mersenneforum.org

mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Conjectures 'R Us (https://www.mersenneforum.org/forumdisplay.php?f=81)
-   -   Bases 501-1030 reservations/statuses/primes (https://www.mersenneforum.org/showthread.php?t=12994)

henryzz 2010-01-11 10:01

[quote=gd_barnes;201450]David, Mark's intent was to only search the above to n=25K. Therefore, if you'd like you can leave R589 reserved on up to n=100K. For now, that is how I'll show it on the reservations page.


Gary[/quote]
I was sort of planning to stop there anyway although I would like to leave a good sieve file for the next person. Testing is getting too slow now.

rogue 2010-01-11 13:58

[quote=gd_barnes;202397]Mark,

Did you read post 858? If not, here is a link:

[URL="http://www.mersenneforum.org/showpost.php?p=200902&postcount=858"][COLOR=#800080]http://www.mersenneforum.org/showpost.php?p=200902&postcount=858[/COLOR][/URL]
[/quote]

I did, but I didn't re-read before my post. Had I done so, I wouldn't have listed those bases and primes in my post. Again, I apologize to those I whose ranges I poached.

Note that I completed to n=25K and released these bases. I suspect those with reservations intend to take then much further.

Gary, would it make sense to have a reservation system that could update the webpage immediately instead of requiring you to manually update them?

gd_barnes 2010-01-11 22:57

[quote=henryzz;201465]I was sort of planning to stop there anyway although I would like to leave a good sieve file for the next person. Testing is getting too slow now.[/quote]

OK, great. Thanks for running a sieve file to n=100K. Testing such a large base to n=100K could give some pretty big top-5000 primes.

Mini-Geek 2010-01-13 18:54

Reserving Riesel base 605 (conj. k is 100).

MyDogBuster 2010-01-13 22:01

BTW all, If a Riesel base ends in 4 or 9 and shows as not being started(ie 604, 609, 774, 819) then I have probably completed it. Those are the algebraic factors I'm working on. Gary doesn't want me to reserve more than 3 at a time so that I don't swamp him with updates.

Hey Gary, only about 45 to go and they are all tested to n=25K.

gd_barnes 2010-01-13 23:43

[quote=MyDogBuster;201817]BTW all, If a Riesel base ends in 4 or 9 and shows as not being started(ie 604, 609, 774, 819) then I have probably completed it. Those are the algebraic factors I'm working on. Gary doesn't want me to reserve more than 3 at a time so that I don't swamp him with updates.

Hey Gary, only about 45 to go and they are all tested to n=25K.[/quote]

I'm well caught up. Go ahead and reserve 10 at a time and post them 10 at a time if you want. Since that many updates will only last about a week, I should be able to keep up.

Mini-Geek 2010-01-14 00:39

Riesel Base 605
Conjectured k = 100

Found Primes:[code]2*605^188-1
4*605^3-1
6*605^2-1
8*605^4-1
10*605^2379-1
12*605^3-1
14*605^2-1
16*605^1-1
18*605^1-1
20*605^28-1
22*605^1-1
24*605^1-1
26*605^4-1
28*605^9-1
30*605^1-1
32*605^4-1
34*605^163-1
36*605^5-1
38*605^1598-1
40*605^3-1
42*605^1-1
44*605^3210-1
46*605^19-1
48*605^4-1
50*605^1910-1
52*605^13569-1
54*605^3-1
56*605^6-1
58*605^1-1
60*605^1-1
62*605^2-1
64*605^205-1
66*605^1-1
68*605^36-1
70*605^1-1
72*605^15-1
74*605^5268-1
76*605^1-1
78*605^1-1
80*605^322-1
82*605^27-1
84*605^63-1
86*605^7788-1
88*605^1-1
90*605^1-1
92*605^2-1
94*605^49-1
96*605^12-1
98*605^66-1
[/code]Conjecture Proven

MyDogBuster 2010-01-15 08:53

More algebraic factors

Reserving Riesel bases 514, 534, 544, 549, 564, 574, 579, 604 & 609

kar_bon 2010-01-16 01:21

2*1019^n-1 at n=100k, no prime yet, continuing

KEP 2010-01-16 10:59

Reserving Sierp base 955, to help get another one of the untested list :smile:

KEP!

gd_barnes 2010-01-18 07:02

I'm filling in some holes in Batalov's reservations from November. Some of these had higher primes reported but no lower primes. Some had only k's remaining reported with no primes. Details are now shown on the web pages.

S1002: Previously reported as 10 k's remaining at n=9.8K with no primes posted. Started from scratch, filled in all primes, found 1 more prime for n=9.8K-25K, 9 k's remaining at n=25K. Released back to Batalov, who didn't specify if he wanted it any higher.

Serge, if you see this, are you planning any more work on all of your November reservations? If so, I need all primes so I don't have to do double work. Thanks.


Gary

MyDogBuster 2010-01-19 17:46

Riesel Base 514
 
Riesel Base 514
Conjectured k = 104
Covering Set = 5, 103
Trivial Factors k == 1 mod 3(3) and k == 1 mod 19(19)

Found Primes: 62k's File attached

Remaining k's: Tested to n=25K
9*514^n-1 <------ Proven composite by partial algebraic factors
30*514^n-1

Trivial Factor Eliminations: 38k's

Base Released

MyDogBuster 2010-01-19 17:50

Riesel Base 534
 
Riesel Base 534
Conjectured k = 106
Covering Set = 5, 107
Trivial Factors k == 1 mod 13(13) and k == 1 mod 41(41)

Found Primes: 86k's File attached

Remaining k's: Tested to n=25K
4*534^n-1 <------ Proven composite by partial algebraic factors
9*534^n-1 <------ Proven composite by partial algebraic factors
11*534^n-1
29*534^n-1
46*534^n-1
49*534^n-1 <------ Proven composite by partial algebraic factors
59*534^n-1
64*534^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 10k's

Base Released

MyDogBuster 2010-01-19 17:54

Riesel Base 544
 
Riesel Base 544
Conjectured k = 219
Covering Set = 5, 109
Trivial Factors k == 1 mod 3(3) and k == 1 mod 181(181)

Found Primes: 142k's File attached

Remaining k's:
9*544^n-1 <------ Proven composite by partial algebraic factors
144*544^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 73k's

Conjecture Proven

Nice high conjectured k proven

MyDogBuster 2010-01-19 17:57

Riesel Base 549
 
Riesel Base 549
Conjectured k = 34
Covering Set = 5, 11
Trivial Factors k == 1 mod 2(2) and k == 1 mod 137(137)

Found Primes: 14k's File attached

Remaining k's: Tested to n=25K
4*549^n-1 <------ Proven composite by partial algebraic factors
6*549^n-1

Base Released

MyDogBuster 2010-01-19 18:04

Moving right along

Reserving Riesel 619,634,669,679,684,724,739,759,774,814

I see a light at the end of the tunnel. :toot:

MyDogBuster 2010-01-20 15:46

Riesel Base 564
 
Riesel Base 564
Conjectured k = 114
Covering Set = 5, 113
Trivial Factors k == 1 mod 563(563)

Found Primes: 106k's File attached

Remaining k's: Tested to n=25K
4*564^n-1 <------ Proven composite by partial algebraic factors
9*564^n-1 <------ Proven composite by partial algebraic factors
39*564^n-1
49*564^n-1 <------ Proven composite by partial algebraic factors
64*564^n-1 <------ Proven composite by partial algebraic factors
86*564^n-1

Base Released

MyDogBuster 2010-01-20 15:53

Riesel Base 574
 
Riesel Base 574
Conjectured k = 24
Covering Set = 5, 23
Trivial Factors k == 1 mod 3(3) and k == 1 mod 191(191)

Found Primes: 14k's File attached

Remaining k's:
9*574^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 7k's

Conjecture Proven

MyDogBuster 2010-01-20 15:56

Riesel Base 579
 
Riesel Base 579
Conjectured k = 204
Covering Set = 5, 29
Trivial Factors k == 1 mod 2(2) and k == 1 mod 17(17)

