for anyone interested by this here's the results I got when I limited the exponents to the known exponents:

[CODE]? a=[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, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161];for(w=1,#a,print(a[w]"("w"):");d=prod(x=1,primepi(w),prime(x));f
or(y=1,#a,if((2^a[y]-2)%d==0,print1(a[y]",")));print())
2(1):
2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,4423,9689,9941,11213,19937,21701,23209,44497,86243,110503,132049,216091,756839,859433,1257787,1398269,2976221,3021377,6972593,1346691
7,20996011,24036583,25964951,30402457,32582657,37156667,42643801,43112609,57885161,
3(2):
2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,4423,9689,9941,11213,19937,21701,23209,44497,86243,110503,132049,216091,756839,859433,1257787,1398269,2976221,3021377,6972593,1346691
7,20996011,24036583,25964951,30402457,32582657,37156667,42643801,43112609,57885161,
5(3):
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,
20996011,24036583,25964951,30402457,32582657,37156667,42643801,43112609,57885161,
7(4):
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,
20996011,24036583,25964951,30402457,32582657,37156667,42643801,43112609,57885161,
13(5):
5,13,17,61,89,521,2281,3217,4253,9689,9941,11213,19937,21701,23209,44497,132049,859433,1398269,2976221,3021377,6972593,13466917,30402457,32582657,42643801,43112609,57885161,
17(6):
5,13,17,61,89,521,2281,3217,4253,9689,9941,11213,19937,21701,23209,44497,132049,859433,1398269,2976221,3021377,6972593,13466917,30402457,32582657,42643801,43112609,57885161,
19(7):
13,61,2281,3217,23209,44497,132049,13466917,30402457,42643801,
31(8):
13,61,2281,3217,23209,44497,132049,13466917,30402457,42643801,
61(9):
13,61,2281,3217,23209,44497,132049,13466917,30402457,42643801,
89(10):
13,61,2281,3217,23209,44497,132049,13466917,30402457,42643801,
107(11):
61,2281,42643801,
127(12):
61,2281,42643801,
521(13):
61,2281,42643801,
607(14):
61,2281,42643801,
1279(15):
61,2281,42643801,
2203(16):
61,2281,42643801,
2281(17):
2281,42643801,
3217(18):
2281,42643801,
4253(19):
42643801,
4423(20):
42643801,
9689(21):
42643801,
9941(22):
42643801,
11213(23):
19937(24):
21701(25):
23209(26):
44497(27):
86243(28):
110503(29):
132049(30):
216091(31):
756839(32):
859433(33):
1257787(34):
1398269(35):
2976221(36):
3021377(37):
6972593(38):
13466917(39):
20996011(40):
24036583(41):
25964951(42):
30402457(43):
32582657(44):
37156667(45):
42643801(46):
43112609(47):
57885161(48):
[/CODE]

[QUOTE=Brian-E;354434]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.
[/QUOTE]

You have a very good question and I should be prepared to have a very good
answer. If that would be the case, I would probably not post this in misc.
math. IOW, I don't have a good answer. At least not an answer I'm satisfied
with. :blush:

----

Related or not:

Let me define BAC (beer ad hoc class): Mersenne's number Mq BAC is p when Mq-1
is divisible by each prime up to and including p but not by the next prime.

BAC distribution for prime q's <= 112797829 (6454203 primes):

[CODE]
BAC | # | %
-------+----------+---------
2 1 0.0%
3 3227268 50.0%
5 1613858 25.0%
7 1209946 18.7%
11 0
13 201624 3.1%
17 134388 2.1%
19 60368 0.9%
23 5595 0.1%
29 0
31 0
37 0
41 0
43 1104 0.0%
47 47 0.0%
53 4 0.0%
[/CODE]

As has already been remarked, the weak version of the "conjecture" is more clearly stated as asserting that for every Mersenne prime Mp there is a Mersenne prime Mq satisfying p#|Mq-1.

Observation 1. For sufficiently large p, q>p.

This follows from the fact that [url=http://en.wikipedia.org/wiki/Primorial#Definition_for_natural_numbers]ln(p#) ~ p[/url] while ln(Mp) ~ p ln(2) < p, together with the fact that p#|Mq-1 implies p# < Mq.

