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Old 2003-05-16, 14:41   #1
ron29730
 

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Default The first (non-merseinne) 10 million-digit prime number!!!

If this qualifies for the contest for $100,000, then let GIMPS do with the money as they see fit.

Since by adding 1 to N! gives a prime number (N! + 1 cannot be factored by any number =<N), this is a simple matter of finding a factorial which has 10 million digits - the first of which being 1,737,441. This gives a prime of 10 million and 1 (I think) digits, with the last digit a 1. (Still computing the value, could take a couple of months :().

1,737,411!+1 = xxx...xx 1.

If, however, you need a Merseinne prime, then I have none.

Good luck in the search!!!
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Old 2003-05-16, 14:55   #2
eepiccolo
 
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Well, you have to look at all numbers <= to N!+1 to try to find a factor, not just numbers <=N.
Ex) 4! + 1 = 2*3*4 + 1 = 25, and 5 divides 25.

In fact, I'm not even sure if a prime number exists of the form N! + 1 for N >= 4.

However, in reference to the other part of your post, a 10M digit prime does not need to be a Mersenne prime to qualify for the prize.

EDIT: 11! + 1 is prime, but who knows if there are a finite numbers of primes in this form or not? (I'm sure someone knows.)
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Old 2003-05-16, 15:24   #3
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OK, here's the link for Factorial prime information:
http://www.utm.edu/research/primes/l...Factorial.html
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Old 2003-05-19, 23:11   #4
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Default Re: The first (non-merseinne) 10 million-digit prime number!

Quote:
Originally Posted by ron29730
Since by adding 1 to N! gives a prime number (N! + 1 cannot be factored by any number =<N)
Simply because (N! + 1) (or (N! - 1), for that matter) is not divisible by any prime <= N does not make it prime. Even if you modify the factorial and instead use the product of all primes <= N (a la Euclid in his beautiful proof of the infinitude of primes), N! +- 1 need not be prime, e.g. (2*3*5*7*11*13+1) = 59*509 and (2*3*5*7*11*13*17+1) = 19*97*277.

Now, if one knew every prime <= R (the current record holder), one could form the product of all these, add or subtract one, and the result would be guaranteed to have no factors <= R, i.e. one would have implicitly found a new record-size prime. One can do similar stuff like this with Mersennes: if R is the largest-known Mersenne prime, then 2^R - 1 has no factors < 2*R+1,
i.e. is either an even bigger prime or decomposes into factors bigger than R. But these types of constructions don't qualify for record-prime status: that requires an EXPLICIT prime. Looks like the EFF gets to hold on to their $100K for a little while longer.
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Old 2003-06-07, 21:34   #5
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Yeah it's harder than that you'd probably have to try Prime95 on that exponent.
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Old 2004-02-25, 10:28   #6
mfgoode
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Default Factorial primes

Quote:
Originally Posted by eepiccolo
Well, you have to look at all numbers <= to N!+1 to try to find a factor, not just numbers <=N.
Ex) 4! + 1 = 2*3*4 + 1 = 25, and 5 divides 25.

In fact, I'm not even sure if a prime number exists of the form N! + 1 for N >= 4.

However, in reference to the other part of your post, a 10M digit prime does not need to be a Mersenne prime to qualify for the prize.

EDIT: 11! + 1 is prime, but who knows if there are a finite numbers of primes in this form or not? (I'm sure someone knows.)
See Wilsons Theorem and Corollary on link
http://mathworld.wolfram.com/WilsonsTheorem.html
Mfgoode
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Old 2004-02-25, 11:07   #7
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It is not always true that (2^M)-1, where M is a mersenne prime, is itself prime. MM15 and MM31 are not prime.
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Old 2004-02-26, 23:26   #8
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Nice try Ron.

It would be nice to win $100K so keep at it.
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Old 2004-02-27, 23:20   #9
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Default Not so fast

Quote:
Originally Posted by Digital Concepts
Nice try Ron.

It would be nice to win $100K so keep at it.



It appears by a limiting technique on an algorithm of finding a kp=2^(p-1)-1,
(try finding k of kp=2^(p-1)-1, with 5,7,11,13,...by a method of accumulating a continuous sum of powers of two for the product, and step by step, filling in the powers of two for k. The limit involves p.)
ie fermat test , that 10,001,631 is prime, unless pseudo prime:
Anyone wish to find a witness,etc., or show (10001630)!/(10001631) is even,
if no witness occurs(tough) or the above division is not even, then the number IS prime.
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Old 2004-02-28, 08:17   #10
Cyclamen Persicum
 
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Вut (10^10,000,000)!+1 defenetly contains a prime factor > 10,000,000 digits long.
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Old 2004-02-28, 15:41   #11
mfgoode
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Default non mersenne primes.

Quote:
Originally Posted by jinydu
It is not always true that (2^M)-1, where M is a mersenne prime, is itself prime. MM15 and MM31 are not prime.
:



You have brought up an interesting point on Mersenne primes.
However your comment bears no relevance to the topic under discussion. Ron clearly states about “non mersenne” primes and primes formed from factorials.
That’s why I have referred him to Wilsons theorem and given the link to explore further.
As a ready reference Wilsons theorem states that for any prime one has the formula
(p-1)! = -1 (mod p). This is not true if p is composite and must be prime.
For larger primes this formula is not practical and involves a lot of computation even for a computer. That’s why Mersenne prime formulae are preffered over Wilsons. At the same time Wilsons theorem is both necessary and sufficient for primality. As the number of primes is infinite and this formula involves primes it gives an infinite number of results.

Mally.
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