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
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
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, 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
STATUS
approved