We are interested whether there is a nonnegative integer $u_0$ and an infinite sequence of digits $u_1,u_2,u_3,\cdots$ in base $b$ such that the numbers $u_0{b}^n+u_1{b}^{n-1}+\cdots +u_{n-1}b+u_n,$ where $n=0,1,2,\cdots ,$ 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 exists in base $b=5.$ Now, we prove...

We establish two truncations of Gauss’ square exponent theorem and a finite extension of Euler’s identity. For instance, we prove that for any positive integer $n$, $$\sum _{k=0}^n{(-1)}^k\left[\begin{array}{c}2n-k\\ k\end{array}\right]{(q;q^2)}_{n-k}{q}^{\left(\genfrac{}{}{0pt}{}{k+1}{2}\right)}=\sum _{k=-n}^n{(-1)}^k{q}^{k^2},$$ where $$\left[\begin{array}{c}n\\ m\end{array}\right]=\prod _{k=1}^m\frac{1-q^{n-k+1}}{1-q^k}\text{and}{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k).$$