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Let $f(n)$ be the number of all partitions of $[n]$ with no single blocks. Prove that $B(n) = f(n) + f(n+1)$. Where $B(n)$ is the Bell number. Can anyone help me start a bijection?

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up vote 3 down vote accepted

Hint: If $\sim$ is an equivalence relationship on $[n+1]$ with no singleton blocks, how can you find a partition on $[n]$ which must have singleton blocks?

You basically want to find a bijection showing $f(n+1)=B(n)-f(n)$, where $B(n)-f(n)$ is represented by the obvious set of partitions that do have at least one singleton block.

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Thank you for the help! – MITjanitor Jan 28 '13 at 16:15
So, in a way you consider the $n+1$th element as a kind of ideal element and collect all the singletons to join its class. – Berci Jan 28 '13 at 16:44
Yes, precisely... – Thomas Andrews Jan 28 '13 at 17:15

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