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Euclid's proof of the infinitude of primes goes like this:

Let $p_i$ be the $i$-th prime number.

Then for every $n$, $1+\prod_{1\leq i\leq n}p_i$ is not divisible by any of the first $n$ primes, so its prime divisors must be greater than $p_n$. Thus, we have shown that for every prime, there is a larger prime.

Let $P_n$ be the smallest prime divisor of $1+\prod_{1\leq i\leq n}p_i$.

Question: Are there any interesting results on how close (asymptotically) $P_n$ is to $p_{n+1}$? For instance, is there some constant $c$ such that $\lim_{n\rightarrow\infty}(p_{n+1}/P_n)=c$?

semisimpleton
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    In some cases it should be expected that $P_n$ would be the only divisor of the listed quantity... – abiessu Mar 07 '24 at 02:36
  • The product $\prod_{i\le i\le n}p_i$ is called the $n$-th primorial, denoted by $p_n#$. Its successor $p_n#+1$ is called a Euclid number. If it is prime, it’s called a primorial prime. It’s known neither whether infinitely many Euclid numbers are prime, nor whether infinitely many are composite, nor whether they’re all squarefree (Is there an infinite number of primes constructed as in Euclid's proof?, ... – joriki Mar 07 '24 at 09:43
  • ... Wikipedia, MathWorld).

    That makes it seem unlikely that there are results of the kind you’re looking for. A value $c\ne0$ would be incompatible with this state of ignorance (since the limit for the subsequence of prime Euclid numbers is $0$). The value $c=0$ could in principle be known, but this would mean that while we can’t exclude all Euclid numbers being composite, we can exclude that they have prime factors anywhere near $p_n$.

     – joriki Mar 07 '24 at 09:43

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