I added the FAQ to the following search engines:
Lycos
Excite
Yahoo
Altavista
AOL netfind
Webcrawler
HotBot
InfoSeek
-Lucas
_
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasa
Alex,
The group you seek always has 2^n elements. All bit combinations are
possible.
(P = 2^p-1 is "minus one" in n-bit words. 2*P is minus two, etc. up to
2^n*P which is 0. All bit patterns occur.)
Todd Sauke
>Now, I'm going to toss out an idea. I thought about this a few minutes
>after r
Disclaimer: All my commentary is based on recollections of what I did a
few years ago. I believe that there are comments in an archive somewhere
by Peter Montgomery and George Woltman conatining much more accurate
info.
Lucas Wiman wrote:
>
> Sorry if this is sending out twice, I wasn't sure tha
Thanks. I was also considering a base other than base 2. But I'm afraid
the same problem arises as long as the base is realatively prime to the
Mersenne number we are considering. For example if you look at 2047 (i.e.
2^11 - 1) in base 23 or base 87 you'll see the algorithm I outline below
actu
Mersenne Digest Monday, July 19 1999 Volume 01 : Number 600
--
Date: Sun, 18 Jul 1999 23:03:52 +0100
From: "Brian J. Beesley" <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Setu for Dual processor NT
> Is there
Now, I'm going to toss out an idea. I thought about this a few minutes
after reading the previous message and I want to see if you all think its
worthwhile or not, or whether its even correct or not.
Here goes:
1) If we know the last n bits of a number x, then we can (easily) determine
the last n
On 19 Jul 99, at 3:57, [EMAIL PROTECTED] wrote:
>*) Somebody finds how to parallelize the FFT using little
>communication. The wall-clock time might be reduced 10-fold,
>but the CPU time increased 16-fold. This could be
>great for verifying a new Mer
Sorry if this is sending out twice, I wasn't sure that it got sent the
first time...
>> As I understand the Pollard P-1 algorithm, the time required for 1 bit of
>> the exponent would be equal to 1 iteration of the LL test (one squaring
>> mod Mp), so values wouldn't be that hard to create.
(timi
Chris Jefferson <[EMAIL PROTECTED]> writes
> It is well known that n is prime if for all prime factors of n-1, a^(n-1)
> = 1 mod n and a^((n-1)/q) is not 1 mod n.
What are a and q? You want to say
`if for each prime factor q of n-1, there exists a base a such that ...' .
> For example, ta
It is well known that n is prime if for all prime factors of n-1, a^(n-1)
= 1 mod n and a^((n-1)/q) is not 1 mod n.
For example, take a=2, and 2^p+1 (p is prime, yes, that IS a plus)
Then we need to check if 2^(2^p)=1 mod (2^p -1) and 2^((2^p)/2) is not 1.
Now, we know that 2^p = 1 mod (2^p - 1
introductory article re: distributed computing, decent GIMPS paragraph.
http://www.sunworld.com/swol-07-1999/swol-07-silicon.html?0719i
_
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ --
> We should try to increase this probability before starting any
> full LL test, perhaps by using better factoring algorithms.
> Will v19 have P-1 code that works with numbers of this size?
> But even with P-1, we might not be able to factor that much
> further. The time consuming part of P-1 are
Hi,
from stuff I saw on the list earlier I put together this estimate:
The chance for Mp being prime if it has no factor below 2^q is
2q/p (if q is reasonably big, at i.e. 2^q much bigger than 2p+1).
If we test the Mersennes in the 10M-digits range for factors
to, say, 2^80, the chance for, say
13 matches
Mail list logo
