 mersenneforum.org > Math infinitely many Mersenne primes???
 Register FAQ Search Today's Posts Mark Forums Read  2003-10-01, 15:27 #1 LoKI.GuZ   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???   2003-10-20, 22:27 #2 ixfd64 Bemusing Prompter   "Danny" Dec 2002 California 1001010011102 Posts I've been wondering that too...   2003-10-21, 02:37 #3 only_human   "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).   2003-10-21, 04:30   #4

"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 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:
 Originally posted by wblipp. From the prime number theorem, we know the probablity that numbers near the value "x" are prime is about 1/ln(x). So the probability that numbers near (2^p-1) are prime is about (1/(p*ln(2)). If we add this up for all primes, the expected number of Mersenne primes would be about (1/ln(2)) * [SUM(1/p)]. Some adjustments are appropriate because of known relationships such as M(p) is never divisible by 3, but this still leaves an estimate of the form c*[SUM(1/p)]. We know the sum increases without bound, so we expect there to be an infinite number of Mersenne primes.
Quote:
 Originally posted by LoKI.GuZ. What does this harmonic series mean??
The harmonic series referred to is SUM(1/p).   2003-10-21, 12:20   #5
TauCeti

Mar 2003
Braunschweig, Germany

2·113 Posts Quote:
 From the prime number theorem, we know the probablity that numbers near the value "x" are prime is about 1/ln(x). So the probability that numbers near (2^p-1) are prime is about (1/(p*ln(2)). If we add this up for all primes, the expected number of Mersenne primes would be about (1/ln(2)) * [SUM(1/p)]. Some adjustments are appropriate because of known relationships such as M(p) is never divisible by 3, but this still leaves an estimate of the form c*[SUM(1/p)]. We know the sum increases without bound, so we expect there to be an infinite number of Mersenne primes.
But that reasoning is only helpful, if the distribution of primes is in a sense 'random'. And that is the million dollar question. The prove or disprove of the Riemann hypothesis (RH) could lead to insights about that problem.

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 error-term) are expected to be prime sometimes.

Tau

--
first not real part one half   2003-10-31, 23:59 #6 clowns789   Jun 2003 The Computer 23·72 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?   2003-11-01, 05:39 #7 Val   201358 Posts All mersenne numbers fall on the diagonal, when using only odd numbers for the spiral. That dont help much!  2003-11-01, 09:07 #8 andi314   Nov 2002 2·37 Posts What is the Ulam Spiral??? greetz andi314   2003-11-01, 17:03   #9
only_human

"Gang aft agley"
Sep 2002

72528 Posts Quote:
 Originally posted by cheesehead See earlier thread How do we know there is another Mersenne prime? at http://mersenneforum.org/showthread.php?s=&threadid=753 There, wblipp posted: From the prime number theorem, we know the probablity that numbers near the value "x" are prime is about 1/ln(x). So the probability that numbers near (2^p-1) are prime is about (1/(p*ln(2)). If we add this up for all primes, the expected number of Mersenne primes would be about (1/ln(2)) * [SUM(1/p)]. Some adjustments are appropriate because of known relationships such as M(p) is never divisible by 3, but this still leaves an estimate of the form c*[SUM(1/p)]. We know the sum increases without bound, so we expect there to be an infinite number of Mersenne primes.
This bothers me because it can also be used as an expectation that there are an infinite number of Fermat primes (assuming c is anything greater than 0).

Last fiddled with by only_human on 2003-11-01 at 17:08   2003-11-01, 17:04   #10
only_human

"Gang aft agley"
Sep 2002

2×1,877 Posts Quote:
 Originally posted by andi314 What is the Ulam Spiral???
Here is an explanation:Ulam spiral - Wikipedia   2003-11-10, 22:54 #11 philmoore   "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^p-1 (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.)   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post wildrabbitt Math 8 2016-06-28 08:15 PawnProver44 Homework Help 1 2016-03-15 22:39 aketilander Operazione Doppi Mersennes 1 2012-09-19 22:24 devarajkandadai Math 1 2012-07-13 06:07 optim PrimeNet 13 2004-07-09 13:51

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