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Here I found a question:

Show that every prime not equal to $2$ or $5$ divides infinitely many of the numbers $1,11,111,1111,\dots$ etc.

which is partly solved here Prime numbers divide an element from a set.

From this the following conjecture seems reasonable:

Given any finite set $S=\{q_1,\dots,q_k\}$ of $k$ primes, then any prime $p\notin S$ divides infinitely many of the numbers $a_1,a_2,a_3\dots$, where $a_1=1$ and $a_{n+1}=1+a_n\prod q_i$.

Can this be proved?

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  • $\begingroup$ Please don't reapply the conjecture tag to personal conjectures that turn out to be (well-)known results. $\endgroup$ – Bill Dubuque May 6 at 7:15
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By induction $\,a_n = \dfrac{q^n-1}{q-1},\ q = \prod q_i.\,$ $a_n$ is a divisibility sequence, i.e. $\,n\mid m\Rightarrow a_n\mid a_m\,$ so $\,p\mid a_n\mid a_{nk}\Rightarrow p\mid a_{nk}\,$ for all $\,k\in\Bbb N$

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Define $Q:=\prod q_i$. When $Q=1$ the result is trivial.

Otherwise, it is clear that $a_n = {\underbrace{11\dots1_Q}_{n\ 1\text{'s}}}=1+Q+Q^2+\dots+Q^{n-1} = \dfrac {Q^n - 1}{Q-1}$.

The rest essentially follows from changing $10$ to $Q$. Details below:

Choose your $p\notin S$.

For $p\nmid (Q-1)$, as by Fermat $p\mid (Q^{k(p-1)}-1)$ for all $k\in \mathbb N$, $p \mid a_{k(p-1)}$ by coprimality.

For $p \mid (Q-1)$, observe that $$a_n = 1+Q+Q^2+\dots+Q^{n-1}\equiv1+1+1^2+\dots +1^{n-1}\equiv n \pmod Q$$ so $a_{kp}\equiv kp\equiv 0 \pmod p$ for all $k\in \mathbb N$.

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