20031001, 15:27  #1 
Sep 2003
Brazil
2·7 Posts 
infinitely many Mersenne primes???
found this on http://www.utm.edu/research/primes/mersenne/:
" Are there infinitely many Mersenne primes? Equivalently we could ask: Are there infinitely many even perfect numbers? The answer is probably yes (because the harmonic series diverges). " What does this harmonic series mean?? iis it SUM 1 / n2 as n goes all the way through infinity?? what does this have to deal with mersenne primes??? 
20031020, 22:27  #2 
Bemusing Prompter
"Danny"
Dec 2002
California
100101001110_{2} Posts 
I've been wondering that too...

20031021, 02:37  #3 
"Gang aft agley"
Sep 2002
2·1,877 Posts 
This is interesting to me too. I can only understand these things somewhat as a spectator reading popularized explanations. I understand that the harmonic series diverges. I see explanations that the sum of the reciprocals of all primes diverges. I can understand that they might be talking about the former and thinking about the latter. What I don't see is how this applies to Mersenne primes. I know I looked at that page before and never gave it any thought...until now (that you mention it).

20031021, 04:30  #4  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
See earlier thread How do we know there is another Mersenne prime? at http://mersenneforum.org/showthread.php?s=&threadid=753 There, wblipp posted:
Quote:
Quote:


20031021, 12:20  #5  
Mar 2003
Braunschweig, Germany
2·113 Posts 
Quote:
The prime number theorem (PNT) only states, that the limit of the quotient of the Prime counting function and x/ln(x) is 1 as x approaches infinity but is does not tell us, if there are regions where primes are 'dense' or 'sparse' IF the RH is NOT true, that would implicate, that primes are NOT randomly distributed but there exist oscillations in the distribution. In that case, mersenne numbers could evade 'primeness' depending on the kind of those oscillations aven if they statistically (regarding the PNT without the errorterm) are expected to be prime sometimes. Tau  sweet taste of madness first not real part one half chance admits defeat 

20031031, 23:59  #6 
Jun 2003
The Computer
2^{3}·7^{2} Posts 
I heard the Mersenne primes are on the diagonals of the Ulam Spiral. Can you just spin through that until you get to a diagonal?

20031101, 05:39  #7 
20135_{8} Posts 
All mersenne numbers fall on the diagonal, when using only odd numbers for the spiral.
That dont help much! 
20031101, 09:07  #8 
Nov 2002
2·37 Posts 
What is the Ulam Spiral???
greetz andi314 
20031101, 17:03  #9  
"Gang aft agley"
Sep 2002
7252_{8} Posts 
Quote:
Last fiddled with by only_human on 20031101 at 17:08 

20031101, 17:04  #10  
"Gang aft agley"
Sep 2002
2×1,877 Posts 
Quote:


20031110, 22:54  #11 
"Phil"
Sep 2002
Tracktown, U.S.A.
3×373 Posts 
I think this was posted previously in another thread, but since I didn't find it, I'll apply wblipp's argument to Fermat numbers: Assume that the probability that a Fermat number Fn = 2^(2^n)+1 is prime is proportional to 1/ln(Fn), approximately equal to 1/(ln(2)*2^n). Then the expected number of Fermat primes is c/ln(2) times SUM(1/2^n) as n goes from zero to infinity. Because this sum is finite (being a geometric series), the expected number of Fermat primes is also finite. The difference with Mersennes is that there are many more Mersenne prime candidates, whereas Fermat prime candidates are much sparser.
Actually, Crandall, Papadopoulos, and Mayer give a formula in their paper for the approximate "probability" that a Fermat (or Mersenne) number is prime: (Probability) approx = (e^(gamma)*ln(B))/(ln(2)*exponent) where gamma is the Euler constant and the exponent is p for 2^p1 (Mersenne numbers) or 2^n for 2^(2^n)+1 (Fermat numbers), and B is the lower bound (i.e., the search limit) on any possible factors. Using the search limits given at www.fermatsearch.org we find a "probability" of F33 = 2^(2^33)+1 being prime to be approximately 1 in 57,000,000. The probability for F34 is only about half of that, and F35 half that again. Since all Fermat numbers up to F32 are known to be composite, we can estimate the "probability" of there being even one more Fermat prime as being around one in 32,000,000. (I put "probability" in quotes because the actual probability is either zero or one.) 
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