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2008-01-22, 19:41   #2
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

10B716 Posts

(quote from "Report all primes here", but I have other questions, so I figured I'd ask all here)
Quote:
 Originally Posted by gd_barnes Step 1: For people who have not submitted top-5000 primes previously, create a prover account: ... Step 2: Create a proof code: ... Step 3: Submit the prime: ...
Am I correct in assuming that Step 3 is the only step you'll need to do for primes after your first prime?
I'm going to be joining this project in a month (a month because that's when my next GIMPS number will finish and open up a core). I've been following it closely since it recently began.

Also, I'm wondering why there's two drives. One that's 260K-320K & 320K-333.2K and another that's 333.2K-600K. What's the significance of those different dividers? Why aren't they all in one drive? (other than that the 333.2K-600K began first) Why is the second drive 260K-320K & 320K-333.2K instead of being called 260K - 333.2K?
What program can generate the exact, full number of a k*2^n-1? Also, how (besides generating it and looking there) can you tell the exact number of digits in a k*2^n-1? I know a rough approximation would be n*log(2) because it's similar to a Mersenne number, but I don't know how to get it exactly.

 2008-01-22, 19:54 #3 em99010pepe     Sep 2004 B0E16 Posts I have to agree with you. It's a nonsense having two mini-drives, better to join both... You can generate the full number with PARI software, in here. Exact number of digits of k*2^n-1 will be log(k)+n*log(2)-log(1).
2008-01-22, 20:14   #4
gd_barnes

"Gary"
May 2007
Overland Park, KS

5×17×139 Posts

Quote:
 Originally Posted by em99010pepe I have to agree with you. It's a nonsense having two mini-drives, better to join both... You can generate the full number with PARI software, in here. Exact number of digits of k*2^n-1 will be log(k)+n*log(2)-log(1).
log(1)=0 so that can be ignored. Also, you need to add 1 to obtain the # of digits after you're done and of course take the integer portion of it. The actual formula if you wanted to plug it into Excel is:

int (n*log(2)+log(k)+1)

Gary

 2008-01-22, 20:29 #5 kar_bon     Mar 2006 Germany 2,999 Posts range questions first: 333k to 600k: to find in first place Top500 primes. every prime to check in the Top5000-list must have at least 100354 digits or n>=~333400. second: 260k-320k: all other n tested resulting small primes not for Top5000. all k's have been tested till n=260k so this is the start range of all (including some double checks). third: 320k-333k: in PrimeSearch to report a range done is set to 20k so there are all possible ranges upto 320k before and this last range is necessary to report 320k to 340k. ok? look also here: http://primes.utm.edu/top20/trends.php Last fiddled with by kar_bon on 2008-01-22 at 20:35
2008-01-22, 20:34   #6
gd_barnes

"Gary"
May 2007
Overland Park, KS

5·17·139 Posts

Quote:
 Originally Posted by Mini-Geek (quote from "Report all primes here", but I have other questions, so I figured I'd ask all here) Am I correct in assuming that Step 3 is the only step you'll need to do for primes after your first prime? I'm going to be joining this project in a month (a month because that's when my next GIMPS number will finish and open up a core). I've been following it closely since it recently began. Also, I'm wondering why there's two drives. One that's 260K-320K & 320K-333.2K and another that's 333.2K-600K. What's the significance of those different dividers? Why aren't they all in one drive? (other than that the 333.2K-600K began first) Why is the second drive 260K-320K & 320K-333.2K instead of being called 260K - 333.2K? What program can generate the exact, full number of a k*2^n-1? Also, how (besides generating it and looking there) can you tell the exact number of digits in a k*2^n-1? I know a rough approximation would be n*log(2) because it's similar to a Mersenne number, but I don't know how to get it exactly.

You are correct, #3 is all you would need to do.

See an explanation in drive #2 as to why we specifically divided up the smaller ranges of n=260K-320K and n=320K-333.2K. It comes down to the fact that the 320K-333.2K range fills in Michael Hartley's Prime Search site reservation range of n=320K-340K for each k-value.

As for why we made them two drives; that was done because the project is potentially so huge as it is and I wanted to divide it up into top-5000 parts and non-top-5000 parts. But I'm flexible and that is a good question so...

What do others think about making it one big drive?

It would just be one long page with different sections of files to search.

Any other suggestions are welcome also.

