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 ${\cal P}'={\cal P}$ ?

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    $\begingroup$ I don't think I would call this "the algorithm of the Greeks", since Eratosthenes produced an algorithm which definitely captures all the primes. $\endgroup$
    – Todd Trimble
    Dec 2, 2018 at 22:28
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    $\begingroup$ 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. $\endgroup$ Dec 2, 2018 at 22:33
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    $\begingroup$ In the spirit of 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. $\endgroup$ Dec 3, 2018 at 0:21
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Dec 3, 2018 at 2:17
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    $\begingroup$ @Acccumulation Which is exactly how empty products are treated, yes. $\endgroup$ Dec 4, 2018 at 1:11

1 Answer 1


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.

  • $\begingroup$ 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.) $\endgroup$ Dec 3, 2018 at 14:07
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    $\begingroup$ @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. $\endgroup$
    – user44191
    Dec 3, 2018 at 19:10
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    $\begingroup$ The OEIS A000945 b-file gives $p_{50}=31$ $\endgroup$
    – Henry
    Dec 3, 2018 at 21:32

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