My Math Forum Conjecture about primes of the form 2^k-1

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 October 15th, 2015, 06:07 AM #1 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Conjecture about primes of the form 2^k-1 Here is a little conjecture not easy to prove. Let us define an odd prime number p=2^k-1 Example : p=31 = 1+2+4+8+16 Now let us compute 5 values (31 is a sum of 5 values) v1=31-1=30 v2=31-2=29 v3=31-4=27 v4=31-8=23 v5=31-16=15 Each time we remove from the sum 31 one the elements 2^0,2^1,2^2,2^3,2^4 2 values are prime numbers : 29 and 23 Conjecture: For any odd prime p=(2^k)-1 (k>2) it always exist at least one prime number v(i) such as v(i)=(2^k-2^i)-1 with i varying from 0 to k.
 October 15th, 2015, 08:17 AM #2 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 No one is forced to read nor to answer. If you are interested by the question then welcome to your comments as long as they are related to the problem. Otherwise stay mum. To Al-Mahed and Charles Im not beggar nor student trying to solve his homework. I posted this conjecture in many forums. I do not care if it is going to be read or not. Im here to exchange points of view. Thank you.
 October 15th, 2015, 08:39 AM #3 Member   Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 2^31 - 1 is a prime that this conjecture does not hold with. I checked 2^31 - 1 - 2^k from k = 0 to 31 and none of the values were prime. Thanks from mobel
October 15th, 2015, 09:00 AM   #4
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Quote:
 Originally Posted by numberguru1 2^31 - 1 is a prime that this conjecture does not hold with. I checked 2^31 - 1 - 2^k from k = 0 to 31 and none of the values were prime.
Thank you.
The conjecture does not hold.
I just checked it.
Why?
Time to take then pencil and a paper.

 October 15th, 2015, 12:15 PM #5 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 All the number v(i) have 1 zero when written in base 2. Can we state that any odd large number written in base 2 using only one zero is probably composite? 111111111011111111 composite or prime? What is the probability that such number picked randomly is prime? Why numbers of the form 11111111111111111111111111111 in BASE 2 are rarely prime?
 October 15th, 2015, 01:39 PM #6 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Once a conjecture is false it seems like everything is solved. Why is it false? What if the inverse was true? What will happen large Mersenne primes? will it have at least one prime or not? Can we prove that the conjecture definitely false for n>31?
October 16th, 2015, 05:42 AM   #7
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 Originally Posted by mobel Once a conjecture is false it seems like everything is solved. Why is it false?
You're looking at k numbers around 2^k. A random odd number of this form is prime with probability roughly 2/log(2^k) = (2/log 2)/k. The odds that all k are composite is roughly (1 - (2/log 2)/k)^k which goes to exp(-2/log(2)) =~ 0.05583. Hence the expectation is that, if there are infinitely many Mersenne primes, infinitely many of them will have no prime of the indicated form.

Quote:
 Originally Posted by mobel What will happen large Mersenne primes? will it have at least one prime or not? Can we prove that the conjecture definitely false for n>31?
Actually it fails for every Mersenne prime I've tested greater than M5 = 2^5 - 1: exponents 7 to 756839. I didn't expect so many failures to be honest (see my heuristic above).

 October 16th, 2015, 06:10 AM #8 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Proving that the conjecture is false for large Mersenne numbers will be useful. In fact Im trying to find a range of values where it is highly unlikely that it will contain a prime but it can be proved that this range contain necessarly at least one prime. You did not remember on old post : There is always one prime between n! and n!+n^2 (I do not remember exactly but you could find it in this forum).
October 16th, 2015, 06:18 AM   #9
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 Originally Posted by mobel Proving that the conjecture is false for large Mersenne numbers will be useful. In fact Im trying to find a range of values where it is highly unlikely that it will contain a prime but it can be proved that this range contain necessarly at least one prime.
That would be a major discovery! It was quite hard to prove that there are infinitely many primes of the form $m^3+2n^3$ (Heath-Brown) and that's thick enough that you'd expect many primes of that form.

October 16th, 2015, 06:40 AM   #10
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 Originally Posted by CRGreathouse That would be a major discovery! It was quite hard to prove that there are infinitely many primes of the form $m^3+2n^3$ (Heath-Brown) and that's thick enough that you'd expect many primes of that form.
Primes are not random.
If you start by this obvious observation then it surely exist a thick range containing at least one prime.
Primes have nothing to do with randomness even they look like random.

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