Given is the following sequence: $a_1 = 1$ and $a_n$ equals the biggest prime divisor of $1+ a_1*\dots*a_{n-1}$ . It is then claimed: $11$ does not occur in this sequence.

How can one approach this problem? I first thought about trying a proof by contradiction. Let's then be $a_m$ the first occurrence of $11$ in this sequence. Then there exist $a,b,c,d, e \in \mathbb{N_0}$ such that: $2^a * 3^b * 5^c * 7^d * 11^e = 1+ a_1*\dots*a_{m-1}$ .

From there however, it looks me like a dead end. It seems like I am missing a key observation. I would be happy, if anyone with more knowledge could offer an advice/suggestion how to begin here.

EDIT: Related: Euclid Mullin Sequence

  • $\begingroup$ I am not a number theory specialist, but this does not look “elementary” to me. There are some references in oeis.org/A000946, oeis.org/A216227, and Euclid–Mullin sequence. – Where did you encounter the problem? $\endgroup$ – Martin R Nov 25 '18 at 21:43
  • $\begingroup$ It's from a previous exam preparing for the IMO - so it should be solvable within maybe 1 hour and not require university level mathematics. I tried finding more values of the sequence with a python script, but the values become too big way for the program to handle. $\endgroup$ – Imago Nov 25 '18 at 21:46
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    $\begingroup$ Note: it's easy to see that no prime can appear twice in the sequence. As to $11$, and the other missing primes, the wiki article gives a reference for the claim $\endgroup$ – lulu Nov 25 '18 at 21:46
  • $\begingroup$ @lulu, if this is true, this says indeed a lot and probably just nails the problem. I will look into that. EDIT: Ok the wiki article spoilers quite a lot. $\endgroup$ – Imago Nov 25 '18 at 21:50
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    $\begingroup$ Cox and van der Poorten had a short paper in 1967 that solves this in a fairly elementary way. doi.org/10.1017/S1446788700006236 $\endgroup$ – B. Goddard Nov 25 '18 at 21:59

Note: Personally, I find my solution too short and "too easy" to be a solution to this problem. Therefore, feel free to make a note or comment, when there is an error.

I would "solve" this specific problem stated above like this:

The first elements of the sequence can be computed fairly quickly: $a_1 = 1, a_2 = 2, a_3 = 7, a_4 = 43, a_5 = 139$

Now following lulu's advice, we show that a prime number occurs at most twice in this sequence.

Given that $a_n = 1+a_1 * \dots * a_{n-1}$ it follows that $a_n \equiv 1\pmod {a_i} $ for $1 \le i \le n-1$, thus there is no $a_i$ with $a_i| a_n$ $\Rightarrow a_i \neq a_n \ \forall i$ with: $ \ 1\le i \le n-1$

Now: Let $a_n = 11$ and $b_n = 1+ a_1* \dots \ * a_{n-1}$ then by definition $11|b_n$ and if $k|b_n$ and $k \neq 11$, then $k = 1 \lor k= 5$, in particular: $2,3,7 \nmid b_n$ $\Rightarrow b_n \le 55$, but: $\frac{b_n}{55} \ge \frac{a_1 * \dots * a_5 +1}{55} \gt 1 \Rightarrow$ $b_n$ has at least one prime factor greater than $11$, thus: $11$ can't occur in this sequence.

EDIT: Proving that does not occur in the sequence can be doen quickly: If 5 occurred, then we had for some $b_n = 2^a * 3^b * 5^c = 1+2* \dots * a_{n-1} $.

$a,b$ must be equal to $0$. If they weren't, then $2^a * 3^b * 5^c \equiv 0 \pmod {2,3} $, but $b_n \equiv 1 \pmod {2,3} $

Thus $b_n$ would be of the form: $b_n = 5^c$ for some $c$ however: $5^c - 1 \equiv 0 \pmod 4 \forall \ c \in \mathbb{N}$ and $a_1* \dots \ * a_{n-1} \equiv 2 \pmod 4 $

Thus $5$ cannot occur in the sequence.

However there is still some way to go to prove $11$ does not occur and I actually don't see how one would do that. The same argument/trick, used for $5$, doesn't seem work.

EDIT2 : $5^a * 11 ^b \equiv 3 \pmod 4$ as $2 | a_1 * \dots* a_{n-1} $ $\Rightarrow 1^a * 3^b \equiv 3 \pmod 4 \Rightarrow b$ is odd.

$3 | 5^a * 11^b -1 \Rightarrow (-1)^a *(-11)^b \equiv (-1)^{a+b} -1 \equiv 0 \pmod 3 \Rightarrow a+b $ even$ \Rightarrow a$ is also odd.

Then: $7 | 5^a * 11^b -1 \Rightarrow (-2)^a * 4^b \equiv (-2)^{a+2*b} \equiv 1 \pmod 7 $

This however is only the case: if $a+2*b$ is a multiple of $6$ (Little Fermat), which can't be the case. Hence: $11$ cannot occur in the series and we are done. Quite a journey over the past few days, but I believe the proof is now complete.

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    $\begingroup$ $b_n$ might be $5^y11^z$ for integers $y,z$ $\endgroup$ – Empy2 Nov 26 '18 at 14:27
  • $\begingroup$ Valid point. I need more time to think about it. $\endgroup$ – Imago Nov 26 '18 at 14:40

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