Largest nonmersenne prime

[dash1729]

Originally Posted by a1call

Can you please continue with the numeric example (without actually calculating the values) from then on. How does that result in non-double-Mersenne which is different from 2, 3, 5 and 31.

I'm not sure if you still need my help or not but if so I'm happy to answer the question above.

But before I do--are you familiar with the traditional proof that there are infinitely many primes? My proof is ultimately just a small revision of the traditional proof that there are infinitely many primes to show that there must also be infinitely many non-Mersenne primes.

So I want to be sure you are familiar with that traditional proof to make sure we are starting from the same place. Are you?

[JeppeSN]

There are infinitely many primes of the form 4*x + 1. (This is more elementary than the more general Dirichlet theorem on arithmetic progressions, and certainly more elementary than Bertrand's postulate and the prime number theorem.)

Since no Mersenne prime is of form 4*x + 1, it follows that infinitely many primes are not Mersenne primes (A138837).

(Not to be confused with the unsolved problem: Are there infinitely many primes p such that 2^p - 1 is a composite Mersenne number? (A054723))

Edit: I also like literka's second proof above, about the convergence of the reciprocals of the Mersenne primes.

/JeppeSN

[Uncwilly]

Originally Posted by rajula

The correct answer would be that there is no upper bound, but largest known nonmersenne prime has 3918990 digits.

To date it is: 9383761 digits. (follow the same link)

[BudgieJane]

Originally Posted by retina

Got it. Thanks.

You're welcome.

[a1call]

Originally Posted by dash1729

I'm not sure if you still need my help or not but if so I'm happy to answer the question above.

But before I do--are you familiar with the traditional proof that there are infinitely many primes? My proof is ultimately just a small revision of the traditional proof that there are infinitely many primes to show that there must also be infinitely many non-Mersenne primes.

So I want to be sure you are familiar with that traditional proof to make sure we are starting from the same place. Are you?

I am familiar with this:
https://primes.utm.edu/notes/proofs/...e/euclids.html
And please feel free to substitute 2^n-1 with Mn to make things easier to digest and easier to type.

Thanks for your patience.

[a1call]

So I had a few days to think about this.
If we assume that there are a finite n non-Mersenne Primes, then there can be a finite n' Mersenne primes where n'<=n
And
There can be a finite n'' double Mersenne primes where n''<=n'

There can be a finite n''' triple Mersenne primes where n'''<=n''

There can be a finite n'''' quadrupole Mersenne primes where n'''<=n''
....

So unless we can somehow eliminate the "=" from (more than n of) the above inequalities, there could be argumentaly finite non-Mersenne primes.

ETA Now if multiply any subset of these primes together and add or subtract 1 from it we will end up with a number that is coprime to all of them but not necessarily with prime factors that are smaller than the largest one of them. I still fail to see the contradiction here.

[dash1729]

Originally Posted by a1call

I am familiar with this:
https://primes.utm.edu/notes/proofs/...e/euclids.html
And please feel free to substitute 2^n-1 with Mn to make things easier to digest and easier to type.

Thanks for your patience.

OK--so my proof is essentially the same proof with only minor modifications.

Let's find all primes strictly less than MM13 (that is a double Mersenne prime), multiply them together, add 1, and see what the divisor's are--exactly the same approach that Euclid used. Well, what are those primes? They include the non-double Mersenne primes 2,3,5,31. We were assuming those are the only non-double Mersenne primes in existence. Of course that assumption is false but this is just an example to illustrate the concept.

Then there are the potential double Mersenne primes MM2,MM3,...,MM12. Those numbers--2,3,5,31,MM2,MM3,...,MM12 are the only possible primes less than MM13. In fact, some of them aren't prime, but if we multiply all the primes together and add 1 (call that number L), we will get the following:

L<=1+2*3*5*31*MM2*MM3*...*MM12

A little bit of arithmetic will then allow us to say:

L<=2^2^2*2^2^3*2^2^4*...*2^2^12

Remember that when you multiply two powers with the same base, you get the base to the sum of the exponents. So this becomes:

L<=2^(2^2+2^3+2^4+...+2^12)

Adding up the elements of a geometric series this becomes:

L<=2^(2^13-4)

So by the definition of MM13,

L<MM13

Remember that L was defined to be the product of all primes strictly less than MM13 plus one. So L must be either prime itself, or have a prime divisor--in either case the prime involved must be at least MM13 based on the way L was defined. L cannot have any prime divisor less than MM13. But we just showed L<MM13, so we have a contradiction.

