Phone Book Numbers from 103,374,001 to 103,374,989 that could be worth big bucks

[tuckerkao]

M102,374,113 just passed Kriesel's premium P-1 test -
https://www.mersenne.org/report_expo...exp_hi=&full=1

[Uncwilly]

So what?

Do you think that makes a big difference in the likelihood of it being prime?

[masser]

Why isn't this thread in Misc. Math?

[Viliam Furik]

Originally Posted by tuckerkao

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.

[tuckerkao]

Originally Posted by Viliam Furik

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?

[Uncwilly]

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

[Viliam Furik]

Originally Posted by tuckerkao

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.

[xilman]

Originally Posted by Viliam Furik

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.

[Viliam Furik]

Originally Posted by xilman

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 2^p-1 may be smooth, while the ks in 2kp+1 decompositions of factors may require higher bounds?

[kruoli]

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.

[tuckerkao]

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.