PARI's commands - Page 29

CRGreathouse · 2010-08-12, 14:55

Assumptions: m, t, p, n are positive integers, p > 1.

Quote:

Originally Posted by science_man_88

$24m= 6tp+(p-7)\right m=px+c$
$24m=6tp-(p+7)\right m=px+c ?$

So we have 24m + 7 being an integer with $24m+7=p(6t\pm1)$ as appropriate.

Quote:

Originally Posted by science_man_88

$if(px+c==(4^n-1)/3,(4^\strike n -1)/3$

So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

science_man_88 · 2010-08-12, 15:21

Quote:

Originally Posted by CRGreathouse

Assumptions: m, t, p, n are positive integers, p > 1.

So we have 24m + 7 being an integer with $24m+7=p(6t\pm1)$ as appropriate.

So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

something tells me you are getting at a trivial answer ?

3.14159 · 2010-08-12, 15:32

Quote:

Originally Posted by CRGreathouse

What script are you using to check? My isSPRP gives

Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.

3.14159 · 2010-08-12, 15:36

Quote:

Originally Posted by CRGreathouse

So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

Those are all trivially correct. There is no grand discovery here.

CRGreathouse · 2010-08-12, 15:41

Quote:

Originally Posted by 3.14159

Those are all trivially correct. There is no grand discovery here.

The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".

CRGreathouse · 2010-08-12, 15:42

Quote:

Originally Posted by 3.14159

Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.

Really? Interesting.

kar_bon · 2010-08-12, 15:45

Quote:

Originally Posted by CRGreathouse

That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!

3.14159 · 2010-08-12, 15:54

Quote:

Originally Posted by CRGreathouse

The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".

Wow. That was all that it was? A tautology/circular argument?

!

Quote:

Originally Posted by CRGreathouse

Really? Interesting.

You can get it for yourself here.

I suspect it of being a kook site, but, seeing as their applets work correctly to some extent, I am undecided on that matter.

Try testing it out quickly (Note: Small integers only, please), and if you can catch 25326001 as an error, please note it!

3.14159 · 2010-08-12, 15:56

Quote:

Originally Posted by kar_bon

But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!

Karsten, I found that you were right about the differences in testing times based on the k-values used.

CRGreathouse · 2010-08-12, 16:01

Quote:

Originally Posted by kar_bon

But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!

The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.

Or rather, it was a restricted case of this: if n > 3 is an odd number such that 2^n - 1 is prime, then 2^n - 1 is prime.

3.14159 · 2010-08-12, 16:03

Quote:

Originally Posted by CRGreathouse

The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.

Call the press! We're going to be filthy rich!

Amirite?

Bullshit aside..

CRG, have you managed to do some testing on 25326001 using that app? Does it say that it's a 7-SPRP?

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