Found Primes: 89k's File attached

Remaining k's: Tested to n=25K
4*579^n-1 <------ Proven composite by partial algebraic factors
64*579^n-1 <------ Proven composite by partial algebraic factors
104*579^n-1
106*579^n-1
114*579^n-1
144*579^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 6k's

Base Released

MyDogBuster 2010-01-20 15:59

Riesel Base 604
 
Riesel Base 604
Conjectured k = 21
Covering Set = 5, 11
Trivial Factors k == 1 mod 3(3) and 1 mod 67(67)

Found Primes: 12k's File attached

Remaining k's:
9*604^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 6k's

Conjecture Proven

MyDogBuster 2010-01-20 16:10

Riesel Base 609
 
Riesel Base 609
Conjectured k = 184
Covering Set = 5, 61
Trivial Factors = k == 1 mod 2(2) and k == 1 mod 119(19)

Found Primes: 83k's File attached

Remaining k's:
4*609^n-1 <------ Proven composite by partial algebraic factors
64*609^n-1 <------ Proven composite by partial algebraic factors
144*609^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 5k's

Conjecture Proven

Another nice high CK base proven

MyDogBuster 2010-01-20 16:21

Riesel Base 619
 
Riesel Base 619
Conjectured k = 216
Covering Set = 5, 31
Trivial Factors k == 1 mod 2(2)andk == 1 mod 3(3) and k == 1 mod 103(103)

Found Primes: 66k's File attached

Remaining k's: Tested to n=25K
6*619^n-1
138*619^n-1
144*619^n-1 <------ Proven composite by partial algebraic factors
206*619^n-1

Trivial Factor Eliminations: 37k's

Base Released

MyDogBuster 2010-01-21 14:41

Riesel Base 634
 
Riesel Base 634
Conjectured k = 126
Covering Set = 5, 127
Trivial Factors k == 1 mod 3(3) and k == 1 mod 211(211)

Found Primes: 80k's File attached

Remaining k's: Tested to n=25K
9*634^n-1 <------ Proven composite by partial algebraic factors
20*634^n-1
29*634^n-1

Trivial Factor Eliminations: 41k's

Base Released

MyDogBuster 2010-01-21 14:44

Riesel Base 669
 
Riesel Base 669
Conjectured k = 66
Covering Set = 5, 67
Trivial Factors k == 1 mod 2(2) and k == 1 mod 167(167)

Found Primes: 30k's File attached

Remaining k's:
4*669^n-1 <------ Proven composite by partial algebraic factors
64*669^n-1 <------ Proven composite by partial algebraic factors

Conjecture Proven

MyDogBuster 2010-01-21 14:47

Riesel Base 679
 
Riesel Base 679
Conjectured k = 186
Covering Set = 5, 17
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 113(113)

Found Primes: 58k's File attached

Remaining k's:
144*679^n-1 <------ Proven composite by partial algebraic factors
174*679^n-1

Trivial Factor Eliminations: 32k's

Base Released

MyDogBuster 2010-01-21 14:50

Riesel Base 684
 
Riesel Base 684
Conjectured k = 46
Covering Set = 5, 29, 73
Trivial Factors k == 1 mod 683(683)

Found Primes: 41k's File attached

Remaining k's: Tested to n=25K
4*684^n-1 <------ Proven composite by partial algebraic factors
9*684^n-1 <------ Proven composite by partial algebraic factors
39*684^n-1

Base Released

MyDogBuster 2010-01-21 14:53

Riesel Base 724
 
Riesel Base 724
Conjectured k = 59
Covering Set = 5, 29
Trivial Factors k == 1 mod 3(3) and k == 1 mod 241(241)

Found Primes: 36k's File attached

Remaining k's: Tested to n=25K
9*724^n-1 <------ Proven composite by partial algebraic factors
48*724^n-1

Trivial Factor Eliminations: 19k's

Base Released

MyDogBuster 2010-01-21 14:56

Riesel Base 739
 
Riesel Base 739
Conjectured k = 36
Covering Set = 5, 37
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 41(41)

Found Primes: 11k's File attached

Trivial Factor Eliminations: 6k's

Conjecture Proven

MyDogBuster 2010-01-21 14:59

Riesel Base 759
 
Riesel Base 759
Conjectured k = 56
Covering Set = 5, 19
Trivial Factors k == 1 mod 2(2) and k == 1 mod 379(379)

Found Primes: 25k's File attached

Remaining k's: Tested to n=25K
4*759^n-1 <------ Proven composite by partial algebraic factors
34*759^n-1

Base Released

MyDogBuster 2010-01-21 15:02

Reserving Riesel Bases 819, 829, 834, 849, 859, 864, 879, 919, 924, 934

gd_barnes 2010-01-23 07:22

Here is the split out of the applicable portion of an Email from Serge (Batalov) for the status of his bases 501-1024:

[quote]
Here's the summary:
1. grep PRP in all files. Don't miss the k=316 prime in S961. It is done to 50K.
[/quote]
Gary

MyDogBuster 2010-01-23 17:25

Riesel Base 774
 
Riesel Base 774
Conjectured k = 61
Covering Set = 5, 31
Trivial Factors k == 1 mod 713(713)

Found Primes: 54k's File attached

Remaining k's: Tested to n=25K
4*774^n-1 <------ Proven composite by partial algebraic factors
9*774^n-1 <------ Proven composite by partial algebraic factors
25*774^n-1
30*774^n-1
49*774^n-1 <------ Proven composite by partial algebraic factors

Base Released

MyDogBuster 2010-01-23 17:28

Riesel Base 814
 
Riesel Base 814
Conjectured k = 164
Covering Set = 5, 163
Trivial Factors k == 1 mod 3(3) and k = 1 mod 271(271)

Found Primes: 102k's File attached

Remaining k's: Tested to n=25K
9*814^n-1 <------ Proven composite by partial algebraic factors
14*814^n-1
44*814^n-1
128*814^n-1
134*814^n-1
144*814^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 54k's

Base Released

MyDogBuster 2010-01-23 17:32

Riesel Base 819
 
Riesel Base 819
Conjectured k = 124
Covering Set = 5, 41
Trivial Factors k == 1 mod 2(2) and k == 1 mod 409(409)

Found Primes: 59k's

Remaining k's:
4*819^n-1 <------ Proven composite by partial algebraic factors
64*819^n-1 <------ Proven composite by partial algebraic factors

Conjecture Proven

MyDogBuster 2010-01-23 17:35

Riesel Base 829
 
Riesel Base 829
Conjectured k = 84
Covering Set = 5, 83
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 23(23)

Found Primes: 26k's File attached

Trivial Factor Eliminations: 15k's

Conjecture Proven

MyDogBuster 2010-01-23 17:42

Riesel Base 834
 
Riesel Base 834
Conjectured k = 166
Covering Set = 5, 167
Trivial Factors k == 1 mod 7(7) and k == 1 mod 11(11)

Found Primes: 129k's File attached

Remaining k's:
4*834^n-1 <------ Proven composite by partial algebraic factors
9*834^n-1 <------ Proven composite by partial algebraic factors
49*834^n-1 <------ Proven composite by partial algebraic factors
144*834^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 31k's

Conjecture Proven

MyDogBuster 2010-01-23 17:50

Riesel Base 849
 
Riesel Base 849
Conjectured k = 16
Covering Set = 5, 17
Trivial Factors k == 1 mod 2(2) and k == 1 mod 53(53)

Found Primes:
2*849^1-1
6*849^19-1
8*849^1-1
10*849^21-1
12*849^2-1
14*849^4114-1

Remaining k's:
4*849^n-1 <------ Proven composite by partial algebraic factors

Conjecture Proven

MyDogBuster 2010-01-23 17:59

Riesel Base 859
 
Riesel Base 859
Conjectured k = 44
Covering Set = 5, 43
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 11(11) and k == 1 mod 13(13)

