The beer conjecture

[jnml]

For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime].

With/without bracketed text == strong/weak version.

[LaurV]

This is trivial, because for every odd number n, there exists a p such as 2^p-1 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^x-1 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

[TimSorbet]

Originally Posted by jnml

For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime].

With/without bracketed text == strong/weak version.

(Mq)-1 is a power of 2. It can't have any prime factors besides 2. M(q-1) is one less than a power of 2, it can't have 2 as a prime factor. So, basically, no.

[jnml]

Originally Posted by Mini-Geek

(Mq)-1 is a power of 2. It can't have any prime factors besides 2. M(q-1) is one less than a power of 2, it can't have 2 as a prime factor. So, basically, no.

I'm not sure if I understood the above correctly, but AFAICS, Mq-1 is a power of 2 iff q is 2.

[jnml]

Originally Posted by LaurV

This is trivial, because for every odd number n, there exists a p such as 2^p-1 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^x-1 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...
...
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

Mq-1 is (Mq)-1 is (2^q-1)-1 is 2^q-2. I don't know how numbers of such form relate to your analysis of numbers of the form 2^x-1.

[LaurV]

What Mp has to do here? Your gibberish can be formulated as "for any p, there is a Mq such as Mq-1 is divisible by all primes up to p". That is what I shown you, and not only for primes but for ANY ODD number. Mq-1 is 2 multiplied with an odd number. This is still very trivial.

[jnml]

Originally Posted by LaurV

What Mp has to do here? Your gibberish can be formulated as "for any p, there is a Mq such as Mq-1 is divisible by all primes up to p". That is what I shown you, and not only for primes but for ANY ODD number. Mq-1 is 2 multiplied with an odd number. This is still very trivial.

I'm sorry you find something gibberish in a simple sentence of mine. However, I have no idea where the gibberish is to be found. Can you please elaborate a bit? BTW, are you sure you're considering the word 'prime' in the OP? Occurs three times, actually. IOW, both Mp and Mq are prime numbers and that's, being said explicitly, the only domain of the conjecture.

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 Mx-1 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%29-1

PS: Preemptive disclaimer: Conjecture implies unproven, right?

[science_man_88]

Originally Posted by LaurV

This is trivial, because for every odd number n, there exists a p such as 2^p-1 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^x-1 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

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^q-2) 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.

[science_man_88]

Originally Posted by science_man_88

q#|(2^q-2) 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.

sorry the first q should be s.

[Brian-E]

Originally Posted by jnml

For every Mersenne prime Mp there exists at least one Mersenne prime Mq such that Mq-1 is divisible by each prime up to and including p [but not by the next prime].

With/without bracketed text == strong/weak version.

I found the original statement of your conjecture perfectly clear, and it's surprising that it had to be clarified by science_man_88.

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?

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

This is perhaps your intended justification, but I see no reason why this should work for all p.

[science_man_88]

Originally Posted by Brian-E

I found the original statement of your conjecture perfectly clear, and it's surprising that it had to be clarified by science_man_88.

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?

This is perhaps your intended justification, but I see no reason why this should work for all p.

if I did get the explanation correct I believe I found a counterexample.

? for(w=1,10,d=prod(x=1,w,prime(x));forprime(y=2,1200*w^2,if((2^y-2)%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

this is not in the accepted list. edit: nevermind removing the break finds 2281 for 7