On 10 Jul 99, at 14:57, Peter Foster wrote:
> 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 a
> Take 2 Mersenne numbers, m1 and m2 (m1 < m2).
> Do the usual LL series, but use as the modulus m1*m2.
> At the appropriate step, check if the remainder is
> divisible by m1. If so, then m1 is prime.
> At the end, check if the remainder is divisible by
> m2. If so, then m2 is prime.
> This allow
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
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 LL series, but use as the m
Hello,
> thus the numbers can never be the same above one
> when the mersenne number mod 4 is always 3 and
> the odd square mod 4 is always 1.
Your analysis is correct but your conclusion is wrong.
You've proved that all Mersenne numbers greater than 1 cannot be odd
squares but you haven't pro
Mersenne Digest Friday, July 9 1999 Volume 01 : Number 595
--
Date: Thu, 8 Jul 1999 00:52:45 -0400 (EDT)
From: Lucas Wiman <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Repeating LL reminder
> I know it's a
--- Kris Garrett <[EMAIL PROTECTED]> wrote:
> If it has not been proven that all mersenne numbers
> greater than one is
> prime free then here is a proof for you.
>
> 2^n-1 mod 4 = 3 while n > 1 because every 2^n while
> n > 1 is divisible by 4 itself.
>
> On the other hand every odd square mo
Kris Garrett writes:
Has it been proven that all mersenne numbers greater than one are
square free?
Depends on your definition of 'Mersenne numbers'. If you include
composite exponents, M(6) = 2^6 - 1 = 63 = 3*3*7 is not square free.
If you include only prime exponents, then you don't ne
If it has not been proven that all mersenne numbers greater than one is
prime free then here is a proof for you.
2^n-1 mod 4 = 3 while n > 1 because every 2^n while
n > 1 is divisible by 4 itself.
On the other hand every odd square mod 4 is always 1
because:
if the odd number mod 4 is 1 then:
n
At 06:51 PM 7/9/99 +0100, Brian J. Beesley wrote:
>For reasonably small multi-precision numbers, Head's method is
>actually very good, if you're working on a true RISC processor with
>no integer multiply instruction.
I started using Head's algorithm in 1981 on my Apple II. It was better
than
At 10:16 AM 7/9/99 -0700, Kris Garrett wrote:
>Has it been proven that all mersenne numbers greater than one are
>square free?
As far as I know, it has not been proven (and no repeated factors are known
either).
+--+
| Jud "program first and think late
On 8 Jul 99, at 23:33, Lucas Wiman wrote:
> >That is going to be a *lot* slower than FFT convolution, for numbers the size
> >of the Mersenne numbers we're testing! FFT is O(n*log(n)) where n is the
> >number of limbs in the numbers being multiplied. Head's method is O(n^2),
> >with O being sli
Has it been proven that all mersenne numbers greater than one are
square free?
_
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