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-   Aliquot Sequences (https://www.mersenneforum.org/forumdisplay.php?f=90)
-   -   Reserved for MF - Sequence 4788 (https://www.mersenneforum.org/showthread.php?t=11615)

schickel 2009-12-01 07:18

Just walked in the door from work to this:[code][2009-12-01 03:22:29 GMT] Factor returned by schickel@psln.com:server:v2.1a! a4788.c169 / 246867439541623785917452977412081734719211487:probable:32592798509311135674308419327488775229812399824613803365165161633979950190408523081434602797123630458924959977397735854806101:Probable finder: schickel@psln.com:Machine_1:v2.0k B1:43000000 sigma: 863597136
[/code]Not even turned into the the DB yet....

Edit: Only 161 curves!!

Batalov 2009-12-01 07:29

Shoot. I knew that c169 was 3 (mod 4), so the hopes were low, but not as low as they turned out. Still 2^5, man.

schickel 2009-12-01 07:31

Next line is an easy c125 (so far).

Ooops...make that an even easier c104....left as an exercise for the reader :grin:

Batalov 2009-12-01 07:57

The reader is running gnfs... :devil:

jrk 2009-12-01 08:37

[QUOTE=Batalov;197436]The reader is running gnfs... :devil:[/QUOTE]

Stop gnfs'ing. I clicked "quick ecm" a couple of times, before seeing that you were factoring it, and an easy p26 popped out. Sorry about that.

edit: or did you happen to submit it?

jrk 2009-12-01 08:40

[QUOTE=schickel;197431]Just walked in the door from work to this:[code][2009-12-01 03:22:29 GMT] Factor returned by schickel@psln.com:server:v2.1a! a4788.c169 / 246867439541623785917452977412081734719211487:probable:32592798509311135674308419327488775229812399824613803365165161633979950190408523081434602797123630458924959977397735854806101:Probable finder: schickel@psln.com:Machine_1:v2.0k B1:43000000 sigma: 863597136
[/code]Not even turned into the the DB yet....

Edit: Only 161 curves!![/QUOTE]

Nice. I finished 623 curves @ B1=43000000, B2=388112953420 without finding the factor.

fivemack 2009-12-01 09:12

[QUOTE=jrk;197441]Nice. I finished 623 curves @ B1=43000000, B2=388112953420 without finding the factor.[/QUOTE]

I found it twice overnight, but was asleep at the time.

Batalov 2009-12-01 09:18

[FONT=Arial Narrow]Then sigh not so,
But let them go,
And be you blithe and bonny,
Converting all your sounds of woe
Into Hey, nonny, nonny.[/FONT]

On to the c161, guys!

schickel 2009-12-01 09:22

[QUOTE=fivemack;197443]I found it twice overnight, but was asleep at the time.[/QUOTE]Ooops....sorry for stealing your thunder. :blush:

10metreh 2009-12-01 17:28

If I've done my calculations correctly, the good news is that we can't lose 2^4 and we can't pick up a 3 either on the next line.

Raman 2009-12-01 19:42

Who is regularly updating up the factor database at
[URL]http://factorization.ath.cx/search.php[/URL]

I hope that there are many people; my question is that

For an aliquot sequence or home prime sequence
will the database automatically produce the next iteration
once we completely submitted away the factors?
Will it automatically check up if the remaining cofactor
is prime or not? Will it automatically eliminate small factors
of any number or any iteration by using brute force trial
division or the Pollard's Rho algorithm? How does it work
out?

Mini-Geek 2009-12-01 20:08

[quote=Raman;197499]Who is regularly updating up the factor database at[URL="http://factorization.ath.cx/search.php"] http://factorization.ath.cx/search.php[/URL][/quote]
(if I understand your question correctly) Anyone can enter a factor. Various people do, depending on which particular sequence/work you're talking about. For Aliquot sequence 4788 (the subject of this thread), it is whoever happens to be the one who finds a factor.
(or maybe you meant...) A user named Syd is the one who runs the DB, but hasn't had much time to put towards it lately.
By the way, it is also available at a newer and easier-to-remember URL: [URL]http://factordb.com/[/URL]
[quote=Raman;197499]For an aliquot sequence or home prime sequence will the database automatically produce the next iteration once we completely submitted away the factors?[/quote]
Yes.
[quote=Raman;197499]Will it automatically check up if the remaining cofactor is prime or not?[/quote]
Yes. Composite numbers appear blue, prime numbers appear black, and PRP numbers appear dark red.
[quote=Raman;197499]Will it automatically eliminate small factors of any number or any iteration by using brute force trial division or the Pollard's Rho algorithm?[/quote]
No. To have it run GMP-ECM with increasing bounds, (good for factors up to about 20 digits), click the cofactor and click the Quick ECM button.
[quote=Raman;197499]How does it work out?[/quote]
Not sure what you mean by this, but the database is used extensively by this project and is very useful. It's also useful for other things, such as Cunningham Tables.

fivemack 2009-12-02 10:25

2000@4e7 on the C161, no factor.

Running another 2000@4e7.

jrk 2009-12-03 02:04

I've finished 1000 curves @ B1=43000000, B2=388112953420

Batalov 2009-12-03 04:01

On my Phenom, I've finished 1000 curves @ B1=110e6, B2=def
and will run some more (while I am waiting for a Lanczos job)

fivemack 2009-12-03 09:38

Second 2000@4e7 produced no factor either.

