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#1784 |
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Sep 2002
2·3·7·19 Posts |
P-1 found a factor in stage #2, B1=738000, B2=19943000.
UID: Jwb52z/Clay, M102443659 has a factor: 9267323975458581946834423 (P-1, B1=738000, B2=19943000), 82.938 bits. |
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#1785 |
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Sep 2017
USA
3248 Posts |
M103012487 has a 98.476-bit (30-digit) factor: 440774470073643783648685727201 (P-1,B1=2000000,B2=132000000)
I think this is the largest B2 that I've found!!! Not a bad sized factor either ![]() \(k = 2^4 × 5^2 × 7 × 17 × 19 × 151 × 160,087 × 97,859,501\) |
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#1786 |
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Nov 2014
2×3×5 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 Brent-Suyama 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_k-1:
print(f"{factor_of_k} divides ({primorial}*{i}+{j})**{e}-1")
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#1787 |
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Jun 2003
5,051 Posts |
The poly is (2310*i)^12-j^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)^12-j^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 2021-02-05 at 12:53 |
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#1788 | |
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Nov 2014
2×3×5 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 co-prime 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)**12-0**12 195509617 divides (210*0)**12-0**12 195509617 divides (2310*0)**12-0**12 195509617 divides (2310*5183)**12-423**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 2021-02-05 at 13:29 |
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#1789 |
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Jun 2003
10011101110112 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. |
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#1790 | |
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Nov 2014
2×3×5 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 2021-02-05 at 16:12 |
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#1791 |
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"James Heinrich"
May 2004
ex-Northern Ontario
3×5×227 Posts |
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#1792 | |
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"Viliam Furík"
Jul 2018
Martin, Slovakia
2×3×5×19 Posts |
Quote:
M42654757 has a 177.384-bit (54-digit) composite (P24+P31) factor: 249962766200012947561203808756653177525031174447055441 (P-1,B1=2000000,B2=60000000) |
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#1793 | |
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6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
23·1,223 Posts |
Quote:
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