Bases 251-500 reservations/statuses/primes

[MyDogBuster]

Reserving Riesel 493 and Sierp 409 as new to n=25K

[unconnected]

Riesel base 461, k=8
Primes:
2*461^6-1
4*461^3071-1

Trivial factors k=6
Base proven.

[unconnected]

Riesel base 311, k=14
Primes:
2*311^2-1
4*311^5-1
8*311^2-1
10*311^1-1
12*311^146-1

Trivial factors k=6
Base proven.

[unconnected]

Riesel base 317, k=14
Primes:
2*317^10-1
4*317^119-1
6*317^1-1
8*317^2-1
10*317^1-1
12*317^1-1

Base proven.

[rogue]

unconnected, Gary would like everyone to limits themselves to two results per day because too many results at one time becomes more difficult for him to manage. With the limit there are more than a couple of people who would produce a dozen or more results a day (to clean up conjectures with low k).

[gd_barnes]

Thanks, Mark, for being a watchdog. :-) To clarify slightly: That's two new bases per day. Also, reporting 2 completions of previous new base reservations and then reserving 2 more, all in the same day, is fine too. On existing bases, people can reserve and report as much as they want, as long as they aren't biting off too much for their resources. It's easy to manage those. The initial analysis and page/thread updates needed on new bases is what takes much longer.

Here is my general estimate: It takes me an average of ~15 mins. per new base to go through everything and generally about 5 mins. on existing bases. Of course these are only averages. Some new bases take nearly 30 mins. and others take 5-10 mins. So if 4 people do 2 new bases in a day, that's 2 hours, which includes no time for any other admin work. Even the k=8 conjectures with no algebraic factors on the Sierp side can take up to 10 mins. Here is what I do on new bases:

1. Save off or cut-paste any results/primes (most of the time).
2. Sort the primes by n descending for entry on the pages.
3. Spot verify that all k's are accounted for.
4. Open up Robert's Riesel or Sierp conjectures file and enter the base, conjecture, covering set, trivial factors, k's remaining, and top 10 primes (from #2) on the pages. (I don't take people's word for it in posts because sometimes those have typos.)
5. If a Riesel base, analyze it for 2 different kinds of algebraic factors. If a Sierp base, look for GFN primes. (Algebraic factor analysis can take ~10-15 mins. all by itself.)
6. Remove the base from one of the untested bases threads.
7. If 1k remaining, add it to the 1k thread.
8. Sometimes for new people or a tough effort, start a quick double check effort up to n=1000 or 2500.
9. Sometimes follow up with a post if new algebraic factors are found or if the prime distribution looks unusual.

That's for each and every new base that comes through. And if I have to let it go for ~5 days, I've had it take me ~8 hours, which is what happened on my last business trip. I was caught totally off guard because few new bases were coming before I left and then they came in droves starting about 2 days after I left.

I hope this clarifies a little more about why I ask for only 2 new bases at a time per day. Of course it's OK to do the occassional 3 new bases at once especially if it completes a big group of them or a bunch of them below some big round number. Of course I'll list the 3 that unconnected did (he started that original base that makes a total of 4 the day before). I'd only kindly ask that most of the time, we keep it at 2.

BTW, good job, unconnected, on finding that final prime on R307. Like Mark said, most of the time, people stop at n=25K but of course that's just because it's a nice round number that doesn't take terribly long. (I admit that I started the trend and it just kind of stuck.) Some people like KEP and Batalov like to search up to n=50K, 100K, or higher right away.


Thanks,
Gary

[rogue]

Primes found:

2*503^860-1
4*503^1-1
6*503^22-1

2*538^8-1
3*538^1-1
5*538^1-1
6*538^14-1

The other k have trivial factors. With a conjectured k of 8, these conjectures are proven.

[rogue]

Originally Posted by gd_barnes

Here is my general estimate: It takes me an average of ~15 mins. per new base to go through everything and generally about 5 mins. on existing bases. Of course these are only averages. Some new bases take nearly 30 mins. and others take 5-10 mins. So if 4 people do 2 new bases in a day, that's 2 hours, which includes no time for any other admin work. Even the k=8 conjectures with no algebraic factors on the Sierp side can take up to 10 mins. Here is what I do on new bases:

1. Save off or cut-paste any results/primes (most of the time).
2. Sort the primes by n descending for entry on the pages.
3. Spot verify that all k's are accounted for.
4. Open up Robert's Riesel or Sierp conjectures file and enter the base, conjecture, covering set, trivial factors, k's remaining, and top 10 primes (from #2) on the pages. (I don't take people's word for it in posts because sometimes those have typos.)
5. If a Riesel base, analyze it for 2 different kinds of algebraic factors. If a Sierp base, look for GFN primes. (Algebraic factor analysis can take ~10-15 mins. all by itself.)
6. Remove the base from one of the untested bases threads.
7. If 1k remaining, add it to the 1k thread.
8. Sometimes for new people or a tough effort, start a quick double check effort up to n=1000 or 2500.
9. Sometimes follow up with a post if new algebraic factors are found or if the prime distribution looks unusual.

