Nathan Russell [EMAIL PROTECTED] comments
--On Wednesday, June 11, 2003 10:58 PM -0400 George Woltman
[EMAIL PROTECTED] wrote:
2) I'll change prime95 to NOT delete the save files when a new prime is
found. When a prime is reported, I'll ask for the save file and rerun the
last 30 minutes of
= From [EMAIL PROTECTED] Fri Feb 7 05:15:26 2003
= I am using mprime 22.12 on a pentium 166 MMX to do trial factoring. For the
= exponents currently being assigned from primenet it takes this machine about
= 12 minutes to factor from 2^57 to 2^58.
=
= I thought I would try factoring some small
A Mersenne number M_p = 2^p - 1, where p is prime and p 1000,
has a prime divisor q with q == 1 (mod 2002) and q == -1 (mod 2001)
[== denotes congruent]. Find q mod 2003.
Peter Montgomery
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Brian J. Beesley [EMAIL PROTECTED] wrote
- From [EMAIL PROTECTED] Sun Nov 10 20:22:33 2002
- On Saturday 09 November 2002 04:45, George Woltman wrote:
- A harder problem is finding some smooth ECM curves to test. I do not
- have tools to compute group orders. If someone can help by finding a
- From [EMAIL PROTECTED] Sat Nov 9 00:21:04 2002
Anurag Garg [EMAIL PROTECTED] writes
- I have been unable to reach his pages using both the domain name
- hogranch.com - which gets a non-existant domain name error - or the IP
- address 209.133.33.182 .
-
- What gives?
- Anurag
-
How many
Griffith, Shaun [EMAIL PROTECTED] asks
Luke Welsh wrote:
The EE times (www.eet.com) reports:
Research by a team of U.K. mathematicians has produced a different
approach to two of the most fundamental aspects of electronic computation:
hardware addition and multiplication.
Can
John R Pierce [EMAIL PROTECTED] writes
- Can anyone come? Announce it on the list if you do it.
- Where abouts are they held?
-
-- I will if I can. Depends on the date and family conflicts and where in the
- rather sprawling SFBA it is... I'm in Santa Cruz, so naturally would prefer
-
To: [EMAIL PROTECTED]
Subject: Mersenne: n-th decimal digit in real number
Date: Tue, 09 Oct 2001 15:09:08 -0700
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Daran [EMAIL PROTECTED] asks
Is M751 now the smallest unfactorised composite Mersenne? What is the
smallest Mersenne not completely factorised?
M751 and M809 are the first Mersennes with no known factor.
The first holes in the 2^n - 1 table are 2,673- c151 and 2,683- c203.
This
From: Nathan Russell [EMAIL PROTECTED]
Hi all,
I'm a bit puzzled. The other day, I donated blood and kept my mind
busy by doing LL tests on a few exponents mentally. I kept getting
the result that the LL test for M(2) ends up resulting in a repeating
value of -1, and certainly cannot
From: Nathan Russell [EMAIL PROTECTED] writes
On Mon, 30 Jul 2001 11:45:26 -0700 (PDT), Pavlos S.
[EMAIL PROTECTED] wrote:
It has been a while since RSA labs published new
factoring challenges..Does anyone know if there is any
project going on at the moment for these challenges?
From: CARLETON GARRISON [EMAIL PROTECTED]
I honestly thought that the long term goal (maybe not by this panel but for
others) was to factor all these numbers and that we were setting/recording a
lower boundary for that effort.
Carleton Garrison
It will be a _long_ time before we
Andy Hedges [EMAIL PROTECTED] asks:
Anyone have any idea why for k = 659 there are very little primes? In fact
for k up to 20 there are none (I haven't found any in this range yet!).
Let k = 659.
If n == 1 (mod 2) then k*2^n == 1 (mod 3)
If n == 2 (mod 4) then k*2^n == 1 (mod 5)
If n ==
Brian J. Beesley [EMAIL PROTECTED] wrote,
in response to my earlier message
The point of course is that there is a formal proof that, if a prime
p is congruent to 3 modulo 4 and 2p+1 is also prime, then 2^p-1 is
divisible by 2p+1 - which makes searching for a factor of M(p) by
trying
On 26 Apr 2001, Brian J. Beasley wrote
On 26 Apr 2001, at 6:34, Hans Riesel wrote:
Hi everybody,
If 2^p-1 is known to be composite with no factor known, then so is
2^(2^p-1)-1.
