20100130, 08:32  #45 
Jan 2010
379 Posts 
Can anyone tell me how do I connected to that scientist who manipulated?

20100130, 10:18  #46 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10011111010010_{2} Posts 
Offtopic stuff about standards in the use of English has been moved, as requested. Anyone who wishes to continue can head over to
http://www.mersenneforum.org/showthread.php?t=13033 Paul 
20100130, 11:30  #47  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5,879 Posts 
MiniGeek already told you.
Quote:


20100130, 18:04  #48 
Jan 2010
379 Posts 
Yes I know that mini greak told me how to do it, but I didn't understand what he said.

20100130, 18:26  #49 
Jan 2010
573_{8} Posts 
n: natural number.
e(m)<: the smallest odd element of m. p: prime element of merssen numbers with an odd exponentiation. e(p1)<: the smallest odd element that accepts: . The problem for me is to explain what is the p elements. I can say that 7,23,31.. are examples of it. If we put them in p: Other ideas about the elements are: 1)The smallest element of p1 which accepts is allways odd. 2)If the smallest element of p1 that accepts isn't an odd number, p isn't an element of merssen numbers with an odd exponential. I know, I didn't write everything legal and understandable, but that what I can do here in English. Please write a non petty answer and be nice.. ( I mean, not bad and not neutral). Last fiddled with by blob100 on 20100130 at 18:49 
20100130, 18:58  #50 
Dec 2008
Boycotting the Soapbox
2·5·61 Posts 
Now we got you, you ... you nonkosherophagian gentile! You are attempting to make all israelians, Jews and other geniuses look like total morons, right?

20100130, 19:00  #51 
Jan 2010
379 Posts 
what?
I just put my conjecture here and want to get a review if it was found before... 
20100130, 20:57  #52 
Jan 2010
573_{8} Posts 
Here I published my conjecture and no one helps me :(

20100130, 21:42  #53  
Nov 2003
7232_{10} Posts 
Quote:
Do you mean that p is a prime such that 2^p1 is also prime i.e. p is a Mersenne exponent? Quote:
Do you really mean "greater than" in you use of ">"? Do you mean that e(p1) is the order of 2 mod p? If you don't know what the order of an element in a group is, then please look it up. The order of a mod p is the smallest integer x such that a^x = 1 mod p. Or do you really mean that e(p1) is the smallest integer x such that 2^x > p, where ">" really means greater than? This can't be what you mean. It would make no sense. You only write "> p", and I see no modular notation here. Quote:
equals 1 mod 7 for all natural n. It never equals 1 mod 7. Similarly 2^(11n) = 1 mod 23 for all n and 2^5n = 1 mod 31 for all n. Quote:
Are you asking whether for all primes p, the smallest integer x such that 2^x > p always has x odd? The answer to that is trivially no. Quote:
is always odd. Please, please please, read abook on number theory. Follow and use its notation. I do not know what you mean by "accepts". I think I understand what you mean by "smallest element of p1". Do you mean the SET of integers [1, p1]? Have you studied sets at all? Quote:
than your English. I have not used it in 40 years. Francais, peutetre? 

20100130, 21:45  #54 
Nov 2003
2^{6}·113 Posts 

20100131, 02:05  #55  
Cranksta Rap Ayatollah
Jul 2003
1010000001_{2} Posts 
Quote:
So, for , because ? and for , because ? So is your conjecture that for a prime p, if e is the smallest factor of p1 such that and e is even, then p does not divide for all primes q > 2? Last fiddled with by Orgasmic Troll on 20100131 at 02:06 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Some ideas regarding NFS...  paul0  Factoring  3  20150314 19:55 
Ideas for the future beyond justkeepencrunching  Dubslow  NFS@Home  13  20150202 22:25 
two ideas for NPLB  MiniGeek  No Prime Left Behind  16  20080301 23:32 
GROUP IDEAS  TTn  15k Search  15  20030923 16:28 
Domain name ideas...  Xyzzy  Lounge  17  20030324 16:20 