Cfrgtkky

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

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 1pmod {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: $ 1le 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.