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Old 2004-03-01, 16:50   #12
mfgoode
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Default 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.
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Old 2004-03-01, 18:18   #13
rogue
 
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[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?
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Old 2004-04-06, 11:53   #14
Cyclamen Persicum
 
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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...
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Old 2004-04-08, 03:47   #15
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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.
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Old 2004-04-16, 14:21   #16
Cyclamen Persicum
 
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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.
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Old 2004-05-02, 05:26   #17
ixfd64
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Most likely a troll. Notice how he only has one post?
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Old 2004-05-15, 20:23   #18
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Default 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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