Reserving:
26^169+1
30^149-1
88^113-1

Chris

I've nearly finished those 3 so reserving:
29^151+1
54^127+1
79^116+1

Chris

Reserving:

98^113+1 c194

Reserving:
39^139+1
42^136+1
45^134+1

Also 2 GNFS targets I've put in the polynomial request thread:
26^166+1
59^136+1

Chris

Reserving:

98^113-1 c211

Hi there,

If it's OK I would like to contribute some factors.
If nobody has any objection, I'll do the C173 cofactor of [URL="http://www.factordb.com/index.php?id=1100000000007943119"]56[SUP]131[/SUP]-1[/URL]
by SNFS (I suppose it has been ECM-ed enough).

I don't know for sure how much ECM has been run, but none of the numbers I've factored have been ECM misses. So it's almost certainly had plenty of ECM.

And welcome to the project.

Chris

[QUOTE=chris2be8;358650]
And welcome to the project.

Chris[/QUOTE]

Thanks;
I used [URL="http://myfactors.mooo.com/"]myfactors.mooo.com[/URL] to get the polynomial:

[CODE]
# MurphyE Score: 1.033e-12 anorm: 1.126e46 rnorm: 2.420e45. (Since anorm is larger, sieve on the algebraic side. ie. "-a")
# SNFS difficulty is 229.858 which is approximately equivalent to GNFS difficulty 158. (Since n has 173 digits, it's recommended to use SNFS.)
# (some extra msieve library info) size: 1.923e-11 alpha: 1.653 rroots: 2
n: 58858763449145861682189209763193295417732963791373596736630811521069708557553014101459353660615483936299841727914109825712549670400384577919658321433057164682108047938754697
skew: 1.25118
type: snfs
c6: 8
c0: -7
Y1: 1
Y0: -144246936514426199369626914514553274368
m: 144246936514426199369626914514553274368
rlambda: 2.6
alambda: 2.6
lpbr: 29
lpba: 29
mfbr: 58
mfba: 58
alim: 38400000
rlim: 38400000
qintsize: 160000
[/CODE]

Since it is a rather big factorization for the limited resources I have,
I would like to run the sieving "by hand" on several boxes instead of
running factmsieve.py on only one box.
The question I have is how to compute the starting value of q?
I took a look at factmsieve.py from which I deduced that
q0=3840000/2=19200000 for the parameters above but I have seen
posts on this forum where people use q0 = alim / 3.

[QUOTE=YuL;358957]Thanks;
I used [URL="http://myfactors.mooo.com/"]myfactors.mooo.com[/URL] to get the polynomial:

[CODE]
# MurphyE Score: 1.033e-12 anorm: 1.126e46 rnorm: 2.420e45. (Since anorm is larger, sieve on the algebraic side. ie. "-a")
# SNFS difficulty is 229.858 which is approximately equivalent to GNFS difficulty 158. (Since n has 173 digits, it's recommended to use SNFS.)
# (some extra msieve library info) size: 1.923e-11 alpha: 1.653 rroots: 2
n: 58858763449145861682189209763193295417732963791373596736630811521069708557553014101459353660615483936299841727914109825712549670400384577919658321433057164682108047938754697
skew: 1.25118
type: snfs
c6: 8
c0: -7
Y1: 1
Y0: -144246936514426199369626914514553274368
m: 144246936514426199369626914514553274368
rlambda: 2.6
alambda: 2.6
lpbr: 29
lpba: 29
mfbr: 58
mfba: 58
alim: 38400000
rlim: 38400000
qintsize: 160000
[/CODE]

Since it is a rather big factorization for the limited resources I have,
I would like to run the sieving "by hand" on several boxes instead of
running factmsieve.py on only one box.
The question I have is how to compute the starting value of q?
I took a look at factmsieve.py from which I deduced that
q0=3840000/2=19200000 for the parameters above but I have seen
posts on this forum where people use q0 = alim / 3.[/QUOTE]

I start with the smallest q that is practical. Although smaller q's
produce more duplicates, they also have a much higher yield.

For me, "practical" means the following: After sieving one side I
compute the actual factorizations of the smooth relations via a
re-sieve. I store the factors that are found during the re-sieve in a
hash table. Smaller q's have higher yield. Some q's are small enough
that they produce enough potential smooth candidates as to overflow the
hash tables. I start with the smallest q that will not produce an overflow.

[QUOTE=YuL;358957]Thanks;
I used [URL="http://myfactors.mooo.com/"]myfactors.mooo.com[/URL] to get the polynomial:

[CODE]
# MurphyE Score: 1.033e-12 anorm: 1.126e46 rnorm: 2.420e45. (Since anorm is larger, sieve on the algebraic side. ie. "-a")
....
[/CODE]

[/QUOTE]

I'll second the "Welcome to the project".:smile:

re: the anorm & rnorm values, my level of expertise is not that great, so consider them as rough guidelines only. For your particular values, I believe that they are close enough that you should probably sieve on both sides.

Sorry, I don't have any experience with trying a lower q0 value.

Good luck with your factorization.

Thank you all for your help;

I lack the knowledge necessary to understand the various posts above especially
the one from Mr Silverman. I am just a software engineer who graduated in mathematics
more than 13 years ago, I wasn't even into number theory at that time.
BUT I'm willing to learn because I love mathematics and it is a lot a fun to factor big numbers.

By the way regarding the number I reserved I'm currently sieving
from 18M to 42M using 17 cores on 3 different boxes. I deduced from
[URL="http://homepage2.nifty.com/m_kamada/math/snfs_rels.png"]M. Kamada's graph[/URL] that I'll need about 72 million relations.