20101206, 16:18  #1 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB_{16} Posts 
better k weights
In thinking about low k's regarding this discussion, I thought I'd calculate the Nash weights of all the S6 k's. I did so, and surprisingly, found that k=97131 had a higher weight (250) than two others that really had about twice the candidates remaining (k=76441, weight 242, and k=50252, weight 243). Nash weights are calculated as the number of candidates remaining in n=100001110000 after sieving to 511. In this case anyway, that isn't nearly enough to make a good representation of how many candidates you'll have to test (on two counts: some k's may have oftrepeated factors over 511, and that's not enough candidates, especially for lower k's, to have very accurate results; I suspect the first makes large inaccuracies like this one and the latter just makes for slightly noisier results). For comparison to try to get more realworld results, I also sieved the k's from n=400k800k to P=1e6 and 10e6. To compensate for different depths and n ranges, I'm considering the ratios between these three different weights. The ratio between my two depths (1M and 10M) was only negligibly different (mean 0.8576, std. dev. 0.0069), but the ratio between the Nash weight and my test results varied wildly, from 8.54 at k=97131, then most around 18, and the highest at 20.17. The mean was 17.4264 and standard deviation in that ratio was 2.5567. This other result more closely matched the realworld result of k=97131 being about half the weight of 76441 and 50252.
In short, I submit that instead of Nash weights calculated as: number of candidates remaining in n=100001110000 after sieving to 511, for example, the srsieve command: srsieve n 100001 N 110000 P 511 "97131*6^n+1" It would be far better to calculate weights as: number of candidates remaining in n=400000800000 after sieving to 1000000, (or perhaps even larger bounds for both, e.g. 100K1M sieved to 10M or 100M) for example, the srsieve command: srsieve n 4e5 N 8e5 P 1e6 "97131*6^n+1" If desired, we could divide the new weight by 18 to get a number on the approximately same scale as a Nash weight. Last fiddled with by MiniGeek on 20101206 at 16:31 
20101206, 17:23  #2  
"Mark"
Apr 2003
Between here and the
2^{3}·5^{2}·29 Posts 
Quote:
I suggest a range size of 10,000 for the last item because that makes computations much easier as users typically do ranges in multiples of 10,000. For the fourth item I see no reason why srsieve is the one program to generate such a number. Sieving of other forms (with their respective programs) will give an idea as to density of that form. 

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