I don't know the smallest n for which q must be > p for all p > n, but as 100# is about a million times larger than 2^100, it's pretty clearly much less than 57,885,161. Therefore the "conjecture" implies an infinite sequence of Mersenne primes Mq[sub]i[/sub] satisfying q[sub]i[/sub]#|Mq[sub]i+1[/sub]-1.

Observation 2. The weak version of the "conjecture" is equivalent to the proposition that for any prime p (or equivalently any n) there is a Mersenne prime Mq satisfying p#|Mq-1 (equivalently n#|Mq).

This follows from the infinity of Mersenne primes Mq, together with the fact that if p#|Mq-1 then any smaller n# will also divide Mq-1.

I would expect the weak form, (and therefore the strong form) of the "conjecture" to be false. Although we do believe on [url=http://primes.utm.edu/mersenne/heuristic.html]heuristic grounds[/url] that there are infinitely many Mersenne primes whose exponents range over the entire space of primes q, your "conjecture" requires infinitely many Mersenne primes with exponents in each of a diminishing sequence of subsets of the prime numbers, namely the sets Sp = {q : p# | Mq-1} (Mq not assumed to be prime). Someone with more mathematical knowledge than I may be able to produce a heuristic argument to support (or refute) this.

I put the word "conjecture" in quotes, because mathematicians usually only use that word if they have good reason (short of proof) to believe the proposition true.

[FONT=Calibri,sans-serif][SIZE=2][QUOTE=jnml;354360]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).[/QUOTE]

I totally forgot about this topic, but it popped up in the list, once the newer posts.

Well, I missed the "prime" qualifier related to the second mersenne number, and I am very sorry I accused you of telling gibberish. That was me who read your text incorrectly. Without that "prime" qualifier, the "conjecture" was indeed trivial.

[U]With[/U] the qualifier, the "conjecture" is most probably false.

What you say above related to 11 [B][U]is [/U][/B](this time seriously!) gibberish. Sorry. One don't need to "hint" nor "search" for such example, because the proof that no such number exists is trivial. I.e. all mersenne numbers which have 3, 5, 7, and 11 as factors, will also have 13 as a factor. From here, all you have to do to get your identity, is multiply by 2.

Let's denote p(x) the smallest n such 2^n-1 is divisible by x. This function is defined for all odd positive integers. The value is easy to compute for all positive odd integers: for example p(3)=2, p(5)=4, p(7)=3, p(9)=6, p(11)=10, p(13)=12, p(15)=4, p(17)=8, etc. There is a simple algorithm to compute this [/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2](and it is equivalent of factoring eulerphi(n))[/SIZE][/FONT], but for now you have to trust me. :smile:

Also, please remark that if p(x)=n, then (by definition) 2^n-1 is divisible by x, AND for every positive integer k, 2^(kn)-1 is divisible by x.

Example:
p(17)=8, therefore 2^8-1, 2^16-1, 2^24-1, etc, are all divisible by 17. No other mersenne numbers are divisible by 17.
p(21)=6, therefore 2^6-1, 2^12-1, 2^18-1, etc, are all divisible by 21. [/SIZE][/FONT]
[FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]No other mersenne numbers are divisible by 21.
[/SIZE][/FONT]
This function has a very interesting property: if p(x)=a and p(y)=b, and gcd(x,y)=1, then p(x,y)=lcm(a,b). This is simple to prove. Homework.

That is why, if p(7)=3 and p(3)=2, then p(21)=lcm(2,3)=6.

(this function has another interesting property which is not the object of our proof, but just for the sake of exposing it: if n is not a Wieferich prime, then p(n^k)=n^(k-1)*p(n), ex: p(3)=2, p(9)=3*2=6, p(27)=3*3*2=18, etc. Things are getting more complicate for Wieferich primes, but fortunately only two of those are known).

Therefore, if p(x) have the values enumerated above, then we have p(3*5*7*11)=lcm(2,4,3,10)=60. So, every mersenne number of the form 2^(60*k)-1 is divisible by 3*5*7*11. You can try and see. There are no other mersenne numbers with this property.