Much as I hate to nitpick a far better mathematician than myself,
this is seems to be an
Spike Jones [EMAIL PROTECTED] comments;
--953E7B41754C35108B97B346
Nathan I liked your comment about the largest genuine composite:
a number known to be composite but for which none of the factors
are known. I suppose we could set up a computer to arbitrarily
generate a few
From: [EMAIL PROTECTED]
In a message dated 05/04/2001 05:19:38 GMT Daylight Time, "Gary Untermeyer"
[EMAIL PROTECTED] writes:
Let x be a prime number. Consider the series of numbers that take the
following form:
x, x + n, x + 2n, x + 3n, x + 4n, x + 5n, x + 6n, where n is
- From: "Hoogendoorn, Sander" [EMAIL PROTECTED]
- But what if the mod M comes out to 1 on one of
- the intermediate steps? Then 1^2 - 2 = -1
- Then what? spike
-Then you will be stuck in a loop, same thing happens when the outcome is
- -2, -1, 0 and 2
This case never occurs. If M =
From [EMAIL PROTECTED] Tue Mar 13 16:06:33 2001
I wonder if there is possible to find primitive roots in GF(q^2), say a
+ ib, with the condition
a^2 + b^2 = 1 mod q
If this were possible, then
(a + ib)*(a - ib) = 1 + 0i (mod q)
i, e. the inverse of the root would be the
Jason Stratos Papadopoulos [EMAIL PROTECTED] writes
Hello again. After working out how to do integer FFTs on the Alpha, I'm
considering additional code that uses Fast Galois Transforms (FGTs),
e.g. for complex integers in GF(p^2) for p a prime congruent to 3 mod
4. Bill Daly mentioned this
Fix a length n. If (a0, a1, ..., a_(n-1)) and
(b0, b1, ..., b_(n-1)) are sequences of length n
(in some ring or field), then their circular convolution
is (c0, c1, ..., c_(n-1)) where
c_i = sum_{i == j + k mod n} a_j * b_k
For example, when n = 5, the circular convolution of
(a0,
Nathan Russell observes:
Well, unless there is an announcement within the next few hours, 2000
will be the first calender year in the history of GIMPS without a
Mersenne prime.
Is the number of searchers, and the power of their hardware, increasing
enough to keep up with the
If p is a prime which for which 2^22-th roots of unity exist,
then p == 1 (mod 2^22). All such primes have 2 as a square, but 2 will
be a 2^22 power only one time in 2^21.
Of the 2^61/ln(2^61) primes between 2^61 and 2^62,
only one in 2^42 will meet both requirements.
You should find
Chris Nash proposed:
We could let prime95 decide the next election grin.
Give everybody a different prime number. Multiply the primes for
candidate A together, likewise for B.
And like in this election, unless we can completely factor the results,
we wouldn't know who won, either.
Hi all,
The mailing list has been quiet. I hope everyone enjoyed
a happy Thanksgiving (or at least a good weekend for non-U.S. readers).
The focus has been on double-checking, by a different mechanism
than that which made the original count, alas in a non-Mersenne context.
If you are trying to factor Mp = 2^p - 1 where p is prime and
p 2^58, then Mp can have no 59-bit factors
[All factors are == 1 (mod 2*p).]
But when factoring a random integer n, we estimate the probability
of a factor between 2^51 and 2^59 as
1 - product (1 - 1/q)
Nathan Russell asks
To start with, thanks to all those who answered my first question.
What I want to know is as follows: If no factors are found for an exponent
after running the needed curves for a 55 digit factor, what is done? Are
additional curves run, or is that exponent put
"Nathan Russell" [EMAIL PROTECTED]
Nathan Russell observes:
This is something I've been toying with the last few days.
I can prove that all even positive integers except 2, 4 and 8 can be written
as the sum of Mersenne primes.
