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Dragonfall
Jun24-09, 09:21 PM
n^n

or

\prod_{r=0}^{n-1}\left(\begin{array}{c}\sum_{j\leq r}\left(\begin{array}{c}n\\ j\end{array}\right)\\2^r\end{array}\right)

Asymptotically, I mean.

CRGreathouse
Jun25-09, 01:37 AM
Since
\prod_{r=0}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}={0\choose1}\prod_{r=1}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}=0
I'd have to go with n^n.

Dragonfall
Jun25-09, 01:58 AM
How about now?

CRGreathouse
Jun25-09, 03:39 AM
It seems that the product grows faster.