Running 4000@4e7 and GPU polysearch in parallel.

jrk 2009-12-03 09:53

So if I'm keeping count correctly, that is 5000 curves at ~4e7 and 1000 curves at 11e7 total so far.

That is roughly a t50.

[QUOTE=fivemack;197669]Running 4000@4e7 and GPU polysearch in parallel.[/QUOTE]
I am also running a GPU poly search.

Andi47 2009-12-03 10:10

[QUOTE=jrk;197672]So if I'm keeping count correctly, that is 5000 curves at ~4e7 and 1000 curves at 11e7 total so far.

That is roughly a t50.
[/QUOTE]

So maybe we need several curves at B1=11e7 to go for t55?

jrk 2009-12-03 10:15

[QUOTE=jrk;197672]I am also running a GPU poly search.[/QUOTE]
And I have adjusted my parameters, so Tom & I should be searching different spaces. I am using small A5 values this time instead of choosing large values.

fivemack 2009-12-03 10:59

Umm, I am just running 'msieve -v -np' so starting with A5=0, so I may be stepping on your toes. Can you suggest better parameters?

10metreh 2009-12-03 11:36

[QUOTE=Andi47;197678]So maybe we need several curves at B1=11e7 to go for t55?[/QUOTE]

The t55 estimate was for the c169, not the c161. I think maybe another t50 would be in order.

bsquared 2009-12-03 14:13

[QUOTE=jrk;197672]So if I'm keeping count correctly, that is 5000 curves at ~4e7 and 1000 curves at 11e7 total so far.

[/QUOTE]

Add another 800 @ 11e7; done last night.

jrk 2009-12-03 16:53

[QUOTE=fivemack;197676]Umm, I am just running 'msieve -v -np' so starting with A5=0, so I may be stepping on your toes. Can you suggest better parameters?[/QUOTE]
You are OK. Please continue. I meant that I changed msieve's parameters for selecting the rational coefficient, so it should not choose the same as a normal msieve.

Although now I am not sure that very small A5's are much better...

jrk 2009-12-03 17:31

+150 curves @ B1=110000000, B2=1589211473866

henryzz 2009-12-03 18:47

i think based on jasonp's recent posts in the msieve gpu thread that there is sufficient randomness in the selected search space that we can search the same A5 values without much loss of efficiency
i would check with him first though

jrk 2009-12-03 23:32

[QUOTE=henryzz;197697]i think based on jasonp's recent posts in the msieve gpu thread that there is sufficient randomness in the selected search space that we can search the same A5 values without much loss of efficiency
i would check with him first though[/QUOTE]
No, it isn't random. Msieve doesn't do what Jason was suggesting. If you want to run msieve, make sure you start with a higher A5 than where Tom ends. Something above 100K should be high enough.

So far I have searched through to 6K, but with a higher rational coefficient which does not overlap the normal msieve selection. I'm not entirely convinced that this is worth doing yet, though. But I have found some good polys.

jrk 2009-12-03 23:33

Tom, what do you suggest is a good target score for this number?

So far my best is a 1.035e-12.

jrk 2009-12-03 23:41

Once the ECM is finished it might be worth doing some poly searching with pol5 as well.

fivemack 2009-12-04 12:59

Looks as if 1e-12 is a pretty reasonable score, I've got a couple above that level after a day of polsel, my best is

[code]
# norm 9.159378e-16 alpha -6.373628 e 1.028e-12
skew: 15249475.33
c0: 5865211302555522024730119243238438762464
c1: 2222587031863375855063420818065564
c2: 82026281057189973786655296
c3: -9332835943493042911
c4: -2896035406838
c5: 6960
Y0: -20007761313361394254221339534949
Y1: 350230344983122741
[/code]

with (14e, 30-bit lpa/lpr, alim=rlim=50e6)

[code]
total yield: 1737, q=50001037 (0.14975 sec/rel)
[/code]

I would go with the best polynomial that we've found by the time people are satisfied there's been enough ECM; not sure it's worth also doing pol51m.

This is a six-CPU-month sieving job, we'll probably get it done inside two weeks, so it's not worth delaying the start of sieving by two days even to find a polynomial that makes the sieving 10% quicker.

Andi47 2009-12-04 15:31

[QUOTE=fivemack;197768]Looks as if 1e-12 is a pretty reasonable score, I've got a couple above that level after a day of polsel, my best is

[code]
# norm 9.159378e-16 alpha -6.373628 e 1.028e-12
skew: 15249475.33
c0: 5865211302555522024730119243238438762464
c1: 2222587031863375855063420818065564
c2: 82026281057189973786655296
c3: -9332835943493042911
c4: -2896035406838
c5: 6960
Y0: -20007761313361394254221339534949
Y1: 350230344983122741
[/code]

with (14e, 30-bit lpa/lpr, alim=rlim=50e6)
[/QUOTE]

I guess we'll need ~100M relations?

Edit: your [URL="http://www.mersenneforum.org/showpost.php?p=155821&postcount=77"]formula[/URL] gives a "good score" as 1.276e-12.

10metreh 2009-12-04 15:52

c161 is near the crossover between 14e and 15e. Which is better with the current best poly?

