Bases 2 & 4 reservations/statuses/primes

[mdettweiler]

Reserving Riesel base 2 odd-n k=39687 from n=1.25M-2M. I'll start this on my quad when Gary and I finish our collaborative S9 reservation up to 750K and estimate that it should take no more than two weeks or so after that.

[gd_barnes]

Originally Posted by mdettweiler

Reserving Riesel base 2 odd-n k=39687 from n=1.25M-2M. I'll start this on my quad when Gary and I finish our collaborative S9 reservation up to 750K and estimate that it should take no more than two weeks or so after that.

2 weeks on a quad to go n=1.25M-2M?? That sounds like a very fast quad or a very low-weight k, even if it is only the odd n's.

I'd be curious to see what your test time is on an n=~2M candidate.

[mdettweiler]

Originally Posted by gd_barnes

2 weeks on a quad to go n=1.25M-2M?? That sounds like a very fast quad or a very low-weight k, even if it is only the odd n's.

I'd be curious to see what your test time is on an n=~2M candidate.

According to Jean Penné's page it's the lowest-weight of the R2-odds. But, you're right, I'm beginning to wonder if that estimate is a tad off...I'll recheck that and try to get a better idea of how long it will take.

[mdettweiler]

Originally Posted by mdettweiler

According to Jean Penné's page it's the lowest-weight of the R2-odds. But, you're right, I'm beginning to wonder if that estimate is a tad off...I'll recheck that and try to get a better idea of how long it will take.

Yeah, you were right--it's going to take a LOT longer than that. It seems I probably won't get any farther than 1.4M or maybe 1.5M in two weeks. I could probably make it to 2M in a month or two at the latest, which still isn't too bad compared to the other bases I've been doing lately, so I'll still go ahead with it.

[MyDogBuster]

A couple of statuses (or stati or what ever the plural is for status):

Sierp Odd n - k= 85287 tested to n=1474035 base 2

Riesel Odd n - k= 155877 tested to n=1429120 base 2

Continuing both

[mdettweiler]

BTW, I noticed that for these tests, which need to be sieved at the equivalent base 4 n-level for 2*k, are automatically converted to base 2 by LLR, but don't have their k halved as well. This is consistent with what we've seen in PFGW as well; when there's multiple things to be converted, they can do one but not the other.

Just to verify, such a number can be converted as follows, right?
k*4^n-1 = .5k*2^(n+1)-1
Since this will be run through a v2.4.6 PRPnet setup, which doesn't automatically convert power-of-2 bases, I'll need to preconvert the sieve file before loading it. (With the base preconverted LLR should be able to handle the k conversion on its own, though for consistency I'd kind of prefer to have the whole thing entirely converted ahead of time.)

[gd_barnes]

Originally Posted by mdettweiler

BTW, I noticed that for these tests, which need to be sieved at the equivalent base 4 n-level for 2*k, are automatically converted to base 2 by LLR, but don't have their k halved as well. This is consistent with what we've seen in PFGW as well; when there's multiple things to be converted, they can do one but not the other.

Just to verify, such a number can be converted as follows, right?
k*4^n-1 = .5k*2^(n+1)-1
Since this will be run through a v2.4.6 PRPnet setup, which doesn't automatically convert power-of-2 bases, I'll need to preconvert the sieve file before loading it. (With the base preconverted LLR should be able to handle the k conversion on its own, though for consistency I'd kind of prefer to have the whole thing entirely converted ahead of time.)

Math class time. Plug some actual small numbers in for k and n and see what you get. For example, try k=10 and n=6. Then fiddle with it until you know what you have is correct. I'd suggest using an Excel spreadsheet.

[mdettweiler]

Originally Posted by gd_barnes

Math class time. Plug some actual small numbers in for k and n and see what you get. For example, try k=10 and n=6. Then fiddle with it until you know what you have is correct. I'd suggest using an Excel spreadsheet.

Yeah, I did try plugging some actual numbers into LLR before seeing your post here; I discovered that it actually doesn't autoconvert k's regardless of whether it's converting a base. However, I was still able to confirm that k*2^n-1 = 2k*2^(n-1)-1 because the residues matched.

