of ordered pairs. Think about empty graph on $n$ vertice, each vertex is named $1$ to $n$. Draw directed edge $i\rightarrow j$ if there is pair $(i, j)$. By conditions of our pairs, number of edge is $n-k$ and each vertex has in(out)degree at most $1$.
This means each connected component of this graph is single vertex and directed path. And since those two kinds of graph has $v-1$ edges if it has $v$ vertice, total number of edges will be $n-\#($connected components$)$. This shows number of connected component is $k$. Now it's easy to find out original sets. Make block $\{b_1, b_2, ..., b_m \}$ from directed path $b_1\rightarrow b_2\rightarrow...\rightarrow b_m$, and make single element block from single vertex, we're done. $\square$
For our original problem, let's show stronger result;
If $S_d(n, k)$ $(n\geq k>d)$ denotes family of set partitions of $[n]$ into $k$ parts, and two elements difference at most $d$ is in different block. Then $|S_d(n, k)|=|S_0(n-d, k-d)|=S(n-d, k-d)$ holds, where $S(n, k)$ is stirling number of second kind.
Proof. Just turn every elements of $S_d(n, k)$ and $S_0(n-d, k-d)$ into $n-k$ ordered pairs, using our claim. And correspond two set of ordered pairs $A \in S_d(n, k)$, $B \in S_0(n-d, k-d)$ if following holds; $(i, j)\in A \iff (i, j-d)\in B$. Since each pair in $A$ has difference$>k$, we can make corresponding $B$ from each $A$. Same is true for each $B$, and this relationship is bijection. Furthermore, the number of pairs and 'at most $1$' property is conserved in bijection. Hence problem is solved. $\square$
Comment. There is 'proof without words' showing this result. Only you need to know is above claim. Think about large chessboard, and color every cell $(i, j)$ such that $1\leq i\leq n-1$, $2\leq i\leq n$, $i<j$. And prepare $n-k$ chess rooks and place them on chessboard as set of $n-k$ ordered pairs told. Then these rooks can't attack each other.
Likewise, each set partition of $[n]$ into $k$ parts(elements in $S_0(n, k)$) correspond to non-attacking rook placement of $n-k$ rooks in 'stair' board of size $n-1$. This is famous combinatorial meaning of stirling number of second kind.
Then our problem is simple. Each elements in $S_d(n, k)$ correspond to previous rook placement which diagonals $y=x+1, ..., y=x+d$ cannot contain rook. If we get rid of this $d$ diagonals, remaining $n-k$ rooks are in $n-d-1$ stair, forming non-attacking placement. This correspond to each elements in $S_0(n-d, k-d)$.
I think there is a mistake in @Mike Earnest's solution(wrote here because my reputation is below $50$).
Pick any $k\geq 2$ and think about $P$ having two different parts $B_1$, $B_2$ such that maximal/second maximal element of $B_1$ is $l=n-k-1, x=n-k-3$, and maximal/second maximal element of $B_1$ is $l'=n-k-2, x'=n-k-4$.
Since $l-x=l'-x'=2$, $x, x'$ is bad element which is in different bad chain. And leader of their bad chain must be $l$, $l'$ each. Since only $z\in \{n-k+1, ..., n\}$ satisfies $n+1-z\equiv 2\pmod{k+1}$ is $z=n-1$, $x, x'$ must move to new part where $n-1$ belongs to. But this is contradiction, since $x, x', n-1$ is now in same part of $P'$ but $x-x'=1$.