Found Primes: 11k's File attached

Remaining k's:
26*859^n-1

Trivial Factor Eliminations: 9k's

Base Released

MyDogBuster 2010-01-23 18:05

Riesel Base 879
 
Riesel Base 879
Conjectured k = 34
Covering Set = 5, 11
Trivial Factors k == 1 mod 2(2) and k == 1 mod 439(439)

Found Primes: 14k's File attached

Remaining k's: Tested to n=25K
4*879^n-1 <------ Proven composite by partial algebraic factors
24*879^n-1

Base Released

MyDogBuster 2010-01-23 18:08

Reserving Riesel Bases 939, 949, 964, 969, 984

This is the last of b = 4 mod 5 algebraic factors that I have done

gd_barnes 2010-01-24 03:47

Batalov sent me full results file on all of his bases. Here is some additional detail to add to his Sierp base 961 status in the Email that I posted above:

R1021 is at n=50K; released
S1002 is at n=27.7K; released

Here is the additional base 961 prime that had not yet been posted in the forum:
346*961^30374+1 is prime

He no longer has any bases > 500 reserved.

gd_barnes 2010-01-24 11:17

[quote=MyDogBuster;202973]Reserving Riesel Bases 939, 949, 964, 969, 984

This is the last of b = 4 mod 5 algebraic factors that I have done[/quote]

Don't forget Riesel base 799. It shouldn't be too bad.

MyDogBuster 2010-01-24 11:52

[QUOTE]Don't forget Riesel base 799. It shouldn't be too bad. [/QUOTE]

With a conjecture k of 1,885,767,686,976 it should only take me an hour or so to complete (if I don't test it above n=1). Your Excel spreadsheet formula will have a hard time finding the partial algebraic factors anyway. Maybe we can have a race to see who gets there first. Just kidding. :devil:

gd_barnes 2010-01-24 12:11

[quote=MyDogBuster;203037]With a conjecture k of 1,885,767,686,976 it should only take me an hour or so to complete (if I don't test it above n=1). Your Excel spreadsheet formula will have a hard time finding the partial algebraic factors anyway. Maybe we can have a race to see who gets there first. Just kidding. :devil:[/quote]

:missingteeth::missingteeth:

And here's the kicker: I bet you can't test it to n=1 in an hour, even if you put all of your machines on it at the same time. For speed, be sure and set no factoring with the -f0 switch. You can even forget about proving the PRPs. (I'm not sure if running at least 2 regular ABC2 PFGW scripts would be faster [one each for k==(0 mod 6) & (2 mod 6)]. Likely not since there's so many other trivial k's.)

Better yet, I'll give you a day with all of your machines and you still don't have to prove the PRPs. Just divide the k-value by 50 or whatever # of cores you have and split them up that way just like you'd do different P-ranges while sieveing.

Good luck! :smile:

Edit: I got to k=1M in just a few mins. of course that's < 1 one-millionth of the entire k-range. :-)


Gary

Mini-Geek 2010-01-24 13:31

I wonder what would be the fastest way to go about doing something like that...my initial impression would be that sieving might be a good thing to do, since the numbers are so plentiful that anything where you look at them one at a time would be terribly slow, and so small that they might be easily TFd (or proven prime by TF).
I doubt any current sieving programs would be at all optimized for something like that, though.

KEP 2010-01-24 14:09

I've done some testings on Riesel base 633 with a conjectured k=1004, it has by tonight been taken to n=25K, however I need some help on finding out, weather or not the k=64, has to be excluded from further testing (beyond n=25K) or it has to remain in the k's remaining list. Srsieve raised a warning, that k=64 for Riesel base 633, has algebraric factors, however I've never gotten the website used to testing for algebraric factors to work. So can either Gary or Ian tell me weather or not it is composit for all n's for k=64 or it in fact is expected to have a prime eventually?

Btw, consider it reserved, to n=25K, if I want to take it higher I'll let you all know :smile:

Regards

KEP

Mini-Geek 2010-01-24 16:02

My first idea was to do it purely with a sieve program, but considering k's with trivial factors, MOBs, etc. I figured it'd be too complicated to bypass the PFGW script completely. So I decided to try using PFGW to n=0 (essentially just eliminating trivial factors and MOBs) followed by srsieve. Srsieve doesn't know to stop searching a number and declare it prime when it passes the square root of it, (only when you pass the number itself) so you have to manually set the max p to the square root of the largest number. That works, and is significantly faster (not drastically faster, but significant...maybe about 15% faster) than using PFGW to n=1, but doesn't eliminate the k's from the file by itself. And my remove-ks.pl script isn't very good at doing that many different k's. So in the end, I figured it's probably fastest (not to mention way easier!) to just use PFGW with the max n set to 1.
Here are the timings I found:[code]k=1 to 10,000:
PFGW n=0:
7 CPU seconds (checks for trivial factors and MOB)
PFGW n=1:
10 CPU seconds
srsieve:
<1 CPU second

k=1 to 100,000:
PFGW n=0: (no -f)
56 CPU seconds
PFGW n=0: (-f)
57 CPU seconds
srsieve:
24 CPU seconds (56+24=80 CPU seconds)
PFGW n=1: (no -f)
95 CPU seconds
PFGW n=1: (no -f, minimized to tray)
88 CPU seconds
PFGW n=1: (-f)
95 CPU seconds

k=1885767586974 to 1885767686974 (final 100k):
PFGW n=1: (no -f)
115 CPU seconds
PFGW n=1: (-f)
121 CPU seconds
[/code]If the time for each 100,000 range averaged 105 CPU seconds, my two cores could complete all the k's to n=1 in only about 31.4 years! See you June 2041! :smile:
Looks like there is no big difference between using -f and not for numbers this small.

For the record, I ran srsieve with these options:
[code]srsieve -n1 -N1 -P8939 -q -Z -G pl_remain.txt[/code]i.e. min and max n set to 1, min p defaulting to 3 and max p set to sqrt(100000*799-1) (rounded up), quiet so it doesn't fill the screen and so waste CPU time, higher priority, outputting to the .prp format, and reading the list of sequences from pl_remain.txt left from PFGW's run, and its output (in t17_b799.prp) is all the primes.

MyDogBuster 2010-01-24 16:13

[QUOTE]So can either Gary or Ian tell me weather or not it is composit for all n's for k=64 or it in fact is expected to have a prime eventually?[/QUOTE]

I'll let Gary handle that. I'm still learning this stuff and 633 is beyond my knowledge at the moment.:question:

MyDogBuster 2010-01-24 18:32

Riesel Base 919
 
Riesel Base 919
Conjectured k = 24
Covering Set = 5, 23
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 17(17)

Found Primes:
2*919^4-1
6*919^11-1
8*919^1-1
12*919^1-1
14*919^2-1
20*919^1-1

Trivial Factor Eliminations: 5k's

Conjecture Proven

MyDogBuster 2010-01-24 18:35

Riesel Base 924
 
Riesel Base 924
Conjectured k = 36
Covering Set = 5, 37
Trivial Factors k == 1 mod 13(13) and k == 1 mod 71(71)

Found Primes: 30k's File attached

Remaining k's:
4*924^n-1 <------ Proven composite by partial algebraic factors
9*924^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 2k's

Conjecture Proven

MyDogBuster 2010-01-24 18:39

Riesel Base 934
 
Riesel Base 934
Conjectured k = 21
Covering Set = 5, 11
Trivial Factors k == 1 mod 3(3) and k == 1 mod 311(311)

Found Primes: 12k's File attached

Remaining k's:
9*934^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 6k's

Conjecture Proven

henryzz 2010-01-24 18:40

Using maxima i have found the following algebraic factorizations:
64*633^(2*n)-1=(8*633^n-1)*(8*633^n+1)
64*633^(3*n)-1=(4*633^n-1)*(16*633^(2*n)+4*633^n+1)
which leaves 1 and 5 mod 6
n=1 and 5 mod 12 are divisible by 17
which leaves 7 and 11 mod 12
after that there doesnt seem to me anything else obvious that would form a covering set
basically one third of n values are left after algebraic factorizations and half of them are eliminated by the trivial factor 17 so it is possible to test just one sixth of candidates

henryzz 2010-01-24 18:57

after a sieve to 1e8 there arent that many to remove anyway
[CODE]Read 4985 terms for 1 sequence from NewPGen format file `t17_b633_k64.npg'.
64*633^n-1: n = 1*m+0, 4985 terms
n = 3 (mod 12): 338 terms
n = 4 (mod 12): 265 terms
n = 7 (mod 12): 2258 terms
n = 11 (mod 12): 2124 terms[/CODE]
only 12.1% of the file will be removed when removing ns with algebraic factors

henryzz 2010-01-24 19:34

More discoveries:smile:
More sieving without removal decreases the percentage of candidates that have algebraic factors. At 1e10 only 10.2% of the file would be removed.
It is however worthwhile removing them as it doubles the speed of the sieve when using sr1sieve.:smile:
Are there any other bases with complex algebraic factors like riesel 633? If so what bases?

