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Truncatable primes and unavoidable sets of divisors. (English) Zbl 1127.11010

Summary: 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 \(u_0b^n+ u_1b^{n-1}+\cdots+ u_{n-1}b+u_n\), 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 \(b\in \{2,3,4,6\}\), and that no unavoidable set exist in base \(b=5\). Now, we prove that there are no unavoidable sets in base \(b\geq 3\) if \(b-1\) is not square-free. In particular, for \(b=10\), this implies that, for any finite set of prime numbers \(\{p_1,\dots,p_k\}\), there is a nonnegative integer \(u_0\) and \(u_1,u_2,\dots\in \{0,1,\dots,0\}\) such that the number \(u_010^n+ u_110^{n-1}+\cdots+ u_n\) is not divisible by \(p_1,\dots, p_k\) for each integer \(u\geq 0\).

MSC:

11A63 Radix representation; digital problems
11A41 Primes
11B50 Sequences (mod \(m\))
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References:

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