A283455 - OEIS

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

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OFFSET

1,2

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

Table of n, a(n) for n=1..57.

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

MATHEMATICA

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

PROG

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

CROSSREFS

Cf. A046800, A283364.

Union of {1}, A000043, A085724.

Sequence in context: A330500 A030151 A307360 * A283364 A131617 A214321