KEP 2010-01-24 19:35

[QUOTE=henryzz;203088]after a sieve to 1e8 there arent that many to remove anyway
[CODE]Read 4985 terms for 1 sequence from NewPGen format file `t17_b633_k64.npg'.
64*633^n-1: n = 1*m+0, 4985 terms
n = 3 (mod 12): 338 terms
n = 4 (mod 12): 265 terms
n = 7 (mod 12): 2258 terms
n = 11 (mod 12): 2124 terms[/CODE]
only 12.1% of the file will be removed when removing ns with algebraic factors[/QUOTE]

Thanks for your input. Just call me an illiterate, but does all your explanation mean that the 1/6 that is left, has to be tested or does what you stated actually mean that the k has to be left out of the que? Also after sieving, does one have to remove n's manually or will the sieve rule out all n's with algebraric factors?

Again, for now, it is remaining in the test file, at least up to n=25K, so everything should be good and fine for now :smile:

Regards

KEP

henryzz 2010-01-24 19:53

1 Attachment(s)
[quote=KEP;203097]Thanks for your input. Just call me an illiterate, but does all your explanation mean that the 1/6 that is left, has to be tested or does what you stated actually mean that the k has to be left out of the que? Also after sieving, does one have to remove n's manually or will the sieve rule out all n's with algebraric factors?

Again, for now, it is remaining in the test file, at least up to n=25K, so everything should be good and fine for now :smile:

Regards

KEP[/quote]
1/3 needs to be sieved but half of that will quickly be sieved away by the factor 17
the k does need to be tested as far as i can tell although i wouldnt do anything serious until gary posts
to remove the ns with algebraic factors use the perl script in post #64 of this thread to remove ns that are 0 mod 2 and 0 mod 3
i have attached a file sieved to 1e10 for that k with the algebraic factors removed

MyDogBuster 2010-01-24 20:46

Riesel Base 939
 
Riesel Base 939
Conjectured k = 46
Covering Set = 5, 47
Trivial Factors k == 1 mod 2(2) and k == 1 mod 7(7) and k == 1 mod 67(67)

Found Primes: 18k's File attached

Remaining k's:
4*939^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 3k's

Conjecture Proven

MyDogBuster 2010-01-24 20:49

Riesel Base 949
 
Riesel Base 949
Conjectured k = 56
Covering Set = 5, 97
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 79(79)

Found Primes: 18k's File attached

Trivial Factor Eliminations: 9k's

Conjecture Proven

MyDogBuster 2010-01-24 20:53

Riesel Base 964
 
Riesel Base 964
Conjectured k = 194
Covering Set = 5, 193
Trivial Factors k == 1 mod 3(3) and k == 1 mod 107(107)

Found Primes: 123k's File attached

Remaining k's: Tested to n=25K
9*964^n-1 <------ Proven composite by partial algebraic factors
129*964^n-1
141*964^n-1
144*964^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 65k's

Base Released

MyDogBuster 2010-01-24 20:56

Riesel Base 969
 
Riesel Base 969
Conjectured k = 96
Covering Set = 5, 97
Trivial Factors k == 1 mod 2(2) and k == 1 mod 11(11)

Found Primes: 41k's File attached

Remaining k's:
4*969^n-1 <------ Proven composite by partial algebraic factors
64*969^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 4k's

Conjecture Proven

MyDogBuster 2010-01-24 21:00

Riesel Base 984
 
Riesel Base 984
Conjectured k = 196
Covering Set = 5, 197
Trivial Factors k == 1 mod 983(983)

Found Primes: 180k's File attached

Remaining k's: Tested to n=25K
4*984^n-1 <------ Proven composite by partial algebraic factors
9*984^n-1 <------ Proven composite by partial algebraic factors
18*984^n-1
49*984^n-1 <------ Proven composite by partial algebraic factors
64*984^n-1 <------ Proven composite by partial algebraic factors
81*984^n-1
86*984^n-1
99*984^n-1
119*984^n-1
120*984^n-1
121*984^n-1
144*984^n-1 <------ Proven composite by partial algebraic factors
169*984^n-1 <------ Proven composite by partial algebraic factors
191*984^n-1

Base Released

I'll be doing the rest of b=4 mod 5 (larger ck's and stuff I missed) over time. Not reserving anything, for now. Time for a break.
By my count, there are 28 left, most have a ck of >200. I still owe R864.

KEP 2010-01-24 21:52

[QUOTE=henryzz;203098]1/3 needs to be sieved but half of that will quickly be sieved away by the factor 17
the k does need to be tested as far as i can tell although i wouldnt do anything serious until gary posts
to remove the ns with algebraic factors use the perl script in post #64 of this thread to remove ns that are 0 mod 2 and 0 mod 3
i have attached a file sieved to 1e10 for that k with the algebraic factors removed[/QUOTE]

Thanks for the sieve file, and thanks for your input and help. Your skills is far greater than mine currently. However, I'm wondering now, is there a way to determine weather or not a base contain k's with algebraric factors (such that I can avoid choosing bases in the future of wich there is algebraric factors for some k's)?

Thanks again.

Regards

KEP

henryzz 2010-01-25 07:47

[quote=KEP;203116]Thanks for the sieve file, and thanks for your input and help. Your skills is far greater than mine currently. However, I'm wondering now, is there a way to determine weather or not a base contain k's with algebraric factors (such that I can avoid choosing bases in the future of wich there is algebraric factors for some k's)?

Thanks again.

Regards

KEP[/quote]
Now thats beyond me:smile:
I would actually like to find a k which algebraic factors make as low weight as yours myself so i would like to know what the conditions are.

gd_barnes 2010-01-25 09:39

ALL Riesel bases have squared k's with partial algebraic factors. The base doesn't matter (as long as the conjecture is > 4). Just think of k=4 or 9 on any Riesel base. For k=4, all even n's are eliminated by [2*b^(n/2)-1]*[2*b^(n/2)+1]. The only time algebraic factors can eliminate a k is if the rest of the n's are eliminated by a single factor (frequently 5) or a covering set of factors.

Just let srxsieve do its/their thing. If you get the "x*xx^n-1 contains algebraic factors", it means you can manually remove some (actually very few) n-values. If you have a squared k, you can manually remove the few even n's left over. If you have a cubed k, you can manually remove the fewer n==(0 mod 3) left over. It does NOT mean you can fully eliminate the k from testing.

k=64 happens to be one where you can remove both even n's and n==(0 mod 3) since it's a perfect square and cube. But...it's not just Riesel base 633. You can remove those n-values for k=64 on ALL Riesel bases. On the Sierp side, you could only eliminate n==(0 mod 3) for k=64 because k^2*b^2+1 does not algebraically factor whereas k^2*b^2-1 factors to (k*b-1)*(k*b+1).

Personally, I don't bother with manually removing n-values because most of my testing has been n<=25K. I'd suggest the same for others. It's not worth the personal time to do it for a small amount of CPU time savings. I'd suggest doing the above only for n>25K.