Batalov 2009-12-04 18:32

Another 1200 @ 110e6 are done, so fire at will.

jrk 2009-12-04 18:42

[QUOTE=Batalov;197789]Another 1200 @ 110e6 are done, so fire at will.[/QUOTE]
Don't fire yet, I may have a better poly from the overnight search. I'll post again in a bit.

jrk 2009-12-04 18:57

[QUOTE=jrk;197790]Don't fire yet, I may have a better poly from the overnight search. I'll post again in a bit.[/QUOTE]
I do.

[code]n: 22315086009543699881696673140311700365075401903410041991888847307663799610560235574994125636703013521384678857814894314475474921423859163748271799984152025524899
# norm 1.050714e-15 alpha -7.337146 e 1.121e-12
skew: 25520528.70
c0: 31410679515197091100341643412427401243420
c1: 4180601981543089138092261003493340
c2: -2347505003114131780984812301
c3: -51235139041477161474
c4: 3624790086036
c5: 7560
Y0: -19679511428703587012905042402449
Y1: 492226961856379171
rlim: 50000000
alim: 50000000
lpbr: 30
lpba: 30
mfbr: 60
mfba: 60
rlambda: 2.6
alambda: 2.6[/code]

[QUOTE=10metreh;197778]c161 is near the crossover between 14e and 15e. Which is better with the current best poly?[/QUOTE]
14e is faster.

jrk 2009-12-04 19:25

[QUOTE=jrk;197694]+150 curves @ B1=110000000, B2=1589211473866[/QUOTE]
I completed another 240 curves @ those limits. Still no factor.

The total curves done are about 1.8 * t50.

Andi47 2009-12-04 19:37

I have started an overnight run of p-1 (B1=1e10, B2=1e16, resuming from my yesterday's 1e10/1e15 run) a few hours ago, I guess it will finish around midnight (CET)

jrk 2009-12-04 19:58

[QUOTE=fivemack;197768]with (14e, 30-bit lpa/lpr, alim=rlim=50e6)[/QUOTE]
I checked 29 vs 30 bit, 14e vs 15e and alim=rlim=25e6 vs 50e6 vs 75e6.

30 bit, 14e and alim=rlim=50e6 indeed appears to be the winner.

bsquared 2009-12-04 20:07

[QUOTE=jrk;197792]I do.

[code]n: 22315086009543699881696673140311700365075401903410041991888847307663799610560235574994125636703013521384678857814894314475474921423859163748271799984152025524899
# norm 1.050714e-15 alpha -7.337146 e 1.121e-12
skew: 25520528.70
c0: 31410679515197091100341643412427401243420
c1: 4180601981543089138092261003493340
c2: -2347505003114131780984812301
c3: -51235139041477161474
c4: 3624790086036
c5: 7560
Y0: -19679511428703587012905042402449
Y1: 492226961856379171
rlim: 50000000
alim: 50000000
lpbr: 30
lpba: 30
mfbr: 60
mfba: 60
rlambda: 2.6
alambda: 2.6[/code]


14e is faster.[/QUOTE]

Is everyone done poly searching then? I'll start on 10M-20M if so. Andi_47, I'll happily kill the jobs if you turn up a factor.

jrk 2009-12-04 20:16

[QUOTE=bsquared;197802]Is everyone done poly searching then? I'll start on 10M-20M if so. Andi_47, I'll happily kill the jobs if you turn up a factor.[/QUOTE]

I have stopped poly searching, but not sure if Tom has.

Andi47 2009-12-04 21:26

[QUOTE=bsquared;197802]Andi_47, I'll happily kill the jobs if you turn up a factor.[/QUOTE]

It went a bit faster than I thought; no factor to B1=1e10, B2=1e16. I will extend to B2=3e16, ETA ~10:30 a.m. (CET) tomorrow. (i.e. ~12 hours from now)

fivemack 2009-12-04 21:28

I've stopped searching, and interrupted my ECM after about 3500@4e7; will take 20M-40M for the sieving.

10metreh 2009-12-05 07:22

[QUOTE=fivemack;197816]I've stopped searching, and interrupted my ECM after about 3500@4e7; will take 20M-40M for the sieving.[/QUOTE]

Tom, what range of Q should we sieve?

jrk 2009-12-05 07:39

[QUOTE=10metreh;197875]Tom, what range of Q should we sieve?[/QUOTE]

I am not Tom, but,

It will take as much as 50e6 or 60e6 of Q (so since bsquared started at 10e6, end at 60e6 or 70e6) but my sieving test was limited so I may be a bit off.

Andi47 2009-12-05 09:55

[QUOTE=Andi47;197815]It went a bit faster than I thought; no factor to B1=1e10, B2=1e16. I will extend to B2=3e16, ETA ~10:30 a.m. (CET) tomorrow. (i.e. ~12 hours from now)[/QUOTE]

done, no factor.

fivemack 2009-12-05 10:39

Yield looks a bit under 2 relations per Q; we need a bit over 100 million relations, so say 10M - 70M for the range.

Andi47 2009-12-18 17:12

After these factors...

