Comment #23 on issue 1827 by asmeurer: log eval gives more standard form http://code.google.com/p/sympy/issues/detail?id=1827
Now you've got me wondering just how lucky it was. Given a number n, what's the probability that the second largest factor is greater than or equal to k (if I remember correctly, the complexity of integer factorization is roughly exponential in the number of digits in the second largest factor)?
And it wasn't really luck, because I made sure that it was hard to factorize (to make the point).
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