Now, multiply this by 2, on both sides, as I said in the beginning, to get your identity: and you will get that every number of the form 2^(60*k+1)-2 is divisible by 11# (11 primorial, or 2*3*5*7*11). Indeed, 2^61-2, and 2^121-2 are both divisible by 11, and all "multiples" of them. (I use quotes because multiples refers to powers, not the numbers effectively).

Let's calculate p(3*5*7*11*13). This, according with the definition, it is the smallest 2^n-1 which is divisible by the product in parenthesis.

Following the same procedure, we get the lcm of p(x) for all primes x in the product, or: p(3*5*7*11*13)=lcm(2,4,3,10,12)=60. Indeed, both 2^60-1 and 2^120-1 and all "multiples", are divisible not only by 3, 5, 7, 11, but [U]also by 13[/U]. Multiply both sides by 2: we get [B]2^(60k+1)-2 is divisible by 13#[/B], QED! there can be NO mersenne number which fulfill the "strong" condition in the conjecture, for 11.

This "proof" can be repeated for ALL primes q for which t=nextprime(q) is an exponent of a mersenne prime. Because obviously, p(t) is always smaller than q in this case. Ex:
p(t=7)=3, << q=5;
p(t=31)=5, << q=29;
p(t=127)=7, << q=113
p(t=8191)=13, << q=8179
(where "<<" denotes "much smaller than").

To your disappointment, the pair (11,13) is somehow special, it is unique with the mentioned property, and 13=nextprime(11) and nor 11 neither 13 being a factor of a mersenne with prime exponent.

But the rule can trivially be applied to prime preceding exponents of mersenne primes. Always p(29#) will be equal to p(31#), and so on.

That won't be true for 23, 47, and all primes which are proper factors of mersenne numbers. We deal here with the law of small numbers. Compute, and see.

Now, I would extend the definition of p(x) to include numbers of the form 2*(2k+1): I would define "p(x)=p(x/2)" if x is 2*odd. (this means p(x) is define for all the odd numbers, and for all the numbers which are 2*odd_number). This is just a notation, to be able to write p(q#), instead of 2*p(3*5*7*11*....q), or p(q#/2). It is not used in any calculus. Just keep in mind that, if q=2*(2k+1), then p(q)=p(q/2). Just a notation.

With this "extension", for any given prime q, the series {2^(k*p(q#)+1)-1, k in Z, k>0} contains ALL the mersenne numbers M, for which M-1 is divisible by q#, and does NOT contain any number without this property. This means that the set in the accolades contains only mersenne numbers M, primes or not, for which M-1 is divisible by 2*3*5*7*....*q, and THERE ARE NO OTHER mersenne numbers with this property which are not contained in the set.

And back to the conjecture: practically you claim that:

[U]CLAIM 1[/U]. For every prime q for which Mq is prime, the series {2^(k*p(q#)+1)-1, k>0, k in Z}, contains at least one prime. This is (heuristically) false.

AND:

[U]CLAIM 2.[/U] For every prime q for which Mq is composite, the series {2^(k*p(q#)+1)-1, k>0, k in Z}, contains only composite. This is (provable) false.

(It is not clear for me if you claim this second part too, or you only claim the first part. I.e. is the conjecture only with "if", or it is with "if and only if". I only proved that the second part "holds" for 11, and few others, but we deal with the law of small numbers here.)

You have to think how this p(x#) behave: it will "[URL="http://idioms.thefreedictionary.com/mark+time"]mark the time[/URL]" for mersenne primes, for factors of mersenne primes, and for many other primes x for which p(x) (with no sharp sign) is enough small. But it will have large jumps for the other. Here is where you have to look for counterexamples. You might get lucky...
[/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2]


[/SIZE][/FONT]

[FONT=Calibri,sans-serif][SIZE=2][had to break it, too long.. it took me few hours, with many interruptions, to write it]

Numerical examples:

q=3: you claim that the series {2^(2k+1)-1, natural k} contains at least a prime (indeed, we have many "2k+1" here, all known mersenne numbers fall here). For that prime, the M-1 may [weak version], or may not [strong version] be divisible by 5.

