Strategic Double Clicking

[Mark Rose]

.

[Madpoo]

Originally Posted by frmky

Neither matched.

http://www.mersenne.org/report_expon...5064059&full=1
http://www.mersenne.org/report_expon...6497473&full=1

Isn't that kind of fun, in a weird way?

Be sure and pick on James for his test on M36497473 ...

(of course there's always that chance *yours* is bad... )

[Madpoo]

Originally Posted by chalsall

Better generate another list

Okay, here we go.

For this list, I used the criteria of unassigned exponents below 58M (as usual) where:
- The computer has zero good results
- 3 or more bad results

There are 47 in the list. Some of the computers involved have a large # (14 or 28) unknowns... once those start getting double/triple checked it'll change their good/bad count. But those same computers also have a couple suspect results as well, so I'm still not convinced they're all that reliable until more data is collected.

exponent	Bad	Good	Unk	Sus	worktodo
34938793	3	0	2	0	DoubleCheck=34938793,71,1
35006537	3	0	1	1	DoubleCheck=35006537,71,1
35090593	3	0	4	3	DoubleCheck=35090593,71,1
35159191	3	0	3	0	DoubleCheck=35159191,71,1
35379403	3	0	3	0	DoubleCheck=35379403,71,1
35662219	5	0	2	5	DoubleCheck=35662219,71,1
35863103	3	0	2	3	DoubleCheck=35863103,71,1
35963243	5	0	5	0	DoubleCheck=35963243,71,1
36009163	3	0	5	2	DoubleCheck=36009163,71,1
36026687	4	0	4	8	DoubleCheck=36026687,71,1
36086483	3	0	2	1	DoubleCheck=36086483,71,1
36138559	5	0	8	5	DoubleCheck=36138559,71,1
36186791	5	0	5	0	DoubleCheck=36186791,71,1
36273859	3	0	1	0	DoubleCheck=36273859,71,1
36381031	4	0	1	2	DoubleCheck=36381031,71,1
36392009	5	0	4	7	DoubleCheck=36392009,71,1
36435929	5	0	6	3	DoubleCheck=36435929,71,1
36503917	3	0	5	2	DoubleCheck=36503917,71,1
36532913	3	0	1	1	DoubleCheck=36532913,71,1
36536707	4	0	1	0	DoubleCheck=36536707,71,1
36576769	5	0	4	7	DoubleCheck=36576769,71,1
36648539	3	0	5	2	DoubleCheck=36648539,71,1
36670561	5	0	4	7	DoubleCheck=36670561,71,1
36688991	4	0	4	8	DoubleCheck=36688991,71,1
36743743	3	0	3	1	DoubleCheck=36743743,71,1
36892939	4	0	3	2	DoubleCheck=36892939,71,1
36904081	5	0	8	5	DoubleCheck=36904081,71,1
36938339	5	0	6	3	DoubleCheck=36938339,71,1
36955621	3	0	2	1	DoubleCheck=36955621,71,1
37064927	4	0	1	0	DoubleCheck=37064927,71,1
37260571	4	0	4	8	DoubleCheck=37260571,71,1
37359779	3	0	4	3	DoubleCheck=37359779,71,1
40922261	3	0	7	1	DoubleCheck=40922261,72,1
41294441	3	0	2	0	DoubleCheck=41294441,72,1
41364503	4	0	14	2	DoubleCheck=41364503,72,1
41364541	4	0	14	2	DoubleCheck=41364541,72,1
41442407	4	0	14	2	DoubleCheck=41442407,72,1
41485517	4	0	3	2	DoubleCheck=41485517,71,1
41501191	3	0	1	2	DoubleCheck=41501191,71,1
41979727	4	0	28	1	DoubleCheck=41979727,72,1
41992207	3	0	7	1	DoubleCheck=41992207,72,1
42187843	4	0	28	1	DoubleCheck=42187843,72,1
43671757	4	0	28	1	DoubleCheck=43671757,72,1
44876599	3	0	1	1	DoubleCheck=44876599,72,1
55583947	4	0	6	0	DoubleCheck=55583947,73,1
57783881	4	0	6	0	DoubleCheck=57783881,73,1
57859121	4	0	6	0	DoubleCheck=57859121,73,1

[Madpoo]

And here's a shorter list with:
- Good results between 1 and 10
- At least 5 times as many bad as good

As you can see, some of these also have a high # of suspect results which is a pretty good sign (combined with their known bad) that those will turn out bad as well.

Enjoy!

exponent	Bad	Good	Unk	Sus	worktodo
34769731	6	1	2	9	DoubleCheck=34769731,71,1
35200849	6	1	6	0	DoubleCheck=35200849,71,1
35997649	6	1	6	0	DoubleCheck=35997649,71,1
36278761	6	1	6	0	DoubleCheck=36278761,71,1
36589351	6	1	1	0	DoubleCheck=36589351,71,1
36717713	11	2	1	1	DoubleCheck=36717713,71,1
45951173	11	2	6	10	DoubleCheck=45951173,72,1

[chalsall]

Originally Posted by Madpoo

There are 47 in the list.

I've grabbed the first 24 (34938793 to 36688991 inclusive) from this list. Results should be in late Tuesday / early Wednesday.

And you're correct, this is fun! A bit like weeding a garden!

