Will Any Current 100M Digit LL Tests Finish?

[fivemack]

How did I get 94 months?

Well, the calculation takes 960 months now, and we want a point at which it would take 26 months, so speed needs to double log_2(960/26) = 5.2 times, and each doubling takes 18 months.

[joblack]

Originally Posted by henryzz

does anyone have a time estimate on a Q6600 for a 100M digit test

I have overclocked my Q6600 ( around 3 GHZ) - if you start all 4 threads for one number you need around 5 years ... with 4 numbers on 4 cores its around 2019.

But you can imagine if you upgrade your hardware to the coming quantum cpus it will get faster ;).

[joblack]

Originally Posted by retina

If you mean an exponent of 100M then the answer is yes. It was StarQwest that did 100,000,007. See http://mersenneforum.org/showthread.php?t=6389 for some details.

I also found a Factor on one 100M digits number early ... doesn't know which one it was but somewhere you could really see the result.

(have found it:tron 332199223
F
2008-10-15 22:40
1.7
8183174245366243263319
... but its nothing special??)

100,000,007 wasn't prime - they have found a factor (have read it in the forum).

[jinydu]

Originally Posted by joblack

100,000,007 wasn't prime - they have found a factor (have read it in the forum).

I presume you mean M100,000,007. I didn't hear about any factor being found; Starqwest tried TF and P-1 but found no factors.

[joblack]

Originally Posted by jinydu

I presume you mean M100,000,007. I didn't hear about any factor being found; Starqwest tried TF and P-1 but found no factors.

Do you mean with M100,000,007 2^n-1 where n is 100,000,007 or is the 100,000,007 directed to the 100,000,007 digits in the decade system? I´m a little bit confused about the mixture of the terms ...

[Mini-Geek]

Originally Posted by joblack

Do you mean with M100,000,007 2^n-1 where n is 100,000,007 or is the 100,000,007 directed to the 100,000,007 digits in the decade system? I´m a little bit confused about the mixture of the terms ...

When someone says M100,000,007 they're referring to 2^100,000,007-1, not the 100,000,007th mersenne prime or a 100,000,007 digit mersenne candidate. Generally, M(a large number e.g. >100) refers to 2^(a large number)-1, M(a small number e.g. <50) refers to that (a small number)th mersenne prime, while things like 100M+ can be confusing due to referring to 2^(n>100,000,000)-1 or 10^(n>100,000,000) i.e. the mersenne number of the number of decimal digits.

[joblack]

Originally Posted by jinydu

I presume you mean M100,000,007. I didn't hear about any factor being found; Starqwest tried TF and P-1 but found no factors.

Well I have found a factor for a 100M digit number (decade system):

tron

332199223

F

2008-10-15 22:40

1.7

8183174245366243263319

But it´s over two months ago (and I didn´t think it´s a big deal??)

... or what did dominicanpapi82 mean by

"At least one 100M mersenne has been finished by SOMEONE, right?"

[James Heinrich]

Originally Posted by joblack

Well I have found a factor for a 100M digit number (decade system):
But it´s over two months ago (and I didn´t think it´s a big deal??)
... or what did dominicanpapi82 mean by
"At least one 100M mersenne has been finished by SOMEONE, right?"

You're right: finding a factor for a 100-million-decimal-digits number is not a big deal (I've found factors for at least 37 such numbers, and I haven't tested very many in that range).

Finding a factor is a good way to consider a Mersenne number "done", since that proves it's not prime. But this thread is about when will the first Lucas-Lehmer test be completed on a number that large. This test would take anywhere from 2-5+ years on current good computers -- that's the part that's a big deal.

[starrynte]

(poll expires tomorrow)

[jinydu]

Originally Posted by Mini-Geek

When someone says M100,000,007 they're referring to 2^100,000,007-1, not the 100,000,007th mersenne prime or a 100,000,007 digit mersenne candidate. Generally, M(a large number e.g. >100) refers to 2^(a large number)-1, M(a small number e.g. <50) refers to that (a small number)th mersenne prime, while things like 100M+ can be confusing due to referring to 2^(n>100,000,000)-1 or 10^(n>100,000,000) i.e. the mersenne number of the number of decimal digits.

I've never seen M(number) being used to refer to number of decimal digits.

[retina]

Perhaps we could have some form of forum convention.

M_{[i]number[/i]} refers to: number = exponent

Mnumber refers to: number = number^th mersenne prime

eg.

M_43112609 = M46