Let be $S(m,k)$ number of partitions of a $k$ element set into $m$ nonempty parts investigating with generating functions I get this very interesting equation$\sum_{k=0}^ {\infty}\sum_{m=0}^ {\infty}S(m,k)\frac{1}{k!}=e^{e-1} $ Can someone tell me if I am right.
Double sequence convergence
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combinatorics
sequences-and-series
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0How did you arriv$e$ at this conclusion? Off hand, I know only one way to do it: the one that's currently in the Wikipedia article on Dobinski's formula. I actually suspect I've read another derivation of it at some point, but I don't remember any more than that. – 2011-09-08
2 Answers
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The inner sum actually has a finite number of non-zero members: $ \sum_{m=0}^ {\infty}S(m,k)= \sum_{m=0}^ {k}S(m,k)=B_k $ where $B_k$ are the Bell numbers. The exponential generating function for them is $\sum_{k=0}^\infty \frac{B_k x^k}{k!}=e^{e^x-1}.$ Putting $x=1$ gives the required equality.
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0@J. M. They're mentioned in the reference, so I decided not to expand :) – 2011-09-08
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Two comments: (1) There is a wikipedia article about this:
http://en.wikipedia.org/wiki/Dobinski%27s_formula
and (2) the conventional notation for Stirling numbers is that $S(m,k)$ is the number of partitions of a set of size $m$ into $k$ subsets, not the number of partitions of a set of size $k$ into $m$ subsets. http://en.wikipedia.org/wiki/Stirling_number