Are all primes Euclid primes?

Question

I was learning Euclid's theorem. If we repeat his construction (properly modified to give only primes) then will we skip any primes? Formally:

$p_1 = 2$, and $p_{n + 1} = \text{smallest} \; q \in \mathbb{N} - \{1\} \; \text{s.t} \quad q \; | \;(p_1\cdot ... \cdot p_n + 1)$

In other words, $p_{n + 1}$ is the smallest number that divides the number made by Euclid's construction on $\{p_1, ..., p_n\}$.

Call the above Euclid primes (starting at $2$). The first few numbers in this sequence are: $2,3,7,43,13,53,5,...$ Note how it skips $5$, but then comes back to it. Are there any primes that will not appear in this sequence? If so, can we pick a different $p_1$ to avoid this? Can we predict the skipped numbers based on $p_1$?

Related question

Is there an infinite number of primes constructed as in Euclid's proof?

Answer 1

As the references in the answer to this question say, this is an open problem; the sequence $(p_n)$ is known as the Euclid–Mullin sequence.

Answer 2

A related problem has been solved. Instead of taking the smallest prime factor, consider taking the largest prime factor.

It was proven in 2012 (Brooker), that this sequence omits infinitely many primes. An elementary proof by P. Pollack and E. Trevino can be read here: http://pollack.uga.edu/mullin.pdf

Specifically, they proved the following:

Let $q_1 = 2$. Supposing that we have defined $q_j$ for all $1 \leq j \leq k$, let $q_{k+1}$ be the largest prime factor of $1 + \Pi^k_{j=1} q_j$. Then the sequence $\{q_i\}$ omits infinitely many primes.