[QUOTE=fivemack;199218]

[code]
Sat Dec 12 11:42:54 2009 commencing Lanczos iteration (4 threads)
Sat Dec 12 11:42:54 2009 memory use: 2078.2 MB
Wed Dec 16 16:41:13 2009 lanczos halted after 105593 iterations (dim = 6676993)
Wed Dec 16 16:41:27 2009 recovered 26 nontrivial dependencies
Wed Dec 16 16:41:29 2009 BLanczosTime: 364766
Wed Dec 16 16:41:29 2009
Wed Dec 16 16:41:29 2009 commencing square root phase
Wed Dec 16 16:41:29 2009 reading relations for dependency 1
Wed Dec 16 16:41:30 2009 read 3336056 cycles
Wed Dec 16 16:41:39 2009 cycles contain 9688282 unique relations
Wed Dec 16 16:45:43 2009 read 9688282 relations
Wed Dec 16 16:47:15 2009 multiplying 9688282 relations
Wed Dec 16 17:19:03 2009 multiply complete, coefficients have about 510.55 million bits
Wed Dec 16 17:19:11 2009 initial square root is modulo 1448310641
Wed Dec 16 18:24:08 2009 sqrtTime: 6159
Wed Dec 16 18:24:08 2009 prp67 factor: 3080910905336815970632301939361370294511667581515422926255876218391
Wed Dec 16 18:24:08 2009 prp94 factor: 7243015684383815871382996543105076138574662378185857800256367590422019476790807257932033039189
Wed Dec 16 18:24:08 2009 elapsed time 104:53:07
[/code][/QUOTE]

we now have a c141 in iteration 2487.

150 curves at B1=3e6 and counting...

henryzz 2009-12-18 17:21

i have also done 100 @3e6

fivemack 2009-12-18 17:28

I've set 20,000 @ 1e7 going, should have some kind of result by Monday.

fivemack 2009-12-19 14:43

ECM not looking terribly promising; I'll do the GNFS for this one, it'll easily be done by Christmas.

Andi47 2009-12-19 14:44

[QUOTE=fivemack;199290]ECM not looking terribly promising; I'll do the GNFS for this one, it'll easily be done by Christmas.[/QUOTE]

Have you done P-1? I would suggest B1=1e9, B2=1e15 before GNFSing (~ 1-2 hours on a c2d~2GHz, 64 bit-system; make sure to specify the -maxmem flag according to your free RAM.)

henryzz 2009-12-19 15:15

[quote=Andi47;199291]Have you done P-1? I would suggest B1=1e9, B2=1e15 before GNFSing (~ 1-2 hours on a c2d~2GHz, 64 bit-system; make sure to specify the -maxmem flag according to your free RAM.)[/quote]
if no-one has done it i would like to do it
cant do that and p+1 quickly though

mdettweiler 2009-12-19 15:30

[quote=henryzz;199294]if no-one has done it i would like to do it
cant do that and p+1 quickly though[/quote]
What P+1 bounds would be best? I could do that if nobody else wants to.

Andi47 2009-12-19 15:40

[QUOTE=mdettweiler;199295]What P+1 bounds would be best? I could do that if nobody else wants to.[/QUOTE]

Depends on your RAM and your system speed. 3 runs at B1=1e9, B2=1e15 should be doable overnight on a C2D, 2 GHz, 2 GB RAM (64 bit). If you have less than 2 GB RAM, better B2=5e13.

(if you want to spend just an hour or two, do B1=1e8, B2=1e13)

Edit: I would have done it if I had a CPU available (i.e. not busy with something else) *now*. I had planned to do p+/-1 tomorrow (GMT+1), but as GNFS is about to start, it should be better run today.

mdettweiler 2009-12-19 15:54

[quote=Andi47;199297]Depends on your RAM and your system speed. 3 runs at B1=1e9, B2=1e15 should be doable overnight on a C2D, 2 GHz, 2 GB RAM (64 bit). If you have less than 2 GB RAM, better B2=5e13.

(if you want to spend just an hour or two, do B1=1e8, B2=1e13)

Edit: I would have done it if I had a CPU available (i.e. not busy with something else) *now*. I had planned to do p+/-1 tomorrow (GMT+1), but as GNFS is about to start, it should be better run today.[/quote]
Okay, I've started a run with 3 curves at B1=1e9, B2=5e13. The machine running it is a C2D @ 2.2Ghz with 1.2GB of RAM allocated (I have 2GB, but want to leave a bit of headroom). There's a bit of thermal throttling in play so it might take a little longer than normal, but I expect it will be done by tomorrow morning (EST).

henryzz 2009-12-19 16:16

i got bored and started the p-1 anyway
it is finished and only took 20 minutes
would it help if i took 1 of the p+1s off you medettweiler? i should be able to do it quite quickly

mdettweiler 2009-12-19 21:58

[quote=henryzz;199304]i got bored and started the p-1 anyway
it is finished and only took 20 minutes
would it help if i took 1 of the p+1s off you medettweiler? i should be able to do it quite quickly[/quote]
Go ahead and do all the P+1's if you like. :smile: It seems there was a problem when I tried to do it:
[code]$ echo "59066643232529177994432826503455645911804472822311052434786614015983698
8457920218063108883526241212940105498096765045722201184707349955127817" | ./ecm
.exe 1e9 5e13 -c 3 -pp1 -maxmem 1228 | tee -a pp1-4788.txt
GMP-ECM 6.2.3 [powered by GMP 4.2.1_MPIR_1.1.1] [ECM]
Input number is 5906664323252917799443282650345564591180447282231105243478661401
59836988457920218063108883526241212940105498096765045722201184707349955127817 (1
41 digits)
Using B1=1000000000, B2=51984971408388, polynomial Dickson(30), sigma=2038232620

Step 1 took 14294375ms
GNU MP: Cannot allocate memory (size=64)

This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's support team for more information.[/code]
The machine has about 1400-1600 MB of free memory, on average, according to Task Manager, so I figured 1.2GB should be quite safe. It would seem that either that was too much, or there was something else that caused it to have a problem allocating memory. Does anyone know what might have caused this?

