# Mersenne Primes using Mersenne Primes as n

I have been exploring the fascinating world of prime numbers, particularly Mersenne Primes, and have noticed an interesting pattern. It seems to me that $2^n - 1$ is prime as long as $n$ itself is a Mersenne prime.

I already know that $2^n - 1$ is not necessarily prime just because $n$ is a normal prime number, but I have not found a counterexample that to the claim that $2^n - 1$ is a prime whenever $n$ is a Mersenne prime.

Is there a proof that this claim is false?

• True for the first four Mersenne primes, and false for the fifth one already - see here. Commented Nov 6, 2017 at 16:23
• I know some facts don't related directly with your question. Maybe you know those, I cited these because I think that are important. You can see from this The Prime Page, by Professor Chris Caldwell the graph for Mersenne primes in section 6 Where is the next larger Mersenne prime?. On the other hand I like very much also several conjectures due to Professor Farideh Firoozbakht, you can read these in The Prime Puzzles & Problems Connection, by Carlos Rivera. I say for example the Puzzle 434 and Puzzle 517.
– user243301
Commented Nov 6, 2017 at 21:33

$2^{13}-1 = 8191$ is a Mersenne prime, but $2^{8191}-1$ is composite.
• @KernelStearns But this is on the internet; the above list says that $2^{8191}-1$ is not a (Mersenne) prime. Commented Nov 6, 2017 at 16:25
• In the past I tried to combine some problems as the mentioned in the top, Puzzle 434 and Puzzle 517 by Firoozbakht from The Prime Puzzles & Problems Connection, or Puzzle 774 by Noahiro Nomoto with the equation for integers $n>1$ involving the sum of remainders function (see the article [1] or [2]), but I didn't any remarkable deduction. Maybe you or @DietrichBurde can do experiments in your home about it. [1] T. Cross, A note on almost perfect numbers, Mathematics Magazine, 47 (1974), 230–231. [2] Z. Spivey, The Humble Sum of Remainders Function, Mathematics Magazine, 78, No. 4 (2005).