# Are primes $p$ such that $2^p-1$ is prime for seemingly random?

Are the prime exponents $p$ such that $2^p-1$ is prime for seemingly random?

So suppose you were choosing two 300-digit primes $p, p_2$ for encryption purposes. If these $p, p_2$ were chosen from, say a list of primes $p$ such that $2^p-1$ is prime for, would these be cryptography secure? (I know finding primes these large and also $2^p-1$ is prime would be basically impossible, but assume this is the case that only you know the list.) Thanks for explanations and help.

• At the risk of sounding snarky, I think if there were a pattern to such primes, someone would have surely noticed it by now... – 2012ssohn Aug 4 '17 at 3:08
• I don't understand the question. If your rsa key is $2^{30}$ bits long then the attacker will doubt you used Mersenne primes, as the Miller Rabin pseudo-prime generation needs something like $\mathcal{O}(2^{120})$ operations at that range. – reuns Aug 4 '17 at 3:36
• The list of such $p$ starts 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667. Seems fairly random to me. – Gerry Myerson Aug 4 '17 at 3:44
• Regardless of whether $2^p-1$ is sufficiently random, you don't want your prime numbers for cryptography to come from any, at least to an outsider, discernible family of choices. – Chickenmancer Aug 6 '17 at 2:57
• To date it is not known whether or not there are infinitely many Mersenne primes. – DanielWainfleet Aug 6 '17 at 4:27