Batalov 2009-12-20 00:42

Segmentation of memory, probably? (process needed a contiguous chunk?)
People get the same symptoms while filtering a large NFS job on a Windows machine: memory seems to be enough, but sooner or later an allocation fails.

Andi47 2009-12-20 07:07

[QUOTE=mdettweiler;199322]Go ahead and do all the P+1's if you like. :smile: It seems there was a problem when I tried to do it:
[code]$ echo "59066643232529177994432826503455645911804472822311052434786614015983698
8457920218063108883526241212940105498096765045722201184707349955127817" | ./ecm
.exe 1e9 5e13 -c 3 -pp1 -maxmem 1228 | tee -a pp1-4788.txt
[/QUOTE]

What does the "| tee -a pp1-4788.txt" flag do? directing the output to the .txt file?

jrk 2009-12-20 07:46

[QUOTE=Andi47;199363]What does the "| tee -a pp1-4788.txt" flag do? directing the output to the .txt file?[/QUOTE]

It is not a flag. The "|" is a Unix pipe, which takes the output of one command and directs it as the input to another command. In this case the ecm output is sent to the command "tee" rather than being written on the console.

[quote="man 1 tee"]NAME
tee - read from standard input and write to standard output and files

SYNOPSIS
tee [OPTION]... [FILE]...

DESCRIPTION
Copy standard input to each FILE, and also to standard output.
[/quote]
So you can see what is happening as well as record it to a file at the same time. Very useful command. The -a option in tee is for append mode, so the file is not clobbered if it exists.

Andi47 2009-12-20 08:33

[QUOTE=jrk;199364]It is not a flag. The "|" is a Unix pipe, which takes the output of one command and directs it as the input to another command. In this case the ecm output is sent to the command "tee" rather than being written on the console.


So you can see what is happening as well as record it to a file at the same time. Very useful command. The -a option in tee is for append mode, so the file is not clobbered if it exists.[/QUOTE]

Thanks. (I used to do the same with ecm <...some flags...> [B]>>outputfile.txt[/B] (without a pipe; two ">"s indicating append mode))

Edit: my GNFS-run on a c121 from alq10212 is now finished, so I can take the p+1 of alq4788.2487

henryzz 2009-12-20 09:42

brilliant i have always wanted the functionality of tee:party:

Mini-Geek 2009-12-20 13:39

[quote=henryzz;199368]brilliant i have always wanted the functionality of tee[/quote]
Actually, >> doesn't have the same functionality as tee. tee writes to standard output as well as a file, >> only writes to the file. If you download cygwin and make sure that its bin folder is in your system's PATH variable, you can use tee, along with many other *nix apps. You can even use the exact line given on Windows, (the one beginning with an echo and piping through ecm and tee) minus the "./" before "ecm.exe". :smile:
Edit: Or was I reading that wrong, and you meant "I have always wanted functionality that I now know tee has, and can now use tee", as opposed to "I have known about the functionality of tee and now can use >> which has the same functionality" (as I assumed it meant)?

Andi47 2009-12-20 14:16

[QUOTE=Andi47;199365]
Edit: my GNFS-run on a c121 from alq10212 is now finished, so I can take the p+1 of alq4788.2487[/QUOTE]

done (3* B1=1e9, B2=1e14), no factor.

henryzz 2009-12-20 19:01

[quote=Mini-Geek;199382]Edit: Or was I reading that wrong, and you meant "I have always wanted functionality that I now know tee has, and can now use tee", as opposed to "I have known about the functionality of tee and now can use >> which has the same functionality" (as I assumed it meant)?[/quote]
your edit was correct:smile:

fivemack 2009-12-22 07:55

prp62 factor: 20516414210656347188393718767755485866219379699286605968224717
prp80 factor: 28789944785697304394120313244130116901674378646444316426898886893821870292864301

(381 CPU-hours on core2/2400 and K10/2500 quads to sieve, 9 wallclock-hours linalg on the K10/2500x4; small primes up to 2^24, large primes to 2^28, siever 13e, sieved 3*2^20 .. 22*2^20)

but factorization.ath.cx appears to be down. [color=blue][b]10metreh:[/b] Not anymore. Here we go...[/color]

10metreh 2009-12-22 08:17

Next line has a c165. ECM in progress.

10metreh 2009-12-22 09:50

Someone must be running the sequence further - it is now on a c136 at line 2490.

fivemack 2009-12-22 10:30

It's probably me. 8000@1e7 running, and a polsel.

fivemack 2009-12-22 11:54

I'll do the GNFS for this one; it'll be over by Christmas. Probably even by Christmas Eve.

fivemack 2009-12-23 23:54

sieved 2M-16.5M with
[code]
# norm 4.431184e-13 alpha -7.190056 e 3.778e-11
skew: 1381889.42
c0: 13935895326125594374428769951334655
c1: 6832990194435681055086947697
c2: -76268592758638424003951
c3: -29988262496226737
c4: 49802168836
c5: 660
Y0: -353206674018659220179872166
Y1: 561766051832977
n: 3628612227069559920455070746827882655714694792773312751322418651421798373899250166140599596858424814418370603853288254380696130063491713
lpbr: 27
lpba: 27
mfbr: 54
mfba: 54
alambda: 2.5
rlambda: 2.5
alim: 10000000
rlim: 10000000
[/code]

and should probably have used 28-bit lp to get better yield (IE this run was a bit slower than a slightly larger number that I did a year ago with 28/13)

