Are there infinitely Mersenne primes?

[Guest]

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n = 2^{i}}^{\infty }x_{n } = 0[/tex] if i >> 26 and if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

If [tex]\prod_{n = 2^{i}}^{\infty }x_{n } = 0[/tex] where i >> 26 and where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there are infinitely many Mersenne primes.


If [tex]\prod_{n = 2^{i}}^{\infty }x_{n } > 0[/tex] where i >> 26 and where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there is only a finite number of Mersenne primes.

Moreover,

[tex]x_{n } = \frac{\pi(\sqrt{M_{n}})}{\pi(\sqrt{M_{n}}) + 1}[/tex] where [tex]M_{n } = 2^{n} - 1[/tex]. ([tex]M_{n }[/tex] is called a Mersenne number/prime.)

Remark: [tex]\pi()[/tex] is the prime-counting function.

We strongly believe there is only a finite number of Mersenne primes.

Relevant Reference Link:

'Mersenne prime',

https://en.wikipedia.org/wiki/Mersenne_prime#Mersenne_numbers_in_nature_and_elsewhere.

[Guest]

An Update: Are there infinitely many Mersenne primes?

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n = 2^{i}}^{\infty }x_{n } = 0[/tex] if i >> 26 and if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

If [tex]\prod_{n = 2^{i}}^{\infty }x_{n } = 0[/tex] where i >> 26 and where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there are infinitely many Mersenne primes.


If [tex]\prod_{n = 2^{i}}^{\infty }x_{n } > 0[/tex] where i >> 26 and where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there is only a finite number of Mersenne primes.

Moreover,

[tex]x_{n } = \frac{\pi(\sqrt{M_{n}})}{\pi(\sqrt{M_{n}}) + 1}[/tex] where [tex]M_{n } = 2^{n} - 1[/tex]. ([tex]M_{n }[/tex] is called a Mersenne number/prime.)

Remark: [tex]\pi()[/tex] is the prime-counting function.

We strongly believe there is only a finite number of Mersenne primes.

Relevant Reference Link:

'Mersenne prime',

https://en.wikipedia.org/wiki/Mersenne_prime#Mersenne_numbers_in_nature_and_elsewhere.

[Guest]

Hmm. Is your reasoning correct? I am not sure!

[Guest]

Hmm. Is your reasoning correct? I am not sure!

Predicting primes accurately is a very uncertain business. And therefore, the unexpected can happen unexpectedly for very large primes regardless of the form (Fermat, Mersenne, etc.) that they may take. So we encourage the author(s) of this post to review the work and check the calculations too. Good luck!

[Guest]

There may be infinitely many Mersenne primes. But we cannot prove it.

[Guest]

There may be infinitely many Mersenne primes. But we cannot prove it.

We know that the last digit (base-ten positional numeral system) of every odd prime number must be either 1, 3, 7, or 9.

And there are infinitely many prime numbers.

The last digit (base-ten positional numeral system) of every Mersenne number ([tex]M_{n } = 2^{n} - 1[/tex]) must be either 1, 3, 7, or 9.

And therefore, there are infinitely many Mersenne prime numbers.

[Guest]

There may be infinitely many Mersenne primes. But we cannot prove it.

We know that the last digit (base-ten positional numeral system) of every odd prime number must be either 1, 3, 7, or 9.

And there are infinitely many prime numbers.

The last digit (base-ten positional numeral system) of every Mersenne number ([tex]M_{n } = 2^{n} - 1[/tex]) must be either 1, 3, 7, or 9.

And therefore, there are infinitely many Mersenne prime numbers.

Remark: The last digit, 5, of infinitely many Mersenne numbers is extremely unlikely.

[Guest]

FYI: "With the exception of [tex]M_{0 } = 3[/tex], the last digit of a Mersenne number is very likely, 1 or 7."

Source Link:

https://en.wikipedia.org/wiki/Mersenne_prime.