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Old 2009-12-19, 23:11   #6
ThiloHarich
 
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Nov 2005

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No I really mean something O(p^c) where c is something around -.72. instead of 2 * log (p) / (p-1).
You can see it as increasing the knuth-schröppel factors 2*log (p) / (p-1) by a factor (p-1) ^ 0.37 / log (p).
I alway used 2 * log (p)/ (p-1). I did not see why a factor of 1 should work better if p divides a.
This is an simple approximation of the values I observed, and it worked for my sieve.
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