[code]
Wed Dec 23 17:58:15 2009 found 3500504 hash collisions in 15421358 relations
Wed Dec 23 17:59:09 2009 found 3639407 duplicates and 11845280 unique relations
Wed Dec 23 18:09:52 2009 matrix is 1626466 x 1626692 (470.4 MB) with weight 121859512 (74.91/col)
Wed Dec 23 18:09:52 2009 sparse part has weight 107055126 (65.81/col)
Wed Dec 23 18:09:52 2009 matrix includes 64 packed rows
Wed Dec 23 21:55:17 2009 BLanczosTime: 13774
Wed Dec 23 22:12:06 2009 sqrtTime: 1009
Wed Dec 23 22:12:06 2009 prp57 factor: 615246762379545008872051386151610963316901978007615471813
Wed Dec 23 22:12:06 2009 prp79 factor: 5897816045444012068057492946558964042767474423553931349916095671773447719452301
[/code]

running ecm on C125 of step 2493 now

fivemack 2009-12-24 01:10

2493 down, set 8000@1e7 going on C136 of 2494 (factor 519158260471434103953487553 divides C163, factorization.ath.cx not presently responding)

fivemack 2009-12-24 10:19

C121 of 2495 has had 200@1e7, I'll be giving it some more but probably it's GNFS-worthy. Someone else can do this one.

10metreh 2009-12-24 10:32

Some bad news: there is a 1 in 2 chance that 4788 will pick up a 3 on the next line (assuming the c121 splits into 2 primes).

Andi47 2009-12-24 11:13

[QUOTE=fivemack;199768]C121 of 2495 has had 200@1e7, I'll be giving it some more but probably it's GNFS-worthy. Someone else can do this one.[/QUOTE]

Can you please post the c121? The DB seems to be down...

R. Gerbicz 2009-12-24 11:59

[QUOTE=Andi47;199774]Can you please post the c121? The DB seems to be down...[/QUOTE]

Here it is:
[code]
3942887076714143202481073805784961368619711215403426727623016228578779996971144714931046107405659926915911799176550312767
[/code]

10metreh 2009-12-24 12:05

[QUOTE=Andi47;199774]Can you please post the c121? The DB seems to be down...[/QUOTE]

It's up now.

Andi47 2009-12-24 12:31

[QUOTE=R. Gerbicz;199776]Here it is:
[code]
3942887076714143202481073805784961368619711215403426727623016228578779996971144714931046107405659926915911799176550312767
[/code][/QUOTE]

Thanks. I will do a p-1 followed by GNFS

edit: started p-1 and polsel in parallel

Raman 2009-12-24 13:16

[quote=10metreh;199770]
Some bad news: there is a 1 in 2 chance that 4788 will pick up a 3 on the next line (assuming the c121 splits into 2 primes).[/quote]

Just I analyzed about that, simply

c171 = 2[sup]4[/sup] . 3559. 220681 . 592772569 . 7547146554874360724158181221231 . c121

s(c171) = (1+2+4+8+16) (1+3559) (1+220681) (1+592772569) (1+7547146554874360724158181221231) (1+p) (1+q) - c171

c171 = 1 mod 3, all the factors 3559, 220681, 592772569, 7547146554874360724158181221231 are 1 mod 3, 31 is 1 mod 3.

c121 is 1 mod 3. If it splits up into 2 factors, it can split up as
(1 mod 3 × 1 mod 3) or (2 mod 3 × 2 mod 3). Each has equally 50% chance.

When it splits up as (2 mod 3 × 2 mod 3) then (1+p) will be divisible by 3, then sigma(c171) - c171 will not be divisible by 3. When it splits up into (1 mod 3 × 1 mod 3), then
s(c171) = (1 . 2[sup]6[/sup] - 1) mod 3 = 1 - 1 = 0 mod 3, then it will pick up the factor 3.

Suppose that it splits up into 3 factors, then a factor of 2 mod 3, will not allow to pick up a 3 as stated above. Even if all the 3 factors are 1 mod 3, then s(c171) = (1 . 2[sup]7[/sup] - 1) mod 3 = 2 - 1 = 1 mod 3, then it will not pick up a 3 anywhere at all.

To come up to a conclusion, the next iteration [B]will[/B] acquire up the factor 3 if and only if
the [B]c121 splits up into two (1 mod 3) factors[/B].
Does [B]not[/B] pick up a 3, when the [B]c121 splits up into two (2 mod 3) factors[/B].

Right, man?

10metreh 2009-12-24 13:29

[QUOTE=Raman;199785]To come up to a conclusion, the next iteration [B]will[/B] acquire up the factor 3 if and only if
the [B]c121 splits up into two (1 mod 3) factors[/B].
Does [B]not[/B] pick up a 3, when the [B]c121 splits up into two (2 mod 3) factors[/B].

Right, man?[/QUOTE]

Correct. :smile:

Raman 2009-12-24 13:51

[quote=10metreh;199787]Correct. :smile:[/quote]

Meanwhile, pray up to God that it splits up into two (2 mod 3) factors :smile:

I wonder whether you check up the residue class of each factor of each iteration mod 3, 5, 7...?

