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Old 2009-01-22, 17:21   #3
brownkenny's Avatar
Jan 2009

2·3 Posts

Thanks for the suggestion, Dr. Silverman. When I made the substitution I wound up with a log(log(n)) term in the denominator that I'm not sure how to get rid of.

I tried estimating some more, but I still can't figure out where the factor of 8 in the term

<br />
\frac{8 \log \log n}{\log n}<br />

comes from.

Thanks again, Dr. Silverman.
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