Cohen (1982) shows all odd squares are members. The remaining terms shown here are conjectural, based on a search up to 10^20 made by Davis et al. (2013).
Comments from Farideh Firoozbakht, Jan 12 2014: (Start)
1. Mersenne primes are not in this sequence. Because if M=2^p1 is prime then M=sigma(m)2m, where m=2^(p1)*(2^p1)^2=(1/2)*(M+1)*M^2 (please see Proposition 2.1 of FiroozbakhtHasler, 2010).
2. If M = 2^p  1 is a Mersenne prime then M^2 + 3M + 1 = 4^p + 2^p  1 is not in the sequence. Because M^2 + 3M + 1 = sigma(m)  2m where m = M^3 + M^2 = 2^p(2^p1)^2 (please see Proposition 2.5, op. cit.).
Examples:
p = 2, M = 3, 4^p + 2^p  1 = 19, m = M^3 + M^2 = 2^p(2^p1)^2 = 36; sigma(m)  2m = 19
p = 3, M = 7, 4^p + 2^p  1 = 71, m = M^3 + M^2 = 2^p(2^p1)^2 = 392; sigma(m)  2m = 71
3. Note that if r is an even number and if for a number k p = 2^k  r  1 is an odd prime then r = sigma(m)  2m where m = 2^(k1)*p. Namely r is not in the sequence (see Theorem 1.1, op. cit.).
It seems that for each even number r, there exists at least one odd prime of the form 2^k  r  1. This means there is no even term in the sequence.
Moreover I conjecture that, for each even number r, there exist infinitely many primes p of the form 2^k  r  1, or equivalently, I conjecture that: For each odd number s, there exists infinitely many primes p of the form 2^k  s.
Special cases:
(i): s = 1, there exist infinitely many Mersenne primes.
(ii): s = 1, there exist infinitely many Fermat primes.
(iii): s = 3, sequence A050414 is infinite.
(iv): s = 3, sequence A057732 is infinite.
(v): s = 5, sequence A059242 is infinite.
and so on. (End)
