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 2004-03-01, 16:50 #12 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 Posts Strange symbol [QUOTE=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. Pray, tell me what this strange symbol < means? Mally.
2004-03-01, 18:18   #13
rogue

"Mark"
Apr 2003
Between here and the

26×3×31 Posts

[QUOTE=mfgoode]
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. Pray, tell me what this strange symbol < means? Mally.
I believe that the intention is the 'less than' symbol. The factors of N!+1 are all greater than N, but less than or equal to sqrt(N!+1).

So who can find the first factor of 1,737,411!+1?

 2004-04-06, 11:53 #14 Cyclamen Persicum     Mar 2003 34 Posts Why the 10,000,000 digits prime is expected to be Mersenne? As to Generalized Fermat b^(2^n)+1, they are much more and more frequient, raise faster due to base and superexponent, and also quick tested - one can use WDT and so one... However, it seems the largest pseudo-Fermat found is about only 500,000 digits... This is because of lack of public spinup comparing to GIMPS...
2004-04-08, 03:47   #15
Maybeso

Aug 2002
Portland, OR USA

1000100102 Posts

Quote:
 Why the 10,000,000 digits prime is expected to be Mersenne?
The main reason is that the L-L primality test that works for Mersenne numbers is much faster than tests that work for GF numbers.

 2004-04-16, 14:21 #16 Cyclamen Persicum     Mar 2003 34 Posts the L-L primality test that works for Mersenne numbers is much faster than tests that work for GF numbers. I dont' think so. I believe that all classical N +/- 1 tests have the same laboriousness. For instance, a*2^n+1 requires only one powermod to be proved, i.е. about n squares. I don't know if WDT suits here, but for b^n+/-1, i.e. for GF, it suits.
 2004-05-02, 05:26 #17 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 22·577 Posts Most likely a troll. Notice how he only has one post?
2004-05-15, 20:23   #18
David John Hill Jr

Jun 2003
Pa.,U.S.A.

110001002 Posts
probably not prime.

Quote:
 Originally Posted by David John Hill Jr 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.
After suggesting this number, I checked with the Caltech runoff of primes, and it is not listed.How reliable the numbers are ,I don't know, nor how they are calibrated.
What I was doing, having noticed that when composite,c|(c-1)! with
anywhere down to about 1/3 of the number not as zero's as 2 divides, was to pick
a number that would multiply with all zeros cancelled when from its half point
defined,
as to 10001630, as =(a)(a).
Then if a is composite (a+1)|(a-1)! or as 10001631|(10001629)! with anywhere up to but not much greater than 2/3 of the numbers being a zero string.
As a preliminary check, or prime to be found in a region of numbers(as in just above 10000000), it appears as a first step, logical, .......though not a guarantee of primeness, unless it can be shown the 2 division is definitely
impossible with a whole number result, in looking for primes.Of course if the number given
is not composite, it is .............prime.It might be possible to work this method in tandem with the prime number theorem, for detail accuracy,
especially when fourier tranforms are producing ambiguity.
I am sure this approach , as aligning fermat with Wilson, has potential.

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