Oh, and three are back from my previous batch; 0 for 3. Will provide the full list once they're all completed.

[Madpoo]

Originally Posted by chalsall

And you're correct, this is fun! A bit like weeding a garden!

Or popping zits... LOL

I just found 2 more today, like this doozy that will now need a quintuple check:
M35759939

Although technically that one was from a batch I'd picked where exponents needed a 3rd/4th check, and wasn't from the lists of possibly suspect work... but still satisfying somehow.

[Madpoo]

Hey all... flagrantflowers mentioned to me that there's some kind of method for verifying the LL residue if we happen to know a factor. He suggested reaching out to science_man_88 about it, but before I bug him I thought I'd group think this thing.

If this were possible, it'd be nice to go through all of the cases where someone did an LL test and it was later factored... might be nice to split the singular "factored" status of such an LL test into "factored - verified" and "factored - bad".

Basically treating the factor as a confirming or disproving 2nd LL test.

Before that even took place I could at least go through and find all the cases where it was already verified by a 2nd LL test. Just in case the residue confirmation math is less than trivial, saving some time.

Anyone have any thoughts on that, or should I just check with science_man_88 as suggested?

[Mark Rose]

Originally Posted by Madpoo

I just found 2 more today, like this doozy that will now need a quintuple check:
M35759939

I should have it done in a few days.

[science_man_88]

Originally Posted by Madpoo

Hey all... flagrantflowers mentioned to me that there's some kind of method for verifying the LL residue if we happen to know a factor. He suggested reaching out to science_man_88 about it, but before I bug him I thought I'd group think this thing.

If this were possible, it'd be nice to go through all of the cases where someone did an LL test and it was later factored... might be nice to split the singular "factored" status of such an LL test into "factored - verified" and "factored - bad".

Basically treating the factor as a confirming or disproving 2nd LL test.

Before that even took place I could at least go through and find all the cases where it was already verified by a 2nd LL test. Just in case the residue confirmation math is less than trivial, saving some time.

Anyone have any thoughts on that, or should I just check with science_man_88 as suggested?

see my post 43 of this thread and I'm not the only one to suggest this is possible there have been other threads talking about this. here's the easiest example in PARI:

f=23;s=4;b=1736;for(x=3,11,s=Mod(s,f)^2-2);if(lift(s)==lift(Mod(b,f)),print("factor found:"f))

basically all this is using is the fact that assuming we have a number y of form ax+b (that is b mod x) assume we know a factor of x, call this factor f we can rewrite this as y=azf+(b\f)*f +b%f where \ between two numbers indicates the floor of integer division and % means modular arithmetic,this can be simplified by saying all multiples of f that are present sum to c*f we then get y=c*f+b%f and so b%f is our residue mod f of our original number if f is a factor of anything but b in the original equation y=ax+b. of course this isn't only potentially useful to use factors to check residues but if we know the residues are correct we can then use the residues from numbers higher than x ( in this case out mersenne number to be factored), and all residues for numbers greater than the mersenne number have to match their residue mod f for f to be a factor of the mersenne number x.

[Madpoo]

Originally Posted by science_man_88

here's the easiest example in PARI:

f=23;s=4;b=1736;for(x=3,11,s=Mod(s,f)^2-2);if(lift(s)==lift(Mod(b,f)),print("factor found:"f))

basically all this is using is the fact that assuming we have a number y of form ax+b (that is b mod x) assume we know a factor of x, call this factor f we can rewrite this as y=azf+(b\f)*f +b%f where \ between two numbers indicates the floor of integer division and % means modular arithmetic,this can be simplified by saying all multiples of f that are present sum to c*f we then get y=c*f+b%f and so b%f is our residue mod f of our original number if f is a factor of anything but b in the original equation y=ax+b. of course this isn't only potentially useful to use factors to check residues but if we know the residues are correct we can then use the residues from numbers higher than x ( in this case out mersenne number to be factored), and all residues for numbers greater than the mersenne number have to match their residue mod f for f to be a factor of the mersenne number x.

So, pretend I don't know the math very well (okay, enough laughing...)

For the PARI syntax, I followed along with M11 having the factor 23, but what is b= in that example? That's not the residue for M11 (it's 0x48), and that's where I lost ya.

Let's say we have this as an example: http://www.mersenne.org/M2000143:
M2000143 has the factor 6174103888966755151 and a verified residue of 0x7AFB854059F08BEF (well, technically the last 64 bits of it, but you know what I mean).

How would that look if I wanted to use the factor to verify the LL test/residue?

[science_man_88]

Originally Posted by Madpoo

So, pretend I don't know the math very well (okay, enough laughing...)

For the PARI syntax, I followed along with M11 having the factor 23, but what is b= in that example? That's not the residue for M11 (it's 0x48), and that's where I lost ya.

Let's say we have this as an example: http://www.mersenne.org/M2000143:
M2000143 has the factor 6174103888966755151 and a verified residue of 0x7AFB854059F08BEF (well, technically the last 64 bits of it, but you know what I mean).

How would that look if I wanted to use the factor to verify the LL test/residue?

b is the decimal value of the final residue mod 2^11-1 see http://oeis.org/A129220

plug the factor in as f, the 11 gets replaced by the exponent in question, and b gets replaced by the decimal value of the full final residue. PARI/gp can still be quite slow another language might be better to use.