20130927, 09:40  #1 
Feb 2012
Prague, Czech Republ
3·67 Posts 
The beer conjecture
For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq1 is divisible by each prime up to and including p [but not by the next prime].
With/without bracketed text == strong/weak version. 
20130927, 11:57  #2 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10279_{10} Posts 
This is trivial, because for every odd number n, there exists a p such as 2^p1 is divisible by that n (and it is easy to find!). All multiples of p have the same property. Just take n to be x# (the primorial of your mersenne prime).
ex: 2^x1 is divisible by: 3 for x=2, 4, 6, 8, 10, etc 5 for x=4, 8, 12... 7 for x=3, 6, 9... 9 for x=6, 12, 18... 11 for x=10, 20, 30... 13 for x=12, 24, 36... ... 23 for x=11, 22, 33... ... 75 for x=20, 40, 60... .... 256641 for x=3300, 6600, etc and not for other smaller x'es edit: 100280245065 (product of the first 10 odd primes) for x=27720 Last fiddled with by LaurV on 20130927 at 12:16 
20130927, 12:01  #3 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts 
(Mq)1 is a power of 2. It can't have any prime factors besides 2. M(q1) is one less than a power of 2, it can't have 2 as a prime factor. So, basically, no.

20130927, 12:07  #4 
Feb 2012
Prague, Czech Republ
3×67 Posts 

20130927, 12:13  #5  
Feb 2012
Prague, Czech Republ
11001001_{2} Posts 
Quote:


20130927, 12:43  #6 
Romulan Interpreter
"name field"
Jun 2011
Thailand
19·541 Posts 
What Mp has to do here? Your gibberish can be formulated as "for any p, there is a Mq such as Mq1 is divisible by all primes up to p". That is what I shown you, and not only for primes but for ANY ODD number. Mq1 is 2 multiplied with an odd number. This is still very trivial.
Last fiddled with by LaurV on 20130927 at 12:44 
20130927, 13:22  #7  
Feb 2012
Prague, Czech Republ
3·67 Posts 
Quote:
Anyway, wrt "What Mp has to do here?": First of all, the strong version hints that there may be no solution (no such q) for eg. p == 11 because M11 in not prime. (And indeed no known solution exists ATM for p == 11 in the strong version). Secondly, but perhaps more importantly: For p in {2, 3, 5}, smallest q's are {2, 3, 5}. For p > 5, smallest q > p (eg. p = 7 > q > 13, etc). IOW, the conjecture should be equivalent to "There are infinitely many Mersenne primes." because for every Mersenne prime above M5 the conjecture "generates"/claim existence of at least one another, but larger Mersenne prime than is the "generating" one. Apply recursively And to be clear, by Mx1 I mean (Mx)1, but up to now I thought there's no ambiguity in that. My teachers and WolframAlpha seem to agree on this [0]. [0]: http://www.wolframalpha.com/input/?i...+%282%5Ex%291 PS: Preemptive disclaimer: Conjecture implies unproven, right? Last fiddled with by jnml on 20130927 at 13:34 Reason: s/twice/three times/ 

20130927, 16:35  #8  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:
I think what jnml is saying is assume the Mersenne Prime exponents are an infinite set S: ie. S= {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951,...} for p and q both in S the following relation holds by conjecture: q#(2^q2) for at least one s,q pair. so p in your example would have to be a Mersenne prime exponent to fit jnml's conjecture. Last fiddled with by science_man_88 on 20130927 at 16:36 

20130927, 21:52  #9 
"Forget I exist"
Jul 2009
Dartmouth NS
20E2_{16} Posts 

20130928, 10:04  #10  
"Brian"
Jul 2007
The Netherlands
2·11·149 Posts 
Quote:
However I'm intrigued to know why you write this conjecture. There must be all sorts of such strong statements about mersenne prime exponents whose truth or falsehood is out of reach of being established. Do you have reason to believe this particular statement? Quote:
Last fiddled with by BrianE on 20130928 at 10:14 

20130928, 12:08  #11  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:
Code:
? for(w=1,10,d=prod(x=1,w,prime(x));forprime(y=2,1200*w^2,if((2^y2)%d==0,print(w","y);break()))) 1,2 2,3 3,5 4,13 5,61 6,61 7,241 8,1801 9,19801 10,55441 Last fiddled with by science_man_88 on 20130928 at 12:14 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
This conjecture may be useful.  reddwarf2956  Prime Gap Searches  2  20160301 22:41 
Conjecture  devarajkandadai  Math  13  20120527 07:38 
Anyone for a beer tasting?  davieddy  Lounge  7  20120310 20:21 
A New Conjecture  AntonVrba  Math  19  20050726 12:49 
"99 Bottles of Beer" in hundreds of languages  ixfd64  Programming  4  20050329 02:59 