The factorization for the various n-values for 64*633^n-1 is normal. On a fair-sized percentage of k's remaining at CRUS, you'll get down to only 1 n-value out of every 6 or 12 n's remaining where there is not a small factor. That low weight is why those k's are remaining. And the fact that a squared k cannot have an even-n Riesel prime is why more squared k's are remaining than one would expect by chance alone.

In a synopsis: On a squared k, after doing a "mental" elimination of even n's, if you can't quickly determine a covering set of small factors (generally < 10K) for odd n's, then just sieve it normally and test it normally. You'll usually only end up testing about 1-3% more n-values than you would if you manually eliminated the even n's.


Gary

gd_barnes 2010-01-25 09:46

[quote=KEP;203116]Thanks for the sieve file, and thanks for your input and help. Your skills is far greater than mine currently. However, I'm wondering now, is there a way to determine weather or not a base contain k's with algebraric factors (such that I can avoid choosing bases in the future of wich there is algebraric factors for some k's)?

Thanks again.

Regards

KEP[/quote]

That's what the "generallizing algebraic factors on Riesel bases" thread is for. Think Riesel bases that are:

b==(4 mod 5)
b==(12 mod 13)
b==(16 mod 17)
b==(28 mod 29)
b==(36mod37)
b==(40mod41)
b==(52mod53)
b==(60mod61)
b==(112mod113)

These are all of the base modulos that have been found so far that can possibly contain k's with "partial algebraic factors" that make a full covering set. Base 633 does not fit any of them.

For non-math type people, I put this in as "English" of terms as possible. Please see the smiley face in the first post in that thread [URL="http://www.mersenneforum.org/showpost.php?p=153696&postcount=1"]here[/URL].

Ian has been kind enough to test ~85-95% of the bases where b==(4 mod 5) so that others don't run into the problems that they entail. The others are less common and require that the conjecture be higher than the lowest possible k that can contain partial algebraic factors to make a full covering set. Frequently, it is not so even if a base hits one of the above criteria, it won't have k's to be concerned about.

Generally, don't worry about it. Just check and see if your base fits one of the modulos in that thread. If it does, I'll be glad to let you know if any k's fit the criteria.

One further point: Algebraic factors are rarely an issue on the Sierp side since 99% of the time they must be for cubes or higher powers. There, it's GFN's to be concerned about but they are much easier to determine. So if algebraic factors concern you, you might stick with the Sierp side.


Gary

gd_barnes 2010-01-25 11:26

[quote=KEP;203051]I've done some testings on Riesel base 633 with a conjectured k=1004, it has by tonight been taken to n=25K, however I need some help on finding out, weather or not the k=64, has to be excluded from further testing (beyond n=25K) or it has to remain in the k's remaining list. Srsieve raised a warning, that k=64 for Riesel base 633, has algebraric factors, however I've never gotten the website used to testing for algebraric factors to work. So can either Gary or Ian tell me weather or not it is composit for all n's for k=64 or it in fact is expected to have a prime eventually?

Btw, consider it reserved, to n=25K, if I want to take it higher I'll let you all know :smile:

Regards

KEP[/quote]


I missed this original post that brought up all of this.

No, you cannot eliminate the k. You can only eliminate even n-values and n-values that are n==(0 mod 3) since the k is both a perfect square and cube.

The same thing applies for k=64 on any Riesel base if the base is not a perfect square or cube.

For 64*633^n-1, as previously shown by David, after eliminating the above for algebraic factors, you're left with n==(1 or 5 mod 6) and after eliminating a factor of 17, you're left with n==(7 or 11 mod 12). There are no other small and consistent factors for those. Therefore, 1 out of every 6 values remains that doesn't have a small factor and hence could contain a prime at higher n-values.

On another note: Have you acquired a lot more machines lately? You have a tremendous amount of work reserved on base 3. Then there's huge bases 63 and 955 and you're still reserving more work. How is base 63 going?


Gary

KEP 2010-01-25 15:42

Base 3, is going to be completed at the end of march for both the reservations, maybe april. Sierp base 63, is still going, but it is currently only running on the dual core. My recent, base 633 reservation, is not goin to be taken further than n=25K, which it completes tonight. Untill my base 3 reservations complete, the quad will only work on the Riesel base 3 tests on all four cores. Currently less than 1500 k's is remaining at a test depth of n=~40-60K. I have not bought new machines, and I may admit, that the k=3677878 to 1G reservation, I would very much like that to be changed back to n=1M. Once I complete S63, R3, aswell as the current R633, then my focus will move to only s955, which I tend to keep till it goes below 1,000 k's remaining. S955 is currently at n=2500, and 8973 k's remaining. I'm not sure if the s63 reservation can be completely finish by the end of march, but in 20-25 days the k=3677878 for R3, will have been taken to n=1M and then complete my original reservation.

But to sum up, all bases is being worked on, even though I still only has 1 DualCore and 1 QuadCore. But on a sidenote, as soon as All R3 and S63 is completed, then all six cores will be hammered on the S955 conjecture.

Hope I got it all, and that it was clear what I stated :smile:

Regards

Kenneth

Ps. The reason Riesel base 3 k<700M range is tested to n=40K-60K is because it reached optimal sievedepth for n<=60K, while it was only tested to n=40K.

MyDogBuster 2010-01-25 19:14

Riesel Base 864
 
Riesel Base 864
Conjectured k = 174
Covering Set = 5, 173
Trivial Factors k == 1 mod 863(863)

Found Primes: 162k's File attached

Remaining k's: Tested to n=25K
4*864^n-1 <------ Proven composite by partial algebraic factors
6*864^n-1
9*864^n-1 <------ Proven composite by partial algebraic factors
49*864^n-1 <------ Proven composite by partial algebraic factors
64*864^n-1 <------ Proven composite by partial algebraic factors
96*864^n-1
114*864^n-1
134*864^n-1
144*864^n-1 <------ Proven composite by partial algebraic factors
169*864^n-1 <------ Proven composite by partial algebraic factors

Base Released

gd_barnes 2010-01-25 23:16

Thanks for the detailed status Kenneth. I'll change k=3677878 back to n=1M. I thought you might change your mind there. :smile:

I know you were doing a sieve for n=1M-1G. Did you work on that any? I wouldn't post such a huge sieve file on the pages but I would be willing to take part of it and post one for n=1M-10M or n=1M-5M depending on the size of the file. But if you only sieved it to a very low depth, don't worry about it. It wouldn't be worth posting.

As for base 955, I seriously doubt that you can get that to < 1000 k's remaining but I haven't done a detailed analysis. But to get rid of nearly 90% of k's after a test to n=2500 is extraordinarily difficult, especially for a high base. But that's OK. If you can at least get it up to n=10K, I'll gladly show the k's remaining on the pages.

KEP 2010-01-26 10:01

[QUOTE=gd_barnes;203217]Thanks for the detailed status Kenneth. I'll change k=3677878 back to n=1M. I thought you might change your mind there. :smile:

I know you were doing a sieve for n=1M-1G. Did you work on that any? I wouldn't post such a huge sieve file on the pages but I would be willing to take part of it and post one for n=1M-10M or n=1M-5M depending on the size of the file. But if you only sieved it to a very low depth, don't worry about it. It wouldn't be worth posting.

As for base 955, I seriously doubt that you can get that to < 1000 k's remaining but I haven't done a detailed analysis. But to get rid of nearly 90% of k's after a test to n=2500 is extraordinarily difficult, especially for a high base. But that's OK. If you can at least get it up to n=10K, I'll gladly show the k's remaining on the pages.[/QUOTE]

Regarding Riesel base 3, k=3677878, I did only very sporadic sieving in 4 different areas, for p<=1T, so it is not even fully sieved to 1 trillion, and since it was nearing <20M candidates, I decided to abandon it, to do some further testing on lower k's and find some k's that could eventually lead me to a Top5000 prime. Should btw notice, that I don't think I have the sieve file any more at all.