q=5: you claim that the series {2^(4k+1)-1, natural k} contains at least a prime (indeed, we have many "4k+1" here: 13, 17, 61, ... ). For that prime, the M-1 may [weak version], or may not [strong version] be divisible by 7. We have a "weak" in 13, and a "strong" in 17. Again, we deal with the law of small numbers (my opinion, I can't prove this in any way!)

q=7: you claim that the series {2^(12k+1)-1, natural k} contains at least a prime. Indeed, we have here: 13 (strong, 2^13-2 is not div by 11, as we proved above), 61 (weak, 2^61-2 is not only div by 11, but also by 13), etc ...

q=11: as M11 is not prime, you (maybe) claim that the series {2^(60k+1)-1, natural k} contains no prime (in weak version?). We know 2^61-1 is prime (!?!?). Not clear what you claim here, if anything. Strong version may hold, because there is no 2^(61k)-2 which is divisible by 11#, but not divisible by 13. So we can say that "strong version holds". (!?!?). Again: the only exponents with this propertyare in a very restricted set, and the fact is consequence of something like p(11#)=p(13#)=60.

q=13: you claim that the series {2^(60k+1)-1, natural k} contains at least a prime. Indeed, we have here 61, with 2^61-2 divisible to 13#, but not to 17; and we have two others: 2281 and [/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]42643801, which (as expected) are divisible to 17. Generally, M-1 have a lot of small factors.[/SIZE][/FONT]

q=17: p(17#) is 120, so, you claim that the series {2^(120k+1)-1, natural k} contains at least a prime. Searching the known primes to see which one has the exponent 1 (mod 120) returns 2281 and 42643801 (which fit all the other above, too). I let you judge the weakness or strength of those.

q=19: p(19#)=360. There would be at least a prime whose exponent is 1 (mod 360). (We have x=42643801, with 2^x-2 divisible by 19#, and [B][U]not[/U][/B] divisible by 23 nor 29. However, for primes under 100, this is divisible by: 2 to 19, then 31, 37, 41, 61, and 73. As random as expected. The other primes under 100, have p(x) not contained in 360.

q=23: p(23#)=3960. There would be no prime whose exponent is 1 (mod 3960) (in case you claim the "claim 2", not yet clear if you do).

q=29: p(29#)=27720. There would be no prime whose exponent is 1 (mod 27720) (in case you claim the "claim 2", not yet clear if you do). This somehow contradicts the next one:

q=31: p(31#)=27720. We don't know enough primes here. But... did you get what is happening further? [/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2]

[/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]p(29#)=[/SIZE][/FONT]p(31#)=[/SIZE][/FONT]p(37#)=p(41#)=p(43#)=27720

but p(47#)=637560. And [/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2]p(53#)=8288280

[/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]p(59#)=[/SIZE][/FONT]p([U][B]61#[/B][/U])=[/SIZE][/FONT]p(67#)=p(71#)=[/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]p(73#)=[/SIZE][/FONT][/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]p(79#)=[/SIZE][/FONT][/SIZE][/FONT]....=240360120

How to do here? Are you claiming that the set [/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2]{2^([/SIZE][/FONT][FONT=Calibri,sans-serif][SIZE=2][FONT=Calibri,sans-serif][SIZE=2]240360120[/SIZE][/FONT]k+1)-1, natural k} contains at least a prime? (because of 61)? And further?

The reciprocals sum of this series diverge very fast... Heuristically, this claim is somehow absurd... Sorry if I understood you wrong again...:smile:

[/SIZE][/FONT]

[QUOTE=LaurV;355509][FONT=Calibri,sans-serif][SIZE=2]

I totally forgot about this topic, but it popped up in the list, once the
newer posts.

... [/SIZE][/FONT][/QUOTE]

Thanks for the lot of information which I can go through slowly only later
(like next weekend at soonest, I'm afraid).

ATM, I'd like to talk only about one little detail. Wrt where you write
"disappointment: It's actually _excitement_ to learn where a mistake is!
Honestly, thanks for your efforts, very appreciated :smile: (Applies to Mr. P-1
as well, of course.)