20120920, 08:36  #1 
Banned
"Luigi"
Aug 2002
Team Italia
2×5×479 Posts 
Math
Newbie question.
We all know the limitations that a Mersenne factor should accomplish:  form of 2kp+1  factor prime  1 or 7 mod 8  16 definite residual classes out of 120 (or 910 out of 4620) Are there more/different clauses to follow for double Mersenne numbers? Luigi 
20120920, 09:23  #2 
Jun 2003
2^{3}×5×11^{2} Posts 
Only these are the "true" limitations.
This is not a true limitation. This is just a consequence that the candidate factor be prime (i.e. no small factors). Not that I know of. 
20120920, 09:39  #3 
Romulan Interpreter
Jun 2011
Thailand
3×17×179 Posts 
We only know the first 3. There is no theoretical "jailbreak" since longlong time. The rest of the "rules", and all the modern tricks involved in programs like mfaktc, etc, are consequences of how we combine the first 3 rules, they don't bring anything new. The "classes" stuff are just special conditions we put for k depending on p (and BTW, your numbers are wrong).
For example, we say "if p=1 (mod 4), then".... (from the condition 1 and 3 you get that the factor is either 8xp+1 or 8xp+6p+1). Or "if p=3 (mod 4), then".... (from the condition 1 and 3 you get that the factor is either 8xp+1 or 8xp+2p+1). This is the same as renaming k into 4k and respective 4k+3, the factors 1 (mod 8) will be 8kp+1 and the others viceversa. Adding p=1 (mod 3) and so on (working it mod 5, 7, 11, with 2^2 or 2^3), you get the whole lot of "classes" (like 4 from 12, 32 from 120, 96 from 420, or 960 from 4620, etc), because if k is not in one of the remaining classes, then either condition 2 is not satisfied (q is not prime) or condition 3 is not (for where we used 2^3). At the end everything is the first 3 rules and the chinese reminder theorem: if p is in THIS particular class (mod 4620), then k must be in THAT particular class (mod 4620) to get the right factor satisfying (1+2+3). And because p is prime, it can only be in "eulerphi(4620)" possibilities, which is 960. For each of them, the class of k is "computed". Well, somehow. Last fiddled with by LaurV on 20120920 at 09:42 
20120920, 14:37  #4  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20120920, 19:33  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
to clear things up P=2^p1 , second a thing I saw a while ago:
MM_{x} = MM_{x1}*(2^{2[SUP]x1}[/SUP]+2)+1 sorry just realized that the 2nd 2 is not supposed to be in the exponent Last fiddled with by science_man_88 on 20120920 at 19:40 
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