Is $2^p - 1$ always prime when p is a mersenne prime?

Question

First mersenne prime $2^2-1=3 $, ** $2^{(2^2-1)}-1$ is also prime How many far can we go to get first composite? $2^{(2^{...(2^{(2^2-1)}-1)}-1)}-1$

Answer 1

To answer the question in the title, no. $2^{13}-1$ is a Mersenne prime, but $2^{2^{13}-1}-1$ is not.

As for that sequence, who knows? We definitely don't know that they're all prime, since we don't even know if there are infinitely many Mersenne primes. Meanwhile, the fifth term in the sequence is already larger than $10^{10^{37}}$ and it only gets worse from there, meaning we will likely never find a counterexample computationally.

Answer 2

We test them sequentially, because once one is composite, all the rest will be.

Last knowledge I had ( about 6 month out of date ,since I quit mersenneforum), was MMMM2 (aka MM127) had been tested to 185 bits without a factor. It's nearest integer to sqrt, is north of a exbibyte in size. without real advancement in knowledge, you probably won't exceed that size.

We already know divisors must fit $2kP+1$ and be $\pm1\bmod 8$ and for double mersennes with P a mersenne(MMp) we get $$4jkp+2j+1$$ as a form of factors, but this doesn't really help. Neither will: $$8xyzq+4xy+2x+1\\16abcdr+8bcd+4cd+2c+1$$ Which you get iterating out to triple and quadruple mersennes respectively.

Edit okay not completely true. j is 0,3,6 or 9 mod 12 due to P's nature.