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<record version="3" id="6436">
 <title>Bell number</title>
 <name>BellNumber</name>
 <created>2004-11-01 12:54:20</created>
 <modified>2008-07-14 13:18:18</modified>
 <type>Definition</type>
 <creator id="5079" name="aoh45"/>
 <author id="3771" name="CWoo"/>
 <author id="5079" name="aoh45"/>
 <classification>
	<category scheme="msc" code="11B73"/>
	<category scheme="msc" code="05A18"/>
 </classification>
 <related>
	<object name="StirlingNumbersSecondKind"/>
 </related>
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 <content>The \emph{Bell number}, denoted $B(n)$ is the total number of partitions of a set with $n$ elements.  For $n=0$, we have $B(0)=1$.  For $n\ge 1$, we have 
\[
B(n) = \sum_{k=0}^n S(n,k) \qquad \textrm{for } n \ge 1
\]
where $S(n,k)$ are the Stirling numbers of the second kind.

\newtheorem{bell}{Proposition}
\begin{bell}
\[
B(n+1) = \sum_{k=0}^n \binom{n}{k} B(k)
\]
\end{bell}

\begin{proof}
We count the number of partitions of a set of $n+1$ elements, depending on the size of the block containing the $n+1$st element. If the block has size $j$ for $1 \le j \le n+1$ then we have $\binom{n}{j-1}$ choices for the $n$ other elements of the block. The remaining $n+1=j$ elements can be partitioned in $B(n+1-j)$ ways. We have therefore that:
\begin{eqnarray*}
B(n+1) &amp; = &amp; \sum_{j=1}^{n+1}\binom{n}{j-1} B(n+1-j) \\
       &amp; = &amp; \sum_{j=1}^{n+1}\binom{n}{n+1-j} B(n+1-j) \\
       &amp; = &amp; \sum_{k=0}^n \binom{n}{k} B(k)
\end{eqnarray*}
\end{proof}

Using the formula above, one can easily derive the first few Bell numbers.  Starting with $n=0$, the first ten Bell numbers are 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147.</content>
</record>
