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Prime numbers; truncatable primes; integer expansions; square-free numbers
We are interested whether there is a nonnegative integer $u_0$ and an infinite sequence of digits $u_1, u_2, u_3, \dots $ in base $b$ such that the numbers $u_0 b^n+u_1 b^{n-1}+\dots + u_{n-1} b +u_n,$ where $n=0,1,2, \dots ,$ are all prime or at least do not have prime divisors in a finite set of prime numbers $S.$ If any such sequence contains infinitely many elements divisible by at least one prime number $p \in S,$ then we call the set $S$ unavoidable with respect to $b$. It was proved earlier that unavoidable sets in base $b$ exist if $b \in \lbrace 2,3,4,6\rbrace ,$ and that no unavoidable set exists in base $b=5.$ Now, we prove that there are no unavoidable sets in base $b \geqslant 3$ if $b-1$ is not square-free. In particular, for $b=10,$ this implies that, for any finite set of prime numbers $\lbrace p_1, \dots , p_k\rbrace ,$ there is a nonnegative integer $u_0$ and $u_1, u_2, \dots \in \lbrace 0,1,\dots ,9\rbrace $ such that the number $u_0 10^n + u_1 10^{n-1}+\dots +u_{n}$ is not divisible by $p_1, \dots , p_k$ for each integer $n \geqslant 0.$
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