20110407, 12:53  #1 
Sep 2009
2^{2}×3^{2} Posts 
studies on largest prime factor ?
Dear All,
I am looking for papers on the largest prime factor of a mersenne number. As far as I know the largest prime factor has to be less than the square root of the mersenne number. Are there any tighter bounds? Thanx in advance 
20110407, 12:57  #2  
Banned
"Luigi"
Aug 2002
Team Italia
4848_{10} Posts 
Quote:
2^111 = 2047 = 23 x 89 and 89 > sqrt(2^11  1) Otherwise, the upper limit of trialfactoring for a number is sqrt(n). Mersenne factors have the property of being of the form 2kp+1 (k=1,2,3... and p= the exponent of the number), so the upper limit may as well be lowered to 2kp+1 < sqrt(p) (where we have 2kp+1 < sqrt(p) < 2(k+1)p+1) There are other heuristics, like the factor should be 1 or 7 mod 8, or [16 different values] mod 120, or [more different values] mod 420 (or 4620). Those values for the factor automatically rule out multiples of 3, 5, 7, 11. Luigi Last fiddled with by ET_ on 20110407 at 13:06 Reason: Edited after Science_Man_88 post 

20110407, 13:01  #3 
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
not completely accurate the largest can be more but to not be prime it must have one below the sqrt. all factors of prime exponents mersennes must be of the form 2kp+1 and also being 8x1 and 6y1 ( and yes I may be missing something).

20110407, 13:37  #4 
"Forget I exist"
Jul 2009
Dartmouth NS
10000011100010_{2} Posts 
http://www.google.ca/search?sourceid...rsenne+numbers is what I did and the last one on my first page of results seems good.

20110407, 16:41  #5  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·937 Posts 
Quote:
Why do people seem to think that they are special? NO number (trivially) can have two prime factors greater than its square root. Quote:


20110407, 20:57  #6  
Banned
"Luigi"
Aug 2002
Team Italia
12F0_{16} Posts 
Quote:
In the first answer I specified Mersenne numbers because that was the question. You are right when you say that the same apply to all numbers. And about my use of the term heuristics... I used that term when I first approached the problem, many years ago. Now I should have learnt that "Theorem" is the right word: thank you for your making it even clearer to everybody. Finally, I am glad that your answer dealt about terms use, and not about trivial math errors in my answer. Luigi Last fiddled with by ET_ on 20110407 at 20:59 Reason: My keyboard often hides vowels... 

20110418, 17:24  #7  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 
Quote:
you proofread your posts before submitting them? 

20110418, 17:27  #8  
"Bob Silverman"
Nov 2003
North of Boston
1110101001000_{2} Posts 
Quote:
claims before posting? 15 seconds of checking factorizations of 2^n1 (from published tables) would have revealed that your claim about the largest prime factor being less than the square root was nonsense. Finding counterexamples is trivial. 

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