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If I start with a set of consecutive primes, let's say $\{2,3,5,7\}$. Consider $2\times 3\times5\times7+ 1 = 211$. Now, $211$ is prime, so I add it to the set: $\{2,3,5,7,211\}$. If the resulting number is not prime, add the prime divisors of this number which are not in the set.

Just out of curiosity, does this generate all primes for any given starting set? If true, how can you prove that?

ultralegend5385
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    I believe this is an open problem, but I might be wrong. See the discussion here. The usual construction only adds the smallest prime obtained by Euclid's construction, not all the primes. So...maybe the broader problem is known? – lulu Mar 28 '24 at 12:51
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    Heuristically , eventually every prime should appear since it is reasonable to assume that the remainder modulo some prime $p$ is random , so we should have a $\frac{1}{p}$-chance to hit some given prime. This is of cours no proof and I strongly doubt there is one. There is a very similar problem for which it is unknown whether every prime appears. – Peter Mar 28 '24 at 13:04
  • Note that we will sooner or later hit a "brickwall" , a number too difficult to be factored in practice. – Peter Mar 28 '24 at 13:06

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