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Old 2021-01-22, 18:11   #78
tuckerkao
 
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M102,374,113 just passed Kriesel's premium P-1 test -
https://www.mersenne.org/report_expo...exp_hi=&full=1
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Old 2021-01-22, 18:18   #79
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So what?
Do you think that makes a big difference in the likelihood of it being prime?
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Old 2021-01-22, 21:37   #80
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Why isn't this thread in Misc. Math?
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Old 2021-01-22, 21:41   #81
Viliam Furik
 
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Quote:
Originally Posted by tuckerkao View Post
M102,374,113 just passed Kriesel's premium P-1 test -
https://www.mersenne.org/report_expo...exp_hi=&full=1
It's not a test, it's factoring attempt. It did not pass. If anything, it failed. Unless it's prime, with high enough bounds, P-1 WILL produce a factor.
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Old 2021-01-22, 22:42   #82
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Quote:
Originally Posted by Viliam Furik View Post
It's not a test, it's factoring attempt. It did not pass. If anything, it failed. Unless it's prime, with high enough bounds, P-1 WILL produce a factor.
All right, let's use M1277 as the example, how high will the bound be large enough to find the factor?
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Old 2021-01-22, 22:56   #83
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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 2021-01-22 at 22:58
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Old 2021-01-22, 23:01   #84
Viliam Furik
 
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Quote:
Originally Posted by tuckerkao View Post
All right, let's use M1277 as the example, how high will the bound be large enough to find the factor?
What I meant was that if there is a factor, you can always find it by P-1 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.
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Old 2021-01-23, 11:32   #85
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Quote:
Originally Posted by Viliam Furik View Post
What I meant was that if there is a factor, you can always find it by P-1 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.
False.

The factor you find may well be the number itself.

Two distinct prime factors may each be (P-1)-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.
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Old 2021-01-23, 13:52   #86
Viliam Furik
 
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Quote:
Originally Posted by xilman View Post
False.

The factor you find may well be the number itself.

Two distinct prime factors may each be (P-1)-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.
Why? Example didn't give me enough explanation for Mersenne exponents.

Do you suggest, that the k from 2kp+1 decomposition of 2p-1 may be smooth, while the ks in 2kp+1 decompositions of factors may require higher bounds?
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Old 2021-01-25, 15:37   #87
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Yes, let's take an example from the Mersenne numbers:
Let's have \(n := 2^p-1\) with \(p = 1{,}061\). Then P-1 would "find" \(n\) with B1 = 1040 and B2 = 1056. This \(n\) has two large factors, a P143 and a P177. The P143 would need B1 = 1016 and B2 = 1099 (!), so you would need to use an extremely odd multiplier (as in choosing B2 = \(x \cdot\,\)B1) for your P-1 runs to find this factor earlier than \(n\) itself. The P177 needs B1 = 1033 and B2 = 1080. 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.
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Old 2021-02-02, 10:02   #88
tuckerkao
 
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I just noticed that Kriesel finished P-1 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 2021-02-02 at 10:04
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