This gives all of us a new appreciation for the work you do to manage this project. Thank you.

[gd_barnes]

Originally Posted by rogue

Note how odd base 440 is. All primes were found with n=1 and n=2.

That is quite strange for so many k's. I think I've had as many as 4 k's with n=1 and n=2 primes but definitely not 6 of them.

Question: I know you've already worked a bunch of the Riesel k=8 conjectures. Have you by chance worked on R272? It's a much smaller base than the others remaining untested at this point. Perhaps it has a k or two remaining and you are just posting the proven ones right now.

I ask because I thought I'd reserve it in the next couple of days if you haven't already worked on it just to knock out the final CK = 8 for b < 300 (on both sides). Regardless, feel free to take it yourself if you haven't already done so.


Gary

[rogue]

Originally Posted by gd_barnes

Question: I know you've already worked a bunch of the Riesel k=8 conjectures. Have you by chance worked on R272? It's a much smaller base than the others remaining untested at this point. Perhaps it has a k or two remaining and you are just posting the proven ones right now.

I ask because I thought I'd reserve it in the next couple of days if you haven't already worked on it just to knock out the final CK = 8 for b < 300 (on both sides). Regardless, feel free to take it yourself if you haven't already done so.

R272 had remaining k (tested to n=2000), so I have left it alone for now. I did not investigate algebraic factorizations so it is possible that one of those would be needed to prove the conjecture. I was going to take another base before Friday, but hadn't decided on anything specific. It was most likely going to be a conjecture with a single k remaining. R272 would be a good choice, but I'll let you take it. I'll find something else as there is so much to choose from. It's like going to the ice cream shop and seeing more than 100 flavors and having difficulty choosing the one that will taste the best...

[gd_barnes]

Originally Posted by rogue

R272 had remaining k (tested to n=2000), so I have left it alone for now. I did not investigate algebraic factorizations so it is possible that one of those would be needed to prove the conjecture. I was going to take another base before Friday, but hadn't decided on anything specific. It was most likely going to be a conjecture with a single k remaining. R272 would be a good choice, but I'll let you take it. I'll find something else as there is so much to choose from. It's like going to the ice cream shop and seeing more than 100 flavors and having difficulty choosing the one that will taste the best...

OK, I'll reserve R272 to n=25K.

Doing some quick math in my head on the algebraic factors:

272 = 2^4*17
-and-
273 = 3*7*13

To determine which numeric factor could combine with algebraic factors to make a full covering set, one must go to the factorization of b+1; in this case 273. Since 3 and 7 are not possible factors as per the algebraic factors thread and 13 is, that's the only one applicable. Since k must be m^2 and m must be m==(5 or 8 mod 13) when the numeric factor is 13, the lowest possible k that could be eliminated would be k=5^2=25. So with a conjecture of k=8, nothing can be removed here. That's the "old" or "classical" kind of algebraic factors.

For the new kind, you have to remove the squared part of b to get a multiplier. (If there is no squared part, then the new kind are not applicable.) In this case 272 / 2^4 = 17. So we have k must be 17*m^2 and m must be m==(3 or 10 mod 13). This means that the lowest k would be k=17*3^2=153. Once again too high for a conjecture of k=8. That's the "new" kind of algebraic factors found by Serge. (Interestingly we had already found it on bases 24 and 54 long before Serge started participating but we just didn't realize it. I just thought it was specific to only those bases. Serge's analysis showed that it was potentially applicable to all bases.)

Therefore nothing can be eliminated by algebraic factors on R272.

BTW, the modulos for the m-values were determined by nothing more than trial and error on my part over a series of bases with the specific numeric factorizations (in this case 13). I do know that x and y add up to f in all cases and I do know that there are no more and no less than 2 modulos (all shown in the algebraic factors thread for the various factors). That is the 5 and 8 in the m==(5 or 8 mod 13) add up to the 13. I do not have a proof for this but I'd be willing to bet that someone could prove it.

That's the mental process that I go through to show them on the pages. I then do a quick check using Alperton's applet. Sometimes I find that I missed something or miscalculated it.

I mention all this detail in case it helps someone else determine the algebraic factors ahead of time.


Gary