Recurrence Relation for Bell Numbers

Theorem
Let $B_n$ be the Bell number for $n \in \Z_{\ge 0}$.

Then:
 * $B_n = \displaystyle \sum_{k \mathop = 0}^n \dbinom n k B_k$

where $\dbinom n k$ are binomial coefficients.

Proof
By definition of Bell numbers:


 * $B_n$ is the number of partitions of a (finite) set whose cardinality is $n$.

Let $k \in \set {k \in \Z: 0 \le k \le n}$.

Let us form a partition of a (finite) set $S$ with cardinality $n$ such that one component has $n - k$ elements.

We can do this by first choosing $k$ elements from $S$, and let the remaining $n - k$ elements form that single component.

From Cardinality of Set of Subsets and the definition of binomial coefficient, there are $\dbinom n k$ ways to do this.

For the chosen $k$ elements, there are $B_k$ ways to partition them.

Thus there are $\dbinom n k B_k$ possible partitions for $S$:
 * $\dbinom n k$ of selecting one component with $n - k$ elements
 * for each of these, $B_k$ ways to partition the remaining $k$ elements.

Summing the number of ways to do this over all possible $k$:


 * $\displaystyle B_n = \sum_{k \mathop = 0}^n \dbinom n k B_k$

Take $B_3$ for example.

This says:
 * $B_3 = B_0 \dbinom 3 0 + B_1 \dbinom 3 1 + B_2 \dbinom 3 2 + B_3 \dbinom 3 3$

$B_3$ appears both sides, for a start, so this cannot be a recurrence formula.

What is happening is (from working out examples) that there is multiple counting.

Taking $k = 1$: we have $\dbinom 3 1 = \dbinom 3 2$ ways of choosing $3 - 1 = 2$ elements for the component with $2$ elements, and then there is $1$ way of partitioning the final $1$ element, leaving us with:


 * $\set {1, 2 | 3}$
 * $\set {1, 3 | 2}$
 * $\set {2, 3 | 1}$

Then taking $k = 2$: we have $\dbinom 3 2 = \dbinom 3 1$ ways of choosing $3 - 2 = 1$ element, and then there are $2$ ways of partitioning the final $2$ elements, leaving us with:


 * $\set {1 | 2, 3}$, $\set {1 | 2 | 3}$
 * $\set {2 | 1, 3}$, $\set {2 | 1 | 3}$
 * $\set {3 | 1, 2}$, $\set {3 | 1 | 2}$

As can be seen, just by taking $2$ examples of $k$, we have already counted $3$ partitions twice and one $3$ times.