Once it acquires up a factor of 3, in order to lose it up:
s(c) = product of (1+p)'s - c.
(1+p) is replaced up by (1+p+p[sup]2[/sup]+...) for the prime powers...

Since c is divisible by 3, in order to lose the 3, the products of (1+p)'s should be 1 or 2 mod 3. The factor of 3 will not stimulate again a 3 in the next line at all, since (1+3+9+...) is always equal to 1 mod 3.

Any of the others should not be divisible by 3 at all.
A prime of 2 mod 3, will induce up a 3 in the next line by using (1+p)
A prime of 1 mod 3, will not.
However, that all of the primes other than 3, cannot be equal to 0 mod 3, anyway.

For the prime powers, power of 2 mod 3 will trigger up a 3 if the power is odd. Power of 1 mod 3 will trigger 3, if the power is even.

I see that it can lose up the 3 within the next iteration, if for all the primes besides 3, [B]the power of 2 mod 3 primes are all even, and then the power of 1 mod 3 primes are all odd[/B].
Or, in fact it is true that it can lose up the 3 within the subsequent iterations by means of using the factor of 9 = 3[sup]2[/sup]

All these are natural processes only, by itself. We can't do anything to change sequences, or write up a random number of our own. We just simply compute the results, and then go on processing with the single correct number of the next iteration.

Andi47 2009-12-24 21:37

[QUOTE=Andi47;199781]Thanks. I will do a p-1 followed by GNFS

edit: started p-1 and polsel in parallel[/QUOTE]

p-1 to B1=1e9, B2=1e15 and 3* p+1 to B1=1e9, B2=1e15: no factor. Starting GNFS...

Andi47 2009-12-25 21:01

c121 factored
 
[CODE]Fri Dec 25 20:39:10 2009
Fri Dec 25 20:39:10 2009
Fri Dec 25 20:39:10 2009 Msieve v. 1.43
Fri Dec 25 20:39:10 2009 random seeds: e4aca9c8 a1f403b9
Fri Dec 25 20:39:10 2009 factoring 3942887076714143202481073805784961368619711215403426727623016228578779996971144714931046107405659926915911799176550312767 (121 digits)
Fri Dec 25 20:39:12 2009 no P-1/P+1/ECM available, skipping
Fri Dec 25 20:39:12 2009 commencing number field sieve (121-digit input)
Fri Dec 25 20:39:12 2009 R0: -220525782874300519278221
Fri Dec 25 20:39:12 2009 R1: 10561525245991
Fri Dec 25 20:39:12 2009 A0: 13919992851438837656957797344
Fri Dec 25 20:39:12 2009 A1: -7900337548465327330903242
Fri Dec 25 20:39:12 2009 A2: 14318233152048859923
Fri Dec 25 20:39:12 2009 A3: -542592852133532
Fri Dec 25 20:39:12 2009 A4: -1569762856
Fri Dec 25 20:39:12 2009 A5: 7560
Fri Dec 25 20:39:12 2009 skew 191636.60, size 1.651620e-11, alpha -6.880605, combined = 2.901254e-10
Fri Dec 25 20:39:12 2009
Fri Dec 25 20:39:12 2009 commencing relation filtering
Fri Dec 25 20:39:12 2009 estimated available RAM is 1985.8 MB
Fri Dec 25 20:39:12 2009 commencing duplicate removal, pass 1
Fri Dec 25 20:39:28 2009 error -15 reading relation 1477379
Fri Dec 25 20:39:48 2009 error -15 reading relation 3405784
Fri Dec 25 20:40:45 2009 found 1083950 hash collisions in 8438938 relations
Fri Dec 25 20:41:14 2009 added 58146 free relations
Fri Dec 25 20:41:14 2009 commencing duplicate removal, pass 2
Fri Dec 25 20:41:20 2009 found 771129 duplicates and 7725955 unique relations

<snip>

Fri Dec 25 20:46:42 2009 matrix is 636031 x 636258 (183.2 MB) with weight 47730814 (75.02/col)
Fri Dec 25 20:46:42 2009 sparse part has weight 41667595 (65.49/col)
Fri Dec 25 20:46:42 2009 matrix includes 64 packed rows
Fri Dec 25 20:46:42 2009 using block size 65536 for processor cache size 4096 kB
Fri Dec 25 20:46:46 2009 commencing Lanczos iteration (2 threads)
Fri Dec 25 20:46:46 2009 memory use: 181.3 MB
Fri Dec 25 20:46:54 2009 linear algebra at 0.2%, ETA 0h55m
Fri Dec 25 21:46:41 2009 lanczos halted after 10061 iterations (dim = 636028)
Fri Dec 25 21:46:43 2009 recovered 27 nontrivial dependencies
Fri Dec 25 21:46:43 2009 BLanczosTime: 3718
Fri Dec 25 21:46:43 2009
Fri Dec 25 21:46:43 2009 commencing square root phase
Fri Dec 25 21:46:43 2009 reading relations for dependency 1
Fri Dec 25 21:46:43 2009 read 317038 cycles
Fri Dec 25 21:46:44 2009 cycles contain 1023738 unique relations
Fri Dec 25 21:46:58 2009 read 1023738 relations
Fri Dec 25 21:47:06 2009 multiplying 1023738 relations
Fri Dec 25 21:50:41 2009 multiply complete, coefficients have about 44.49 million bits
Fri Dec 25 21:50:42 2009 initial square root is modulo 2443927
Fri Dec 25 21:57:45 2009 sqrtTime: 662
[B]Fri Dec 25 21:57:45 2009 prp49 factor: 1614394407529950231948689326128270361414761568603
Fri Dec 25 21:57:45 2009 prp73 factor: 2442331971867286584011502653002290013163478345321068635697408534391895789
[/B]Fri Dec 25 21:57:45 2009 elapsed time 01:18:35[/CODE]

I can't post the factors to the DB right now - DB seems to be down.