Regarding S955, I'm currently on 1 core, sieving to n=100K, and I think I'll break it off at optimal sievedepth for n=10K or n=25K, dependent on how far it has been sieved once all my non base 955 work completes. So sure I'll take it to at least n=10000, which should leave about 4651 k's remaining.

There is currently a sieve speed of ½ G for Sierp base 955, so I think by the time all my non base 955 work completes, that it should be sieved to somewhere around 40-50G. As tonight, local danish time, I expect the sieve file to go below 70M candidates for 8973 sequences going from n>2500 to n<=100K. S955 is btw reducing the number of k's with ~28% for every doubleing of the n, so I expect somewhere between 1500 and 1600 k's remaining at n=100K :smile:

Well this answer got a little long, thanks for changing my reservation back to n=1M for riesel base 3. Hope I got it all.

Does anyone know, if sr2sieve can handle a Multimillion sievefile, with 8973 candidate sequences going from n=2501 to n<=100000, for s955, such as I can sieve with 18000 p/s in stead of 6500 p/s ?

Regards

KEP

gd_barnes 2010-01-26 10:51

[quote=KEP;203269]Does anyone know, if sr2sieve can handle a Multimillion sievefile, with 8973 candidate sequences going from n=2501 to n<=100000, for s955, such as I can sieve with 18000 p/s in stead of 6500 p/s ?

Regards

KEP[/quote]

Yes, sr2sieve can handle as big of a file size as can fit in your computer's memory but no, I don't think there's any chance that you'll be able to use it. That's because the restriction usually comes in with the initial creation of the Lengendre symbols. IF the k's are quite small (< 10000), then those are not too much problem. You can do 1000's of k's. The problem comes in when the k's get large.

As an example, I could not run sr2sieve to sieve Sierp base 31 for n=10K-25K even though there were only about 1500-1600 k's remaining at n=10K. Very maddening because I couldn't optimally sieve nearly as far with much slower srsieve. The reason? The k's were simply too big (not too numerous) for the Lengendre symbol table to fit in my computer's memory because the conjecture is k=~6.3M. My guess is to do the Legendre symbols on base 31 with 1500 k's would take anywhere from 30G to 75G of memory or more. I remember letting it run as long as it could on a machine with 4G memory. Eventually it just gave a "memory alloc" error and saved nothing and I don't think it was much past k=1M yet. I think it took half a day or a little more before it finally filled up. (I didn't realize it was chewing up memory when I did this or I would have stopped it after an hour or so.)

Here is one thing you might attempt although I think it will be slower than simply running the already slow srsieve on all of your k's: Split your k's up into groups of 1000 k's and try running each one of those batches through sr2sieve individually. Hint: Do NOT put all of the large k's in one file. The symbols become much larger as the k's get larger. (From what I can tell, it's more than a linear increase.) I'm not sure of the best way to divide them up but something like (if you were doing 3 files), k=10, 20, & 30 in file #1, k=13, 23, & 33 in file #2, and k=16, 26, & 36 in file #3 would be a way to even out the memory that would be used for each. That would be your best chance.

Before embarking on anything like this, put all the k's in one file and watch the symbols progress on your screen. Also check your task manager for the memory being used to get a reasonable initial estimate of how much total memory might be needed. That way you'll know if the above is a possible option.

One more suggestion: On the conjectures, I don't advise sieving an n-range where the high n-value divided by the low n-value is so large, especially at the low n-ranges. That's because many k's will be eliminated and much of your sieving will be wasted. My suggestion is that the high n-value be no more than 10 times the low n-value. In your case, in the future, consider two sieves of n=2500-25K & n=25K-100K (or n=2500-10K & 10K-100K). The point being that a lot of k's will be eliminated by n=10K or 25K. Although large n-ranges are highly efficient in srxsieve if you are going to test the entire n-range, there is a limit when you stop testing when a prime is found and that limit is based on the percentage of k's that should be eliminated in the range. As a wild guess, I'd estimate that if you expect that > ~65%-75% of your k's will be eliminated in the range you are sieving that you probably should sieve a somewhat smaller range. For example, I thought your sieving n=100K-1M on k=3677878 was a good range to sieve. But n=1M-1G would be way too much. If going higher, I'd suggest sieving either n=1M-5M or n=1M-10M. Iirc, you had a ~60% chance of prime for n=100K-1M so technically on avg, you should be eliminating 60% of your k's in that range (even though it's only one k, the analogy still applies); hence a good choice for the sieve range. That being the case, you should also have a ~60% chance for a prime in n=1M-10M so that's likely close to the largest range that you could efficiently sieve.

Now...if all that you have is a single or a couple of extremely low weight k's such that their combined percentage chance of prime in an n-range is < 50%, then by all means sieve an n-range with a ratio of a top to bottom n-value that is > 10 if you think that testing can complete in 1-2 years or less. If testing takes longer than that, than frequently future software/hardware improvements still make it better to sieve a smaller range. I mean, look at what happened with PFGW in the past 3 months and with 64-bit srxsieve in the last 2-3 years. One increased in speed by > 5 times and the other doubled in speed (from 32 bit) and that doesn't even take into account hardware improvements. The moral of the story: Don't use too many of today's resources to do too high of a percentage of tomorrow's work. Let a large portion of tomorrow's work be done by tomorrow's resources.

BTW, on your percentage drop in k's remaining with each doubling of the n-range, remember that the percentage drops somewhat as the n-ranges grow larger because the remaining k's are lower average weight. (If they were the same avg. weight then theoretically, it should remain the same.) If it's a 28% drop at n=2500, you'd probably be down to ~20% at n=100K. If it was a straight percentage drop, we'd prove these things a lot more easily.

...and you thought your answer was long. :-)


Gary

kar_bon 2010-01-26 11:38

[QUOTE=gd_barnes;203271]The reason? The k's were simply too big (not too numerous) for the Lengendre symbol table to fit in my computer's memory because the conjecture is k=~6.3M. My guess is to do the Legendre symbols on base 31 with 1500 k's would take anywhere from 30G to 75G of memory or more. [/QUOTE]

have you tried to use sr2sieve with option "-x" (or "--no-lookup")?

gd_barnes 2010-01-26 12:40

[quote=kar_bon;203276]have you tried to use sr2sieve with option "-x" (or "--no-lookup")?[/quote]

No. I wasn't aware of what that meant. So does that not do the Legendre symbols? If it doesn't, I wonder how much faster than srsieve it will be.

KEP 2010-01-26 16:43

Thanks Gary for your answer. I think I'll take a look at my 24h sieved file, and then convert the srsieve.out to a sr_955.abcd file, and then try using "-x" as Karsten suggested (ONLY if memory use becomes too big for my computer with legendre symbol lookup tables).

I actually thought that using the even and odd legendre symbol tables, was what actually made sr2sieve so much faster than srsieve. Apparently I was wrong on that one. On a side note, did a time test on testing the highest k for sierp base 955 at n=100K, it took about 3670 seconds to complete 1 test :smile:

But to sum up, yes your answer was longer, but it contained a lot of great ideas and I may consider to do a smaller n-high range for sieving, but nothing has been decided yet, even though your morale of the story is good, and somehow follows what i feel and think, however I also longs for a Top5000 prime, so sieving a high n may make sence in that way :smile:

Again thanks for your inputs and I'm sure looking forward to follow them as of tonight.

Regards

KEP

KEP 2010-01-27 13:29

1 Attachment(s)
Riesel base 633 is complete to n=25K, with 5 k's remaining :smile:

Regarding S955, I tried reducing the max-n to n=10K, and of course for both n=100K aswell for n=10K, using sr2sieve was impossible, with the legendre symbols. At k=10000, it took up already about 500MB. But using sr2sieve, with the -x flag, could maybe be a solution to sieve the high k's. I must say, that the RAM use was only a bit more than 360MB for 8973 k's, for both n=100K aswell for n=10K. HOWEVER, once I got to the line where sr2sieve claimed that it has started, I got an "the program has stopped working and is terminating" or something like that message, whereafter (hope thats the right word :smile: ) the program shuts down. These termination happens for both n=10K and n=100K, using the -x flag. Havent investigated it further, but maybe there is something Geoff needs to look a bit further into.

