A283455 - OEIS

%I #42 Mar 21 2020 09:41:42

%S 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,

%T 107,109,127,131,137,139,149,167,197,199,227,241,269,271,281,293,347,

%U 373,379,421,457,487,521,523,607,727,809,881,971,983,997,1061

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

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

%C 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.

%C 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.

%C 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.

%C Numbers n such that A046800(n) < 3. - _Michel Marcus_, Mar 08 2017

%H The Cunningham Project, <a href="https://homes.cerias.purdue.edu/~ssw/cun/pmain619.txt">The Main Tables</a>, Table 2- Factorizations of 2^n-1, n odd, n<1300.

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

%o (PARI) is(n)=omega(2^n-1)<3 \\ _Charles R Greathouse IV_, Mar 08 2017

%Y Cf. A046800, A283364.

%Y Union of {1}, A000043, A085724.

%K nonn

%O 1,2

%A _Vladimir Shevelev_, Mar 08 2017

%E More terms from _Peter J. C. Moses_, Mar 08 2017

%E a(48)-a(50) from _Charles R Greathouse IV_, Mar 08 2017

%E a(51)-a(57) from _Amiram Eldar_, Feb 13 2020