Edit: If I see correctly, the factors are 1 mod 3. :ouch2:

10metreh 2009-12-25 21:34

The 3 has disappeared again! :party:

Andi47 2009-12-25 21:37

DB is up again, and I see that someone seems to have clicked Quick-ECM - we now have a c161, [B]and the sequence got rid of the 3 *very* quick[/B]

[CODE]2495. 221665515240712835146632915038654332249677600521866315490705901031882614988052048857984072398217318255693431659468618236169689657416113688944691446324158756073966101682032 = 2^4 * 3559 * 220681 * 592772569 * 7547146554874360724158181221231 * 1614394407529950231948689326128270361414761568603 * 2442331971867286584011502653002290013163478345321068635697408534391895789
2496. 207934041439159666402719289421731533248712421461988300495417637950528851328183231602527567551285460133911968968270227215739684980733388937486979959359115607573614923085968 = 2^4 * 3^3 * 317 * 757 * 50924647 * 3088726995612795721 * 12752024987120885573855872723202403795026326832056826089540602758711687497126775191275849808610417125744238645554854915822786087517285333
2497. 391587407602724606683481473757692838731899590545814526536499415906443230131930260643701621866514302532389288790656571729396302031661198446031414125384098925446994919996272 = 2^4 * 3 * 83257 * 97986607633273229949504114208037772282385562811989413937291412910837134348045374724172747703525004537253446354524887729509306032240832213815708720534033906420029477
2498. 620025545710327153129359405293175458009270700840661687036785632653736058180215403645660032042696216110178993345887532990955937370744895402744252179739502852162138050639904 = 2^5 * 61 * 10404019
[/CODE]

Edit: 10metreh was faster.

R. Gerbicz 2009-12-25 21:56

now only c134:
[code]
Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2243524066
Step 1 took 18205ms
Step 2 took 12262ms
********** Factor found in step 2: 1053510145349455459655189063
Found probable prime factor of 28 digits: 1053510145349455459655189063
Composite cofactor 2897943480588228707823873533626392259113337697364013323014773796420074066675329471987609206871621029590
2512926053853268268773796067441 has 134 digits
[/code]

R. Gerbicz 2009-12-25 22:05

Killed.
[code]
Run 45 out of 948:
Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2841086046
Step 1 took 13931ms
Step 2 took 9625ms
********** Factor found in step 2: 275672488609181335964910240943
Found probable prime factor of 30 digits: 275672488609181335964910240943
Probable prime cofactor 10512269451365601486564964719249900397856401577901033520503299884146118828035871491891733785832743
0257887 has 105 digits[/code]

Mini-Geek 2009-12-25 22:10

Index 2501 has a c166. No idea how much ECM so far, but I've clicked two Quick ECMs.

R. Gerbicz 2009-12-25 22:15

Factored.
[code]
Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=843301289
Step 1 took 18580ms
Step 2 took 12230ms
********** Factor found in step 2: 21576911035234107060567359
Found probable prime factor of 26 digits: 21576911035234107060567359
Probable prime cofactor 27324997159006699095076661613314836247165012351115938582637324350291992335079286058776193292444980
2628020051974483243245518676185344336490951 has 141 digits[/code]

R. Gerbicz 2009-12-25 22:31

Another easy goal on line 2502:
[code]
Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2863442735
Step 1 took 15163ms
Step 2 took 10280ms
********** Factor found in step 2: 62867187091895665602431325293
Found probable prime factor of 29 digits: 62867187091895665602431325293
Probable prime cofactor 10388219128879144885656371803679068375540698889290568480135606623289797480435783087640389370988520
407960455115608852781 has 119 digits[/code]

R. Gerbicz 2009-12-25 22:51

now c137 on line 2503:
[code]
Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=1875955981
Step 1 took 16895ms
********** Factor found in step 1: 65494328231336658101893937
Found probable prime factor of 26 digits: 65494328231336658101893937
Composite cofactor 4074988117659100131366888173846118335910309207028358060447648386494072255970799922316606641147156490827
0442953165453791845420594628304571 has 137 digits
[/code]

R. Gerbicz 2009-12-26 00:52

Finished B1=1e6, 948 curves (on three cores) for the c137, so t35 level. Here I stopped.

10metreh 2009-12-26 08:29

BTW: the most recent loss of the 3 is just about higher than the old one at line 2471, so it is a new record for the highest sequence to become stable.

Andi47 2009-12-26 13:44

ECM anyone?

I have done ~150@11e6 so far and queued some more, but with my smallish ressources (compared to others here) it will take several days to finish something like 5000@11e6...

henryzz 2009-12-26 14:27

lots of progress:smile:
i will join you with ecm
starting some curves at 43e6

Andi47 2009-12-26 15:43

p-1 and 3* p+1 (B1=1e9, B2=1e15 for all of these) done, no factor found.

Andi47 2009-12-27 05:36

~1100@11e6, no factor

jrk 2009-12-27 07:55

+80 @ B1=11e7, B2=1589211473866. no factor


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