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Old 2010-08-22, 23:47   #2
wblipp
 
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"William"
May 2003
New Haven

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You might try an Li version motivated by comparison to the prime case. When k=1 this is n/log(n), which we know to improved upon by integral (1/log(x)). This can be motivated by the argument that 1/log(x) is correct density. The analogous adjustment would be the integral of
\frac{(\log\log x)^{k-1}}{(k-1)!\log x}
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