No I really mean something O(p^c) where c is something around .72. instead of 2 * log (p) / (p1).
You can see it as increasing the knuthschrÃ¶ppel factors 2*log (p) / (p1) by a factor (p1) ^ 0.37 / log (p).
I alway used 2 * log (p)/ (p1). 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.
