20210122, 18:11  #78 
Jan 2020
13^{2} Posts 
M102,374,113 just passed Kriesel's premium P1 test 
https://www.mersenne.org/report_expo...exp_hi=&full=1 
20210122, 18:18  #79 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}×2,339 Posts 
So what?Do you think that makes a big difference in the likelihood of it being prime? 
20210122, 21:37  #80 
Jul 2003
wear a mask
2^{4}×97 Posts 
Why isn't this thread in Misc. Math?

20210122, 21:41  #81  
"Viliam Furík"
Jul 2018
Martin, Slovakia
2·193 Posts 
Quote:


20210122, 22:42  #82 
Jan 2020
169_{10} Posts 

20210122, 22:56  #83 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}·2,339 Posts 
B1=3098509800000000098304
B2=999999960009899986925979033990527451136 Yields ~ 100% chance. B1=2,309,850,729,825 B2=554,364,175,158,000 Gives a 30.% chance for only 765 GHz days Last fiddled with by Uncwilly on 20210122 at 22:58 
20210122, 23:01  #84 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2·193 Posts 
What I meant was that if there is a factor, you can always find it by P1 IF you choose high enough bound  high enough being indeterminate term, depending on the factor itself. In other words if it's not a prime and you put in enough work into it (again, enough being the necessary minimum), you will certainly find a factor.

20210123, 11:32  #85  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
3·3,529 Posts 
Quote:
The factor you find may well be the number itself. Two distinct prime factors may each be (P1)smooth for a particular choice of (B1,B2) but no smaller ones. 4369 = (2^4+1)*(2^8+1) to give an entirely trivial example. 

20210123, 13:52  #86  
"Viliam Furík"
Jul 2018
Martin, Slovakia
2·193 Posts 
Quote:
Do you suggest, that the k from 2kp+1 decomposition of 2^{p}1 may be smooth, while the ks in 2kp+1 decompositions of factors may require higher bounds? 

20210125, 15:37  #87 
"Oliver"
Sep 2017
Porta Westfalica, DE
435_{10} Posts 
Yes, let's take an example from the Mersenne numbers:
Let's have \(n := 2^p1\) with \(p = 1{,}061\). Then P1 would "find" \(n\) with B1 = 10^{40} and B2 = 10^{56}. This \(n\) has two large factors, a P143 and a P177. The P143 would need B1 = 10^{16} and B2 = 10^{99} (!), so you would need to use an extremely odd multiplier (as in choosing B2 = \(x \cdot\,\)B1) for your P1 runs to find this factor earlier than \(n\) itself. The P177 needs B1 = 10^{33} and B2 = 10^{80}. All bound were rounded up to the next power of 10 in this example. There might be examples (for Mersenne numbers) where any combination of B1 and B2 will lead to finding the number itself and not its factors, but I'm not aware of any and did not find any on a quick search. I would assume such a number most likely would only have two distinct prime factors and the size of these factors would be large. 
20210202, 10:02  #88 
Jan 2020
13^{2} Posts 
I just noticed that Kriesel finished P1 factoring on M103,374,361 which has the final 3 and the final 1 swapped in the positions from the exponent I began with 
https://www.mersenne.org/report_expo...exp_hi=&full=1 Should I begin to bet on the exponents with the M103,3**,3*1 format? Kriesel has been correct several times, maybe he can run the PRP tests on M103,343,341 or M103,332,391 while I'm going after another such exponent at the same time. Last fiddled with by tuckerkao on 20210202 at 10:04 
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