All above 21 are either 0, 1 or 2 mod 3, and are therefore
"Ian L McLoughlin" [EMAIL PROTECTED] asks
Since the list is quiet...
Does a Fibonnacci series contain a finite or an infinite number of primes?
From what I understand..
In a gen.F sequence if the first two numbers are divisible by a prime all
its numbers are divisible by the same prime, if
Daniel Grace [EMAIL PROTECTED] asks
I looked at Chris Caldwells page on Wieferich (1909)
primes but I could not see exaclty how p^2|2^(p-1)-1 relates to
Mersennes with square factors? I can see that Mp=3(2^(p-1)-1).
So my question is this "How does one derive Wieferich's result,
from the
[EMAIL PROTECTED] gives a UBASIC solution to Steinar's problem:
First, perhaps I should explain some notation :-) a_11 is the letter `a',
followed by 11 in subscript. x^2 is the letter `x', followed by the letter
2 in superscript (ie. `x^2' would be mathematically the same as `x*x').
Wojciech Florek [EMAIL PROTECTED] notes:
Hi all!
S.A. Fulling from Texas AM Univ. posted two articles at
the Los Alamos Nat. Lab. archive. The first is on
quantum computers and refers to the second one `abstracted' as
"This is a pedagogical article cited in the foregoing research note"
Alexander Kruppa [EMAIL PROTECTED] observes:
Hi,
the Lucas-Lehmer iteration
L_0 = 4
L_{n+1} = L_n ^2 -2
looks suspiciously like an iteration used in Pollard-Rho:
x_0 = a
x_{n+1} = x_n ^2 +b
Can this be used to do a LL test and Pollard-rho factoring attempt at
once?
I remember
On Fri, Oct 15, 1999 at 01:03:57AM -0800, Gordon Bower wrote:
Here's another argument - suppose the largest unfactored composite was C. How
long did it take to determine the factorization (or primality) of C-2? (C-1
would be even.) Then to factor C would only take a marginally longer amount of
Ernst Mayer writes:
- Hi, Jason:
-
- Or, if you just want to cheapen a complex multiply, turn the standard
- form
-
-xr,xi * yr,ri - xr*yr-xi*yi, xr*yi+xi*yr
-
- into
-
-xr,xi * yr,ri - xr*(yr - xi/xr * yi), xr*(yi + xi/xr * yr)
-
- and precompute xr and xi/xr. Presto! 2
Will Edgingtom writes:
If I understand P-1 factoring correctly, then using it to a stage one
bound of k to try to factor M(p) will find all possible factors less
than or equal to 2*k*p + 1. I'm assuming that p is less than k (or p
is always used in the powering) and the convention several
FAQ author Lucas Wiman [EMAIL PROTECTED] writes
Well, this might not be a problem. If P-1 tries more than one base,
all that is required (per Mn) is *one* gcd. We just multiply the
remainders from the bases mod Mn, and then take the gcd on the final
remainders. We could have the
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, take
7) Anyone that makes a mathematical or algorithmic breakthrough that
speeds up the search process. I'm talking about a doubling in search speed
not a 1% speedup in assembly code.
I think that this would be great -- but I seriously doubt that any
improvement will be found. We can't get
Ken Kriesel [EMAIL PROTECTED] writes:
George asks, what about the bit second from the left?
In total, 26 0 bits vs only 12 1 bits. (0 26 of 38 times or ~68.4%;
26/12=2.1...)
Excluding p=2 and p= 3, to look only at interior bits, 25 0 bits vs
only 11 1 bits; 0 25 of 36 times, ~69.4%;
Peter Foster asks:
When doing an LL test we are always calculating the
same series of numbers. The modulus is different, so
we can't use the result of one test to help with
another. I'm wondering why we don't do the following:
Take 2 Mersenne numbers, m1 and m2 (m1 m2).
