Let k be a nonnegative integer such that F(k) = 2^(2^k) + 1 is prime (a Fermat prime A019434), then n = (F(k)1)*F(k)/2 appears in the sequence.
Conjecture: a(1)=3 is the only odd member of the sequence.
Conjecture: All elements of the sequence are of the above form derived from Fermat primes.
The sequence has 5 (known) elements in common with sequences A055708 (n1  sigma(n)) and A056006 (n  sigma(n)+2) since a(n) is a subsequence of both.
The first five members of the sequence are respectively congruent to 3, 4, 4, 4, 4 modulo 6.
There are no further entries after a(5) up 8 * 10^9.
Up to n = 1312 * 10^8 there are no further entries in the class congruent to 4 modulo 6.
a(6) > 10^12.  Donovan Johnson, Dec 08 2011
a(6) > 10^13.  Giovanni Resta, Mar 29 2013
a(6) > 10^18.  Hiroaki Yamanouchi, Aug 21 2018
See A125246 for numbers with deficiency 4, i.e., sigma(n) = 2*n  4, and A141548 for numbers with deficiency 6.  M. F. Hasler, Jun 29 2016 and Jul 17 2016
A term n of this sequence multiplied with a prime p not dividing it is abundant if and only if p < n1. For each of a(2..5) there is such a prime near this limit (here: 7, 127, 30197, 2147483647) such that a(k)*p is a primitive weird number, cf. A002975.  M. F. Hasler, Jul 19 2016
Any term x of this sequence can be combined with any term y of A088831 to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2, which is a necessary (but not sufficient) condition for two numbers to be amicable. [Proof: If x = a(n) and y = A088831(m), then sigma(x) = 2x2 and sigma(y) = 2y+2. Thus, sigma(x)+sigma(y) = (2x2)+(2y+2) = 2x+2y = 2(x+y), which implies that (sigma(x)+sigma(y))/(x+y) = 2(x+y)/(x+y) = 2.]  Timothy L. Tiffin, Sep 13 2016
At least the first five terms are a subsequence of A295296 and of A295298.  David A. Corneth, Antti Karttunen, Nov 26 2017
Conjectures: all terms are second hexagonal numbers (A014105). There are no terms with middle divisors.  Omar E. Pol, Oct 31 2018
