You can obtain a complete heuristic picture of what is going, by supposing that quadratic residues (as suggested by Laurent) are the only likely obstruction to getting the next prime. More precisely, if you let $Q_n = \prod_{k \le n} q_k$ be the product of all of the primes discovered so far, then you can let $p_{n+1} = p$ be the first remaining prime such that $Q_n$ is a quadratic residue mod $p$, assuming Denis' first proposal. Or, assuming Denis' second proposal, you can let $p$ be the first remaining prime such that either $Q_n$ or $-Q_n$ is a quadratic residue mod $p$. (Thus it will simply be the next prime if that prime happens to be 3 mod 4.) Or, assuming that you only use the sum formula, you would only check $-Q_n$. Then experimentally, $q_{n+1} = p_{n+1}$ in all three cases.

This describes the sequence that Charles obtained, which I also obtained (at least the first half of it) with a Sage program.

```
qseq = [2]
for z in range(16):
attain = set(); Q = prod(qseq)
prediction = [p for p in list(primes(500)) if not p in qseq and
(legendre_symbol(-1,p) == -1 or legendre_symbol(Q,p) == 1)][0]
for S in Combinations(qseq):
A = prod(S); attain.update([abs(A-Q/A),A+Q/A])
best = 10000
attain.discard(1)
best = min([min(factor(m))[0] for m in attain])
print prediction,':',qseq,'->',best
qseq.append(best)
```

Assuming both the sum and the difference are used, the validity of this prediction reduces to the following: If $p$ is this next prime, then the question is whether $1$ appears in the set product of the subsets $\{q_k,1/q_k\}$ and $\{\pm 1\}$ in the multiplicative group $(\mathbb{Z}/p)^*$. This is a question in additive set theory if you take a formal logarithm and pass to the additive group $\mathbb{Z}/(p-1)$. A reasonable conjecture is that you are convolving a lot of subsets that don't come close to lying in a subgroup of $\mathbb{Z}/(p-1)$, so you can expect $0$ to appear in the sum. The heuristic has such a large margin that it may be possible with existing methods to prove that it is always so.

The other part of the question is how much you deviate from seeing the primes in order. Again, using a randomness heuristic in number theory, the "probability" that a prime $p$ will be skipped at a given stage is 1/2 (assuming that it is 1 mod 4 in the second case). So you expect the first skipped prime to eventually be included with a "half life" of one round. Thus you expect the sequence to only ever be barely ahead of the list of primes, and you expect $q_n$ to be the $n$th prime infinitely often.