Gary

2008-01-22, 20:40   #7
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by em99010pepe I have to agree with you. It's a nonsense having two mini-drives, better to join both... You can generate the full number with PARI software, in here. Exact number of digits of k*2^n-1 will be log(k)+n*log(2)-log(1).
Thanks. What does PARI use as a decimal logarithm command, if there is one?
Quote:
 Originally Posted by gd_barnes log(1)=0 so that can be ignored. Also, you need to add 1 to obtain the # of digits after you're done and of course take the integer portion of it. The actual formula if you wanted to plug it into Excel is: int (n*log(2)+log(k)+1) Gary
Thanks for the correction to the formula.
Quote:
 Originally Posted by kar_bon first: 333k to 600k: to find in first place Top500 primes. every prime to check in the Top5000-list must have at least 100354 digits or n>=~333400. second: 260k-320k: all other n tested resulting small primes not for Top5000. all k's have been tested till n=260k so this is the start range of all (including some double checks). third: 320k-333k: in PrimeSearch to report a range done is set to 20k so there are all possible ranges upto 320k before and this last range is necessary to report 320k to 340k. ok?
Ok, now it makes sense.
Quote:
 Originally Posted by gd_barnes You are correct, #3 is all you would need to do. See an explanation in drive #2 as to why we specifically divided up the smaller ranges of n=260K-320K and n=320K-333.2K. It comes down to the fact that the 320K-333.2K range fills in Michael Hartley's Prime Search site reservation range of n=320K-340K for each k-value. As for why we made them two drives; that was done because the project is potentially so huge as it is and I wanted to divide it up into top-5000 parts and non-top-5000 parts. But I'm flexible and that is a good question so... What do others think about making it one big drive? It would just be one long page with different sections of files to search. Any other suggestions are welcome also. Gary
Now that I see the differences, I'm fine with them being two drives.

This is one of the faster threads on this forum since I posted...but I think that one about LLRNet that got 35 posts in 2 hours was faster. BTW how does the LLR test work (it stands for Lucas Lehmer Riesel, right? obviously some connection to a normal LL test, but what)?

 2008-01-22, 20:44 #8 kar_bon     Mar 2006 Germany 2,999 Posts
2008-01-22, 21:17   #9
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by kar_bon
Maybe I'm missing something obvious, but I know how the LL test works for 2^p-1. What I don't know is how it works for k*2^n-1.

 2008-01-22, 21:46 #10 kar_bon     Mar 2006 Germany 2,999 Posts i found this for the moment: http://links.jstor.org/sici?sici=002...2-F&size=LARGE "Lucasian Criteria for the Primality of N=h*2^n-1" Mathematics of Computation, Vol. 23 108, pp. 869-875, Oct. 1969 as i know Riesel advanced the LL-Test for any number N with h>1 and proved that the starting parameter for the sequence can found to fit the test for every h. (if my memory is right) only this http://nl.wikipedia.org/wiki/Lucas-Lehmer-Rieseltest but netherlands! found this: Code: The main theoretical fact is contained in the Theorem 5 (Lucas'Criteria for h*2^n-1) : Suppose that n=2, h is odd A =( (a+b*sqrt(D))^2)/r, Jacobi(D,N) = -1, and Jacobi(r,N)*sign(a^2-b^2*D) = -1. Then, a necessary and sufficient condition that N shall be prime is that u(n-2) == 0 (mod. N) if u(n) = u^2(n-1) - 2 with u(0) = A^h + A^-h. How to use that ? The number u(0) can be computed using a well known recursion formula: v(0) = 2, v(1) = A+A^-1, v(k) = v(1)*v(k-1) - v(k-2). So, we obtain u(0) = v(h). The remaining problem is to found a value for v(1) . The numbers A and A^-1 are units of the quadratic field K(sqrt(D))(that is to say units of the ring of the integers of this field...). So, they are powers of the fundamental unit of the field. Instead of choosing a square free integer D and searching for units satisfying the conditions of theorem 5, Riesel takes increasing values for v(1), and, remarking that A and A^-1 are the roots of the equation : A^2 - v(1)*A + 1 = 0 computes D as the square free part of v^2(1) - 4. It remains to verify that the resulting D, a, b and r values satisfy the conditions of theorem 5. The value of v(1) so found is the smallest possible. Regards, Jean Penné Last fiddled with by kar_bon on 2008-01-22 at 21:58
2008-01-22, 23:18   #11
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by kar_bon only this http://nl.wikipedia.org/wiki/Lucas-Lehmer-Rieseltest but netherlands!
From the Google translation of this I was able to understand all except the instructions of what starting number to use (i.e. it's just like an LL test, but with different starting number).

Quote:
 Als k gelijk is aan 1, dan is 4 een goede startwaarde als n oneven is, en als n = 3 mod 4, dan geldt u 0 = 3 . If k is equal to 1, then 4 is a good starting point when n is odd, and when n = 3 mod 4, then u 0 = 3.
Okay, I know that if k=1 it's a Mersenne and 4 is used as a starting point. After that, I don't understand the rest.

Quote:
 Als k = 3, dan moet u 0 gelijk zijn aan 5778 voor n = 0 of 3 mod 4. When k = 3, then u 0 must equal 5778 for n = 0 or 3 mod 4. Als geldt dat k = 1 of 5 mod 6 en 3 deelt N niet, dan geldt $u_0 = (2+\sqrt{3})^h+(2-\sqrt{3})^h$ . If true that k = 1 or 5 mod 6 and 3 shares N not, it is true $u_0 = (2+\sqrt{3})^h+(2-\sqrt{3})^h$ .
Can anyone, perhaps someone that knows Dutch or how to find the starting number, explain this?

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