Do the usual
Jud McCranie [EMAIL PROTECTED] writes:
At 10:47 AM 7/6/99 +0200, Benny.VanHoudt wrote:
Now lets only focus on the set 2^p - 1 with p prime, i.e., the set
of numbers that we are checking out at GIMPS. Has anyone proven that
an infinite number these are NOT prime (this is VERY likely to be
Problem A3 in Richard Guy's `Unsolved Problems in Number Theory'
includes this question, by D.H. Lehmer:
Let Mp = 2^p - 1 be a Mersenne prime, where p 2.
Denote S[1] = 4 and S[k+1] = S[k]^2 - 2 for k = 1.
Then S[p-2] == +- 2^((p+1)/2) mod Mp.
Predict which
[EMAIL PROTECTED] writes
Seeing as how anyone with even the most rudimentary of Internet searching
skills (i.e. me) can find a publicly available Internet page with a certain
highly important number on it, I ask why it is there. I thought that "those
in the know" were *absolutely not*
Page A-12 of the June 30, 1999 San Francisco Examiner
has a story `Math expert confirms a prime discovery'
repeating the Oregonian story abut M38.
The report does not give the precise exponent.
The report mentions only George Woltman by name.
Peter
lrwiman [EMAIL PROTECTED] wrote:
It has been mentioned several times recently that factoring is all integer
work, and LL testing is nearly all floating point.
It is my understanding that on intel CPU's, these are done on separate parts of
the CPU. Would it increase net performance to do
Chris Nash [EMAIL PROTECTED] wrote:
Define L(n)=(2+sqrt(3))^n+(2-sqrt(3))^n. It turns out that what we usually
call *the* Lucas sequence is just
S(n)=L(2^n).
Yes. We can check S(0) = L(1) = 4 and S(n+1) = S(n)^2 - 2
from the definitions. So S defines the familiar Lucas sequence.
"Nicolau C. Saldanha" [EMAIL PROTECTED] writes
Do you have more precise information, or do you know where we can find it?
(the help function inside maple gives only a very superficial explanation)
Strong pseudoprime wrt which base? Probably not random, or the search
for a counterexample
Observant "Foghorn Leghorn" [EMAIL PROTECTED] asks:
I noticed that P587 is no longer on Geroge's ECM factoring page, but there
is no factor listed. Did someone else find a factor?
Robert Silverman sieved it by SNFS. CWI did the linear algebra.
C(2,587+)
2769467 13119770765051463547
"Benny.VanHoudt" [EMAIL PROTECTED] asks
I was looking at the LL-test and thought of the following:
To succeed a LL-test the final value (mod 2^p - 1) has to be zero.
To be able to get this there has to be a number x (between 0 and 2^p - 2)
such that x^2 = 2. This because in the final step of
"Sander Hoogendoorn" [EMAIL PROTECTED] asks
Can sombody please tell me when a number is completely factored?
The prime factors of 2^29 - 1 are 233, 1103, and 2089.
The equation
2^29 - 1 = 233 * 1103 * 2089
gives the _factorization_ or _complete factorization_ of 2^29 - 1.
"Sander Hoogendoorn" [EMAIL PROTECTED] asks
Last weekend when i was testing Prime 95 i noticed that factoring low
numbers took much longer as high numbers.
Factoring M17XXX from 2^52 to 2^54 took minutes per pass while factoring
M9XX from 2^52 to 2^54 took about one minute to complete
Olivier Langlois comments (in response to others):
First,
let me thank you all for your helpful explanations.
Is there someone that could be able to explain me the reasons that are
behind the 2 conditions (q/N 32 and maximum convolution error is 0.5
for every Lucas-Lehmer
Dienst writes (in response to Ernst Mayer)
If one instead does a fast polynomial-evaluation stage 2, even one
with modest circular convolution length (to keep memory requirements
reasonable), the stage 2 depth for a given amount of work should rise
dramatically, perhaps 2-3 orders of
Can someone tell me the details of how the discreet weighted transform
works, or tell me where to find information on this subject.
It is described in an article by Crandall et al in Mathematics
of Computation earlier this decade.
We can use the FFT to multiply two
Bojan Antonovic writes:
(Somebody else wrote)
Bah! 128 bit floats are quite useful. Think about percision.
There are segments within the 64 bit range where you get terrible decmil
precision. (such as with any large number) This becomes a problem with
scaling large
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