[dash1729]

Originally Posted by a1call

So I had a few days to think about this.
If we assume that there are a finite n non-Mersenne Primes, then there can be a finite n' Mersenne primes where n'<=n
And
There can be a finite n'' double Mersenne primes where n''<=n'

There can be a finite n''' triple Mersenne primes where n'''<=n''

There can be a finite n'''' quadrupole Mersenne primes where n'''<=n''

It is certainly true that a similar observation could be made about double, triple, etc. Mersenne primes:

If there are only finitely many non-Mersenne primes, then there must only finitely many Mersenne primes that aren't double.

If there are only finitely many non-Mersenne primes, then there must only finitely many Mersenne primes that are double but not triple.

If there are only finitely many non-Mersenne primes, then there must only finitely many Mersenne primes that are triple but not quadruple.

And so on.

However it isn't necessary for my proof to go beyond talking about double Mersenne primes. We don't need to invoke triple, etc, Mersenne primes for my proof to work. The key argument that I made is that double Mersenne primes grow fast enough for Euclid's original argument, with some modification, to work. Moving up to triple Mersenne primes and beyond doesn't really add anything of value to my proof.

[a1call]

Originally Posted by a1call

So I had a few days to think about this.
If we assume that there are a finite n non-Mersenne Primes, then there can be a finite n' Mersenne primes where n'<=n
And
There can be a finite n'' double Mersenne primes where n''<=n'

There can be a finite n''' triple Mersenne primes where n'''<=n''

There can be a finite n'''' quadrupole Mersenne primes where n'''<=n''
....

So unless we can somehow eliminate the "=" from (more than n of) the above inequalities, there could be argumentaly finite non-Mersenne primes.

ETA Now if multiply any subset of these primes together and add or subtract 1 from it we will end up with a number that is coprime to all of them but not necessarily with prime factors that are smaller than the largest one of them. I still fail to see the contradiction here.

A few corrections are due here:

If we assume that there are a finite n non-Mersenne Primes, then there can be a finite n' Single-Mersenne-Primes (where the exponent is not a Mersenne number/Prime) where n'<=n (We could get rid of the "=" here because we know not it in this case)
And
There can be a finite n'' double Mersenne primes where n''<=n'

There can be a finite n''' triple Mersenne primes where n'''<=n''

There can be a finite n'''' quadrupole Mersenne primes where n'''<=n''
....


So unless we can somehow eliminate the "=" from more than n-1 of the above infinite inequalities, there could be argumentatively finite non-Mersenne primes.

ETA Now if multiply any subset of these primes together and add or subtract 1 from it we will end up with a number that is coprime to all of them but not necessarily with prime factors that are smaller than the largest one of them. I still fail to see the contradiction here.

[a1call]

Originally Posted by dash1729

However it isn't necessary for my proof to go beyond talking about double Mersenne primes. We don't need to invoke triple, etc, Mersenne primes for my proof to work. The key argument that I made is that double Mersenne primes grow fast enough for Euclid's original argument, with some modification, to work. Moving up to triple Mersenne primes and beyond doesn't really add anything of value to my proof.

I beg to differ here. If there are as many triple-Mersenne-Primes as there are Double-Mersenne-Primes as there are quatrupole-Mersenne-Primes as there are .... infinitude-Mersenne-Primes as unlikely as it may be rules out the impossibility of there being a finite number of non-Merseene-Primes (In other words there may as well be only a finite number of non-Mersenne primes among the infinite number of primes). An observation that in any given finite set of numbers Double-Mersenne-Prime-numbers grow less rapidly does not prove anything about an infinite set.

[dash1729]

Originally Posted by a1call

An observation that in any given finite set of numbers Double-Mersenne-Prime-numbers grow less rapidly does not prove anything about an infinite set.

My observation was actually that the double Mersenne primes do grow quite rapidly--rapidly enough to prove what we need.

And that observation does in fact lead to a proof--my proof in fact. I've now invested some time in making several follow up posts trying to explain my proof to you in more detail. If you insist on investing your own time not in trying to understand my proof but instead in exploring your own direction (which isn't fruitful) that is on you.