Regards

KEP

gd_barnes 2010-01-29 13:59

R999 is at n=10K; 87 k's remaining; continuing to n=25K. Details on the pages.

These high bases take a LONG time, even if few k's are remaining, which this one does not! :smile:

rogue 2010-01-29 14:17

[QUOTE=gd_barnes;203692]R999 is at n=10K; 87 k's remaining; continuing to n=25K. Details on the pages.

These high bases take a LONG time, even if few k's are remaining, which this one does not! :smile:[/QUOTE]

You've got that right. Riesel base 928 still has over 1000 k left at n=5000. With a conjectured k at just over 32000, k are not getting removed very quickly.

gd_barnes 2010-01-30 00:24

[quote=rogue;203693]You've got that right. Riesel base 928 still has over 1000 k left at n=5000. With a conjectured k at just over 32000, k are not getting removed very quickly.[/quote]

And then they seem to frequently go into these huge "prime holes" like you found with S811 & S961, which makes them even worse.

gd_barnes 2010-02-06 11:48

Dougal released R1017. Therefore I searched it to n=25K. 15 k's are remaining. Details on the pages. No further work.

MyDogBuster 2010-02-13 23:59

Reserving Riesel 639, 709 and 744 as new to n=25K

rogue 2010-02-14 04:19

Riesel base 803
 
Primes found:
[code]
2*803^48-1
4*803^89-1
6*803^1-1
8*803^4-1
10*803^3-1
12*803^14-1
16*803^31-1
18*803^2-1
20*803^4-1
24*803^4-1
26*803^10-1
28*803^1-1
30*803^48-1
32*803^56-1
34*803^119-1
36*803^7-1
38*803^328-1
40*803^1-1
42*803^3-1
44*803^12372-1
46*803^21-1
48*803^1-1
50*803^8-1
54*803^3-1
56*803^2-1
58*803^1-1
60*803^1-1
62*803^14-1
66*803^2-1
[/code]

Remaining k at n=25000:
[code]
14*803^n-1
22*803^n-1
52*803^n-1
64*803^n-1
[/code]

Since 64 = 8^2, I know that there are algebraic factorizations, but I don't know if they cover all n or if they only cover even n. Can someone answer that question?

Base released.

rogue 2010-02-14 04:20

Sierpinski base 803
 
Primes found
[code]
2*803^1+1
6*803^9+1
8*803^1243+1
10*803^6+1
12*803^13+1
14*803^1+1
[/code]k = 4 remains at n=25000. The base is released. As far as I understand it, k = 4 does not have algebraic factorizations for all n.

Mini-Geek 2010-02-14 12:09

[quote=rogue;205604]Since 64 = 8^2, I know that there are algebraic factorizations, but I don't know if they cover all n or if they only cover even n. Can someone answer that question?[/quote]
Even.

Where n being replaced by 2m shows that n is even...
[tex]64*803^n-1=8^2*803^{2m}-1=(8*803^m)^2-1=(8*803^m+1)*(8*803^m-1)[/tex]
(simple manipulations followed by the difference of squares rule)

64*803^n-1 with even n also trivially has a factor of 3. (visible at [URL="http://factordb.com/search.php?query=64*803%5E%282*n%29-1"]http://factordb.com/search.php?query=64*803^%282*n%29-1[/URL])

I don't see any reason that the odd n's can be eliminated. They are extremely low-weight, though, so the lack of a prime by 25K is not surprising. ([URL="http://factordb.com/search.php?query=64*803%5E%282*n%2B1%29-1&v=n&n=1&EC=1&E=1&Prp=1&P=1&C=1&FF=1&CF=1&of=H&pp=50"]look at them here[/URL])


I don't think Sierp 803 k=4 has any algebraic factorizations, because b^n+1 with n=2 doesn't have an algebraic factorization. See [URL]http://www.mersenneforum.org/showpost.php?p=199678&postcount=814[/URL]. In short, n must be at least 3 and not a power of 2. (3, 5, 6, 7, 9, etc.)
k=8...hm...8*803^n+1...has algebraic factors where n is a multiple of 3.
[tex]8*803^n+1=2^3*803^{3m}+1=(2*803^m)^3+1=(2*803^m+1)*(4*803^{2m}-2*803^m+1)[/tex]
(simple manipulations followed by the addition of cubes rule)
I doubt if it has a full covering set. Looking at its [URL="http://factordb.com/search.php?query=8*803%5En%2B1"]list of factorizations[/URL] in the DB, it seems unlikely.
The algebraic factorization isn't really necessary though, as a few small factors cover all n's divisible by 3.
[URL="http://factordb.com/search.php?query=8*803%5E%283*n%29%2B1"]http://factordb.com/search.php?query=8*803^(3*n)%2B1[/URL]

rogue 2010-02-14 23:10

Sierpinski base 516
 
1 Attachment(s)
I have attached the list of primes.

1 is a GFN, which has not been tested.

k=122 is the only k remaining and has been tested up to n=25000.

The other k have trivial factors.

I am releasing this base.

gd_barnes 2010-02-15 11:58

[quote=Mini-Geek;205623]Even.

Where n being replaced by 2m shows that n is even...
[tex]64*803^n-1=8^2*803^{2m}-1=(8*803^m)^2-1=(8*803^m+1)*(8*803^m-1)[/tex]
(simple manipulations followed by the difference of squares rule)

64*803^n-1 with even n also trivially has a factor of 3. (visible at [URL="http://factordb.com/search.php?query=64*803%5E%282*n%29-1"]http://factordb.com/search.php?query=64*803^%282*n%29-1[/URL])

I don't see any reason that the odd n's can be eliminated. They are extremely low-weight, though, so the lack of a prime by 25K is not surprising. ([URL="http://factordb.com/search.php?query=64*803%5E%282*n%2B1%29-1&v=n&n=1&EC=1&E=1&Prp=1&P=1&C=1&FF=1&CF=1&of=H&pp=50"]look at them here[/URL])[URL]http://factordb.com/search.php?query=8*803^(3*n)%2B1[/URL][/quote]

BTW, I factored 64*803^n-1 a little more on the DB factors page...always entertains me for a few mins. :-)

The reason why k=64 is so low weight on many different Riesel bases (if it's not eliminated fully by trivial or algebraic factors) is that it is both a perfect square and perfect cube. Both n==(0 mod 2) and n==(0 mod 3) are eliminated by algebraic factors.

Here:
Factor of 3 is n==(0 mod 2) leaves n==(1 mod 2)
Cubed algebraic is n==(0 mod 3) leaves n==(1, 5 mod 6)
Factor of 17 is n==(1 mod 4) leaves n==(7, 11 mod 12)

So...same conclusion...

No clear set of factors take out n==(7 or 11 mod 12) so the search must go on.

Interestingly...8*803^n+1 also has all n==(7 or 11 mod 12) without a clear set of factors.

mdettweiler 2010-02-16 06:28

Polished off a few of the untested k=6 Sierp. conjectures:

S559: conjectured k 6, proven, primes:
4*559^1+1

S594: conjectured k 6, proven, primes:
2*594^4+1
3*594^1+1
4*594^1+1
5*594^1+1
k=1 is a GFN.

S664: conjectured k 6, proven, primes:
3*664^1+1
4*664^1+1
k=1 is a GFN; k=2 and k=5 eliminated by trivial factors.

S699: conjectured k 6, proven, primes:
2*699^1+1
4*699^1+1

S769: conjectured k 6, proven, primes:
4*769^3+1
k=2 eliminated by trivial factors.

S804: conjectured k 6, proven, primes:
2*804^1+1
3*804^4+1
4*804^1+1
5*804^1+1
k=1 is a GFN.

S874: conjectured k 6, proven, primes:
3*874^2+1
4*874^77+1
k=1 is a GFN, k=2 and k=5 eliminated by trivial factors.

