A283455 Numbers m such that 2^m - 1 has at most 2 distinct prime factors.

1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 17, 19, 23, 31, 37, 41, 49, 59, 61, 67, 83, 89, 97, 101, 103, 107, 109, 127, 131, 137, 139, 149, 167, 197, 199, 227, 241, 269, 271, 281, 293, 347, 373, 379, 421, 457, 487, 521, 523, 607, 727, 809, 881, 971, 983, 997, 1061

Offset: 1

Author

Vladimir Shevelev, Mar 08 2017

Keywords

• nonn

Comments

The sequence differs from A283364 beginning with a(15). All a(n) > 6 are primes or squares of primes.

As in A283364 one can prove that all a(n) > 6 are odd. It is clear that a(n) is either prime or semiprime. Let us show that in the latter case it is the square of a prime. Indeed, let a(n) = p*q, p < q. Then 2^a(n)-1 is divisible by 2^p-1 < 2^q-1. Thus both of them are Mersenne primes.

Let us show that 2^(p*q)-1 differs from (2^p-1)^u*(2^q-1)^v, u,v >= 1. Indeed the equality is possible only in the case p*u + q*v = p*q. Then p|v and q|u. Let u = q*a, v = p*b. Then a + b = 1, which is impossible for u,v >= 1. Hence, 2^(p*q)-1 has a third prime divisor and p*q is not a member.

Are there terms other than 4, 9 and 49 that are squares of primes? Note that, for prime p, 2^(p^2)-1 differs from (2^p-1)^p, so if p^2 is a term, then for a Mersenne prime 2^p-1 and some t >= 1, the number (2^(p^2)-1)/(2^p-1)^t should be a prime or a power of a prime.

Numbers n such that A046800(n) < 3. - Michel Marcus, Mar 08 2017

Links

• The Cunningham Project, The Main Tables, Table 2- Factorizations of 2^n-1, n odd, n<1300.

Crossrefs

Cf. A046800, A283364.

Union of {1}, A000043, A085724.

Programs

Mathematica

Select[Join[Range@ 6, Range[7, 201, 2]], PrimeNu[2^# - 1] <= 2 &] (* Michael De Vlieger, Mar 08 2017 *)

PARI

is(n)=omega(2^n-1)<3 \\ Charles R Greathouse IV, Mar 08 2017

Extensions

More terms from Peter J. C. Moses, Mar 08 2017

a(48)-a(50) from Charles R Greathouse IV, Mar 08 2017

a(51)-a(57) from Amiram Eldar, Feb 13 2020