20210201, 17:05  #1784 
Sep 2002
3·263 Posts 
P1 found a factor in stage #2, B1=738000, B2=19943000.
UID: Jwb52z/Clay, M102443659 has a factor: 9267323975458581946834423 (P1, B1=738000, B2=19943000), 82.938 bits. 
20210203, 03:53  #1785 
Sep 2017
USA
5·47 Posts 
Massive B2
M103012487 has a 98.476bit (30digit) factor: 440774470073643783648685727201 (P1,B1=2000000,B2=132000000)
I think this is the largest B2 that I've found!!! Not a bad sized factor either \(k = 2^4 × 5^2 × 7 × 17 × 19 × 151 × 160,087 × 97,859,501\) 
20210205, 11:33  #1786 
Nov 2014
30_{10} Posts 
Yes, it did. I tried to understand why this large factor 195509617 of k for my factor of M70553939 has been found with e=12 BrentSuyama and a B2 of only 13222500. As far as I understand it after reading this explanation, I think it must be because 195509617 happens to divide a larger polynomial. So I tried to find this larger polynomial with the Python code below, but the output is empty, so I must have done something wrong ...
Code:
#!/usr/bin/python n=70553939 factor_of_k=195509617 B2=13222500 e=12 primorial = 2310 for i in range(0,B2//primorial+1): for j in range(1,primorial,2): # test if 195509617 divides (2310*i + j)^12  1 (we only need to do it for uneven j) if pow(primorial*i+j, e, factor_of_k) == factor_of_k1: print(f"{factor_of_k} divides ({primorial}*{i}+{j})**{e}1") 
20210205, 12:35  #1787 
Jun 2003
4960_{10} Posts 
The poly is (2310*i)^12j^12, and only j that is relatively prime to 2310 need to be considered. Actually, now that I think about it, could be (4320*i)^12j^12
EDIT: If you've used the new 30.4 version, then j can be much bigger than primorial EDIT2: i=5413, j=19891, D=2310 works Last fiddled with by axn on 20210205 at 12:53 
20210205, 13:28  #1788  
Nov 2014
36_{8} Posts 
Quote:
Code:
factor_of_k=195509617 B2=13222500 e=12 for primorial in [30,210,2310]: for i in range(0,B2//primorial+1): # test 2310*k + 1, 2310*k + 3, 2310*k + 5, ... # (didn't bother to check only coprime j for j in range(0,primorial): if pow(primorial*i, e, factor_of_k) == pow(j, e, factor_of_k): print(f"{factor_of_k} divides ({primorial}*{i})**{e}{j}**{e}") Code:
$ python brent.py 195509617 divides (30*0)**120**12 195509617 divides (210*0)**120**12 195509617 divides (2310*0)**120**12 195509617 divides (2310*5183)**12423**12 Maybe I need to take a look into the source code if I want to 100% understand it. But I think it's fine for now. Last fiddled with by gLauss on 20210205 at 13:29 

20210205, 14:12  #1789 
Jun 2003
1001101100000_{2} Posts 
Hmmm... 423 is not relatively prime to 2310, so that is not a valid combo.
Are you sure it was D=2310, E=12? D=210 is another option, as is E=6. 
20210205, 16:09  #1790  
Nov 2014
30_{10} Posts 
Quote:
Code:
#!/usr/bin/python from math import gcd #n=70553939 factor_of_k=195509617 B1=645000 B2=13222500 e=12 def is_coprime(a,b): return gcd(a,b) == 1 for primorial in [6,30,210,2310,4620]: coprimes = [i for i in range(1,primorial) if is_coprime(i,primorial)] print(f"Trying with primorial {primorial}, phi({primorial})={len(coprimes)} (number of coprimes)") for i in range(B1//primorial,B2//primorial+1): for j in coprimes: for e in [2,6,12]: if pow(primorial*i, e, factor_of_k) == pow(j, e, factor_of_k): Code:
Trying with primorial 6, phi(6)=2 (number of coprimes) Trying with primorial 30, phi(30)=8 (number of coprimes) Trying with primorial 210, phi(210)=48 (number of coprimes) Trying with primorial 2310, phi(2310)=480 (number of coprimes) Trying with primorial 4620, phi(4620)=960 (number of coprimes) Last fiddled with by gLauss on 20210205 at 16:12 

20210205, 17:00  #1791 
"James Heinrich"
May 2004
exNorthern Ontario
2^{2}×839 Posts 

20210211, 16:05  #1792  
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{3}·3·19 Posts 
Quote:
M42654757 has a 177.384bit (54digit) composite (P24+P31) factor: 249962766200012947561203808756653177525031174447055441 (P1,B1=2000000,B2=60000000) 

20210211, 18:32  #1793  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
11^{2}·79 Posts 
Quote:


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