Our previous research was devoted to the problem of determining the primitive periods of the sequences ${(G_{n}mod{p}^t)}_{n=1}^{\infty }$ where ${\left(G_n\right)}_{n=1}^{\infty }$ is a Tribonacci sequence defined by an arbitrary triple of integers. The solution to this problem was found for the case of powers of an arbitrary prime $p\ne 2,11$. In this paper, which could be seen as a completion of our preceding investigation, we find solution for the case of singular primes $p=2,11$.

Our research was inspired by the relations between the primitive periods of sequences obtained by reducing Tribonacci sequence by a given prime modulus $p$ and by its powers $p^t$, which were deduced by M. E. Waddill. In this paper we derive similar results for the case of a Tribonacci sequence that starts with an arbitrary triple of integers.

In order to study the behavior of the points in a tower of curves, we introduce and study trivial points on towers of curves, and we discuss their finiteness over number fields. We relate the problem of proving that the only rational points are the trivial ones at some level of the tower, to the unboundeness of the gonality of the curves in the tower, which we show under some hypothesis.

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 $$\left[\begin{array}{c}n\\ m\end{array}\right]=\prod _{k=1}^m\frac{1-q^{n-k+1}}{1-q^k}\text{and}{(a;q)}_n=\prod _{k=0}^{n-1}(1-a{q}^k).$$