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Old 2005-04-05, 18:08   #1
Vijay
 
Apr 2005

2616 Posts
Thumbs up Hello, I'm new to discovering Mersenne primes

Hello, I've only just joined mersenne.org to help discover further mersenne primes, as I felt my computer to be quite powerful and it seemed it needed exercise.
Personally I'm a mathematician.
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Old 2005-04-05, 18:16   #2
ET_
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"Luigi"
Aug 2002
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Quote:
Originally Posted by Vijay
Hello, I've only just joined mersenne.org to help discover further mersenne primes, as I felt my computer to be quite powerful and it seemed it needed exercise.
Personally I'm a mathematician.
Welcome to the group, Vijay!

Luigi
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Old 2005-04-05, 21:17   #3
moo
 
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Jul 2004
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welcome everyone to welcome Vijay i have made 3 yes 3 danceing bannas
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Old 2005-04-05, 21:38   #4
Primeinator
 
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"Kyle"
Feb 2005
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Welcome Vijay!
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Old 2005-04-06, 07:34   #5
blackguard
 
Jan 2005
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Welcome ! The more the merrier
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Old 2005-04-06, 08:21   #6
mfgoode
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Thumbs up Hello I'm new to discovering Mersenne primes

Welcome Vijay! Nice to see a familiar name.
We wont let you down
Mally
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Old 2005-04-06, 11:23   #7
Vijay
 
Apr 2005

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Default Hello Everyone!

Thank you for a warm welcome everyone!
I'm sure with our raising computer power, the next M prime is not to far before
it is uncovered
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Old 2005-04-06, 13:19   #8
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by Vijay
Thank you for a warm welcome everyone!
I'm sure with our raising computer power, the next M prime is not to far before
it is uncovered
Not quite.

Although a proof is lacking, there are heuristical arguments to suggest
that the number of Mersenne primes between 2^n and 2^(2n) should
be (asymptotically) exp(gamma), where gamma is the Euler-Mascheroni
constant. i.e. there should be a little more than one Mersenne prime
in the exponent range from n to 2n, independent of n [on AVERAGE]

Each Mersenne prime test on M_p = 2^p-1 requires p multiplications mod
M_p. Each multiplication takes time O(p log p loglog p) via a weighted
discrete Fourier Transform. Thus, each test requires O(p^2 log p loglog p)
work. Given a newly discovered Mersenne prime M_p, the *expected*
work to then find the next one is

O( integral from p to 2p of x^2 log x loglog x dx)

For sufficiently large x we have:

O(x^2 log x loglog x) ~ x^(2+o(1))

Thus, the above integral can be approximated by:

int from p to 2p of x^(2 + o(1)) dx and this is at least 8 times the
work to find the previous Mersenne prime.

This project has been very lucky in its last two successes. We should expect
that every new Mersenne prime we find will be "roughly" an order of magnitude harder than the previous one.
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Old 2005-04-06, 15:01   #9
mfgoode
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Thumbs up Hello I'm new to discovering Mersenne primes

[QUOTE=R.D. Silverman]Not quite.

WoW Bob! that was a great introduction. Please keep it up.
Mally
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Old 2005-04-06, 19:38   #10
Vijay
 
Apr 2005

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Red face I see...

Ok, so by the equation you feel that it will take time finding the next prime (<<That rhymes!)

But the thing is, its not too long before you'll see some kind of patterns emerging in the mersenne exponents. The cumbersome way of trying to find primes by testing each exponents, will be replaced by an equation that will calculate the exponent itself.

I just know it!

Currently, I'm working on calculating the face of God by integrals
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Old 2005-04-06, 19:41   #11
akruppa
 
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So, does god converge?

Alex
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