Yes, rank 1, or do the expression on each partition p .


----- Original Message -----
From: Andrew Nikitin <[EMAIL PROTECTED]>
Date: Monday, January 28, 2008 5:17
Subject: [Jprogramming] Re: Perfect powers
To: [email protected]

> > From: Roger Hui 
> >
> > For larger n, you may get a faster solution by
> > generating all partitions p of m for which 2 The actual 
> perfect power would be */p^p:i.#p
> 
> Did you mean "1
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