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Let ${\cal P}$ be the set of prime numbers. Define a subset ${\cal P}'=\{p_1,p_2,p_3,\cdots\}$ of ${\cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${\cal P}$ dividing $1+p_1\cdots p_n$. Is there any obstruction to ${\cal P}'={\cal P}$ ?

Konstantinos Kanakoglou
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Zidane
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    I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes. – Todd Trimble Dec 02 '18 at 22:28
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    This and variations have been considered. There is no nice (and known to me) alternate characterization of any of these variants. You should check the OEIS for more information. Gerhard "Should Always Check The OEIS" Paseman, 2018.12.02. – Gerhard Paseman Dec 02 '18 at 22:33
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    In the spirit of https://mathoverflow.net/questions/45951/sexy-vacuity, let me point out that the special case $p_1 = 2$ is unnecessary here — the single general case “$p_n$ is the smallest prime dividing $1 + p_1 \cdots p_{n-1}$” suffices to define the whole sequence. – Peter LeFanu Lumsdaine Dec 03 '18 at 00:21
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    Note that this is not an algorithm at all, just a recursively defined sequence. An algorithm also needs to specify the way to compute it; especially in this case how to compute the smallest prime divisor. – YCor Dec 03 '18 at 02:17
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    @YCor An algorithm is just a process, so this is still an algorithm. It just has one step whose method is unspecified. Which makes it a not-very-useful algorithm, of course, but still valid for thought experiments. – Graham Dec 03 '18 at 09:08
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    Given that this algorithm is defined in terms of the set of prime numbers, it seems to me that there is little sense in which it "produces" the prime numbers. – Acccumulation Dec 03 '18 at 23:45
  • @PeterLeFanuLumsdaine $p_1=2$ is unnecessary only if one treats the null product as being 1. – Acccumulation Dec 03 '18 at 23:46
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    @Acccumulation Which is exactly how empty products are treated, yes. – Johannes Hahn Dec 04 '18 at 01:11
  • I would not call this "the algorithm of the Greeks" because it is only through a mistake about history that it is widely believed that Euclid did this. (Dirichlet probably started this in his posthumous 1859 book on number theory.) Euclid actually wrote that for any finite set $S$ of primes (e.g. $S={5,7}$) the prime factors of $1 + \prod S$ (in the example $1 +(5\times7) = 36 = 2\times2\times3\times3$) are not in $S.$ Catherine Woodgold and I have a joint paper about the history and why it matter in the 2009 volume of the mathematical intelligencer. – Michael Hardy Dec 28 '22 at 21:43
  • @ToddTrimble $\qquad\uparrow\qquad$ – Michael Hardy Dec 28 '22 at 21:45

1 Answers1

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According to Booker - A variant of the Euclid-Mullin sequence containing every prime, as of 2016, this question remains open.

One of the central questions in this area was posed by Mullin [6] in 1963: Does the Euclid–Mullin sequence contain every prime number? Despite a compelling heuristic argument of Shanks [9] that the answer is yes, even the broader question of whether there is any Euclid sequence containing every prime number remains open.

The OEIS contains a decent amount of information. For example, the primes up to 37 do appear within the first 50 terms of the sequence.

user44191
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  • Any idea what the euristic expected asymptotic growth of the least missed prime $p_n$ at step $n$ would be? (Sorry for this sentence's violence to the English language.) – Yaakov Baruch Dec 03 '18 at 14:07
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    @YaakovBaruch I don't have an answer for you, but I do see multiple references to a heuristic argument by Daniel Shanks in the Bulletin of the Institute of Combinatorics and its Applications (specifically, in 1991). Unfortunately, I can't find the argument itself. – user44191 Dec 03 '18 at 19:10
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    The OEIS A000945 b-file gives $p_{50}=31$ – Henry Dec 03 '18 at 21:32