Therefore, it seems I wasn't quite correct in my earlier statement. The correct chain of equality should be:
k*2^n-1 = 2k*2^(n-1)-1 = 2k*4^.5(n-1)-1
Or reversed, i.e. what I'll need to plug into my Perl script (I'm not so good with Excel but I have a Perl script already set up for base 16 that should be able to be modified easily):
k*4^n-1 = k*2^2n-1 = .5k*2^(2n+1)-1
Testing on actual numbers confirms this--LLR produces matching residues both for the original base 4 and the converted base 2 input files.

[gd_barnes]

Originally Posted by mdettweiler

Yeah, I did try plugging some actual numbers into LLR before seeing your post here; I discovered that it actually doesn't autoconvert k's regardless of whether it's converting a base. However, I was still able to confirm that k*2^n-1 = 2k*2^(n-1)-1 because the residues matched.

Therefore, it seems I wasn't quite correct in my earlier statement. The correct chain of equality should be:
k*2^n-1 = 2k*2^(n-1)-1 = 2k*4^.5(n-1)-1
Or reversed, i.e. what I'll need to plug into my Perl script (I'm not so good with Excel but I have a Perl script already set up for base 16 that should be able to be modified easily):
k*4^n-1 = k*2^2n-1 = .5k*2^(2n+1)-1
Testing on actual numbers confirms this--LLR produces matching residues both for the original base 4 and the converted base 2 input files.

Yeah, going from base 4 to base 2 is much easier. From your original answer, you had only missed that the exponent should be 2n+1 instead of n+1.

IMHO, I think the sieve files should be the correct k and n to avoid such confusion. Another searcher or two has had to ask the same or a similar question. I know why Jean did it that way: So it would effectively automatically only search base 2 odd-n. In other words, it would have been an extra step to search all of base 2 for a bit and then remove the even n's. But once they were removed, it will sieve base 2 at the same speed as base 4.

Not sure why you had to compare residues in LLR. It takes no knowledge of Excel to do this. You pull it up and enter =10*4^6-1 into a cell. It will give you the answer. The same thing with OpenOffice in Linux. For that matter, you can use the calculator that Windows or Ubuntu provides.


Gary

[mdettweiler]

Originally Posted by gd_barnes

Yeah, going from base 4 to base 2 is much easier. From your original answer, you had only missed that the exponent should be 2n+1 instead of n+1.

IMHO, I think the sieve files should be the correct k and n to avoid such confusion. Another searcher or two has had to ask the same or a similar question. I know why Jean did it that way: So it would effectively automatically only search base 2 odd-n. In other words, it would have been an extra step to search all of base 2 for a bit and then remove the even n's. But once they were removed, it will sieve base 2 at the same speed as base 4.

Really? Wouldn't it take longer because sieving scales based on the n-range (and, for instance, 50K-200K is a rather bigger range than 25K-100K)?

Not sure why you had to compare residues in LLR. It takes no knowledge of Excel to do this. You pull it up and enter =10*4^6-1 into a cell. It will give you the answer. The same thing with OpenOffice in Linux. For that matter, you can use the calculator that Windows or Ubuntu provides.

lol--duh, I hadn't thought of that. I was actually using much larger numbers (but still small in the grand scheme of things), in the vicinity of n=50K base 2.

[gd_barnes]

Originally Posted by mdettweiler

Really? Wouldn't it take longer because sieving scales based on the n-range (and, for instance, 50K-200K is a rather bigger range than 25K-100K)?

That's an apples to oranges comparison. The scaling is based on the aggregate size range of the numbers, not the n-value. It should sieve n=800K-1.6M base 2 at the same speed as n=100K-200K base 256. For the same reason, it should sieve the file that you're using at the same speed, regardless of whether it's base 2 or 4. After all, it will find the exact same factors either way.

lol--duh, I hadn't thought of that. I was actually using much larger numbers (but still small in the grand scheme of things), in the vicinity of n=50K base 2.

Didn't I suggest using small #'s like k=10 and n=6 in the "math class" to begin with?