S909: conjectured k 6, proven, primes:
2*909^10+1
4*909^1+1

S979: conjectured k 6, proven, primes:
4*979^1+1
k=2 eliminated by trivial factors.

S1014: conjectured k 6, proven, primes:
2*1014^1+1
3*1014^3+1
4*1014^1+1
5*1014^3+1
k=1 is a GFN.

That should be the last of the untested Sierpinski k=6 conjectures. :smile:

rogue 2010-02-16 16:28

Riesel base 516
 
1 Attachment(s)
The primes are attached.

k=78 and k=87 do not have primes for n < 25000.

The other k have trivial factors.

I am releasing this base.

rogue 2010-02-16 20:22

I discovered that I made some errors in the bases that I had reported in the past week.

I went back through my logs and found these primes:

8*803^1243+1

95*516^726+1
120*516^647+1
121*516^531+1

I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt.

I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows...

This is what I get for breaking up the work across multiple computers. :censored:

gd_barnes 2010-02-16 20:35

[quote=rogue;205833]I went back through my logs and found this prime:

8*803^1243+1[/quote]

OK...added to the 1k remaining list.

rogue 2010-02-16 23:23

[QUOTE=rogue;205833]I discovered that I made some errors in the bases that I had reported in the past week.

I went back through my logs and found these primes:

8*803^1243+1

95*516^726+1
120*516^647+1
121*516^531+1

I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt.

I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows...

This is what I get for breaking up the work across multiple computers. :censored:[/QUOTE]

I just realized that I only need to retest for n < 10000 because I see the other remaining k in my sieve file for 10000 < n. That will save me a few days.

gd_barnes 2010-02-17 00:10

[quote=rogue;205833]I discovered that I made some errors in the bases that I had reported in the past week.

I went back through my logs and found these primes:

8*803^1243+1

95*516^726+1
120*516^647+1
121*516^531+1

I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt.

I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows...

This is what I get for breaking up the work across multiple computers. :censored:[/quote]

One thing you might consider is just running the starting bases script up to n=2500 on one core and then sieving and testing n=2500 to n=25K in whatever manner and across whatever # of cores you deem necessary. That's what I do. If you do that, you can use the starting bases primes as a starting point for adding larger primes to it and you'll avoid missing small primes (n<2500) that are on multiple cores.

Even for conjectures as large as k=100,000, it rarely takes more than a couple of days and most of the time it takes less than a day. In other words, it shouldn't be necessary to split searching for such small primes across multiple cores or machines. IMHO, there are so many small primes on most bases that it's a lot safer to run all k's at once on one core up to n=2500.

One thing that I don't do most of the time anymore since we got the new bases script up to par is check all k's that will ultimately need a prime vs. primes submitted by people to see if the k's remaining balance. Eventually I do that, usually when the base is proven or it gets down to 1 k remaining but not right away anymore. It used to be that I checked everyone on a new base up to n=2500. But with the new bases script and the multitude of new bases being run, I don't do that most of the time anymore. It takes enough time to keep the pages up to date.


Gary

rogue 2010-02-17 01:14

[QUOTE=gd_barnes;205861]One thing that I don't do most of the time anymore since we got the new bases script up to par is check all k's that will ultimately need a prime vs. primes submitted by people to see if the k's remaining balance. Eventually I do that, usually when the base is proven or it gets down to 1 k remaining but not right away anymore. It used to be that I checked everyone on a new base up to n=2500. But with the new bases script and the multitude of new bases being run, I don't do that most of the time anymore. It takes enough time to keep the pages up to date.[/QUOTE]

That is similar to what I did. I ran the script to n=100. I then merged the list of k from pl_remain.txt with the primes that I submitted. Excluding k=1, the "holes" in the list would represent the k that need a prime. I then sieve from n=100 to n=10000 (since I already did the range from 10000 to 25000 for those k). For the primes I listed earlier, I searched up to n=25000 for them when I didn't have to. There are no cases where I found a k that I hadn't searched from n=10000 to n=25000.

gd_barnes 2010-02-17 04:27

[quote=rogue;205868]That is similar to what I did. I ran the script to n=100. I then merged the list of k from pl_remain.txt with the primes that I submitted. Excluding k=1, the "holes" in the list would represent the k that need a prime. I then sieve from n=100 to n=10000 (since I already did the range from 10000 to 25000 for those k). For the primes I listed earlier, I searched up to n=25000 for them when I didn't have to. There are no cases where I found a k that I hadn't searched from n=10000 to n=25000.[/quote]

No, what I mean is: When you very first reserve a base, consider running the starting bases script to n=2500 before sieving remaining k's. Running to only n=100 leaves way too much room for error. There are still too many k's remaining and small primes to be found. IMHO, the script should always at least be used for a search to n=1000 with the possible exception of some large huge-conjectured bases (base 63 might be an example).

I promise it won't take much longer and your administrative hassle will be much less.

Heck, I've gotten to where if I know there is only going to be ~1 to 10 k's remaining on a base by n=2500, I'll run the starting bases script to n=5K or 10K without sieving. The more k's that you have to sieve, the more you have to manually account for stuff and the easier it is to forget primes hanging on a "rogue" machine somewhere (pun intended, lol). The starting bases script automates the accounting for all of the k's (with the exception of k's that have partial algebraic factors to make a full covering set) and so avoids such problems.

mdettweiler 2010-02-17 05:02

1 Attachment(s)
Sierpinski base 780: conjectured k 243, 2 k's remaining, tested to n=5K.
k=1 is a GFN; trivial factors and primes attached. Remaining:
43*780^n+1
230*780^n+1
Not reserved.

rogue 2010-02-17 12:39

[QUOTE=gd_barnes;205875]No, what I mean is: When you very first reserve a base, consider running the starting bases script to n=2500 before sieving remaining k's. Running to only n=100 leaves way too much room for error. There are still too many k's remaining and small primes to be found. IMHO, the script should always at least be used for a search to n=1000 with the possible exception of some large huge-conjectured bases (base 63 might be an example).

I promise it won't take much longer and your administrative hassle will be much less.

Heck, I've gotten to where if I know there is only going to be ~1 to 10 k's remaining on a base by n=2500, I'll run the starting bases script to n=5K or 10K without sieving. The more k's that you have to sieve, the more you have to manually account for stuff and the easier it is to forget primes hanging on a "rogue" machine somewhere (pun intended, lol). The starting bases script automates the accounting for all of the k's (with the exception of k's that have partial algebraic factors to make a full covering set) and so avoids such problems.[/QUOTE]

I see. I typically stop low because I found that starting sieving earlier save a lot of execution time. As you pointed out, the administrative hassle can be costly, as it was to me.

I have completed my retest of the k from the aforementioned conjectures (up to n = 10000). I didn't miss any other primes. :tu:

rogue 2010-02-17 12:41

Sierpinski base 520
 
1 Attachment(s)
I have completed my testing of this base to n = 25000 and am releasing it. The primes are attached.

k=1 and k=520 are GFNs, which have not been tested.

k=369 and k=373 remain.

The other k have trivial factors.

gd_barnes 2010-02-17 20:34

[quote=rogue;205897]I see. I typically stop low because I found that starting sieving earlier save a lot of execution time. As you pointed out, the administrative hassle can be costly, as it was to me.

I have completed my retest of the k from the aforementioned conjectures (up to n = 10000). I didn't miss any other primes. :tu:[/quote]

Hum. Strange. I think most of our long time searchers have found that searching to n=500 or 1000 when starting a base before sieving is the most efficient for CPU time spent. Of course it would vary somewhat depending on the size and weight of the base.

Do you set factoring to 100% with the -f switch when running the new bases script? If not, that would explain things. Without trial factoring, you can't go too far before sieving is more CPU efficient.

Along those lines, I've found for extremely high weight bases with huge contectures such as bases 3/7/15, it makes sense to only trial factor to ~30% with the -f30 switch. Searching to n=500 or 1000 generally still holds.


All times are UTC. The time now is 03:48.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.