0
$\begingroup$

A Mersenne number is a number on the form $2^n-1$. For it to be prime, the number $n$ must be prime. My question is that if $n$ is another Mersenne prime, will $2^n-1$ be always prime?

It seems so, to me anyway.

$\endgroup$
  • $\begingroup$ If this were true, this would generate info itelyinfjnitely many mersenne primes! We currently know of only 48. $\endgroup$ – JavaMan Oct 15 '18 at 10:50
1
$\begingroup$

https://oeis.org/A000043

We only know about finitely many Mersenne primes. If your conjecture was true, then we could generate infinitely many Mersenne primes.

$\endgroup$
  • $\begingroup$ Yes. That would be true. $\endgroup$ – Harshith Vasireddy Oct 15 '18 at 11:22
1
$\begingroup$

My own program found that $2^{2^{13} - 1} - 1$ is not a (probable) prime number, confirmed by Wolfram Alpha and https://oeis.org/A000043. Note that $M_{13} = 8191$ is prime and $M_{8191}$ is not.

$\endgroup$
  • $\begingroup$ Wolfram alpha says result unknown $\endgroup$ – Harshith Vasireddy Oct 15 '18 at 11:21
  • $\begingroup$ For me is says it is not (after a few seconds), query was isprime 2^(2^13-1)-1. Anyway, $M_{8191}$ is not in the list of Mersenne primes. $\endgroup$ – gammatester Oct 15 '18 at 11:23
  • $\begingroup$ Thanks. Doubt clarified $\endgroup$ – Harshith Vasireddy Oct 15 '18 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.