20130512, 21:37  #1 
Dec 2011
After milion nines:)
5CF_{16} Posts 
sr2sieve number of candidates in range
When I start sr2sieve then program automatically calculates number of candidates that will be removed when sieve process is over. Since it is very usefull info, can someone tell me how sr2sieve calculates that number, what formula can do that computation and give result.
Thanks for answer 
20130513, 01:30  #2  
"Mark"
Apr 2003
Between here and the
2^{3}×809 Posts 
Quote:
n*(1log(p0)/log(p1)) where n is the number of terms and p0/p1 are the sieve limits. 

20130513, 09:05  #3  
Dec 2011
After milion nines:)
1487_{10} Posts 
Quote:
As we all know, there is some "rich" sieve with many candidates, and there is some very poor sieve with only few candidates left. I know, you say it is approximation, but in whole time I use this program approximation was always very accurate. 

20130513, 12:41  #4 
"Mark"
Apr 2003
Between here and the
6472_{10} Posts 
n is the number of remaining terms in the input range

20130513, 18:56  #5 
Dec 2011
After milion nines:)
1,487 Posts 
Yes it is ok: but how your formula "know" is sieve rich of factor or ih has very small number of factors ( sieve for K=6883 has only 20 kb and for 2145 has 1200 kb)
Or it is same ratio for all numbers Last fiddled with by pepi37 on 20130513 at 18:56 
20130513, 22:55  #6 
"Mark"
Apr 2003
Between here and the
2^{3}·809 Posts 
It is the same ratio for all numbers. It is suggested that you sieve to a value like 1e6 (or higher) with srsieve. You can then use the formula to estimate the number of factors.

20130514, 15:11  #7 
Dec 2011
After milion nines:)
1,487 Posts 

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