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As pointed out by Jeppe Stig Nielsen, my previous question A question about numbers from Euclid's proof of infinitude of primes actually can be viewed as a duplicate of Are Euclid numbers squarefree?

I then thought - maybe a simplified version could be more accessible. Let us modify Euclid's argument for infinitude of primes like this. Having discovered some primes $p_1,...,p_n$, we look at $p_1\cdot...\cdot p_n+1$; all of its prime factors are new, $p_{n+1},...,p_{n+m}$; we then look at $p_1\cdot...\cdot p_{n+m}+1$, etc.

The point is that this procedure is equivalent to something simpler: define recurrently $N_1=2$ and $N_{n+1}=N_n(N_n-1)+1$. We obtain $$ \begin{aligned} N_1&=2\\ N_2&=3\\ N_3&=7\\ N_4&=43\\ N_5&=1807=13\cdot139\\ N_6&=3263443\\ N_7&=10650056950807=547\cdot607\cdot1033\cdot31051\\ N_8&=113423713055421844361000443=29881\cdot67003\cdot9119521\cdot6212157481 \end{aligned} $$ (by construction, every next number only contains new primes, not previously occurring on the list).

The question now is similar to that in the linked questions: are all these numbers squarefree? Are there any more primes in this list?

And an additional question: will all primes eventually occur as factors somewhere on the list?

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    $\begingroup$ Not all primes are in the list. $N_6 \mod 17 = 4, N_7 \mod 17 = 13$. From there on the modulus cycles through 4 and 13 $\endgroup$ – Michael Stocker Mar 24 '17 at 10:11
  • $\begingroup$ @MichaelStocker Great idea! I checked other primes up to 71, the same applies to all of them - they generate quite short periods, the longest has 31 (length 9). Seems like if a prime $p$ does not appear until $p$th step it does not have any chance. $\endgroup$ – მამუკა ჯიბლაძე Mar 24 '17 at 12:06

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