On a problem of Molluzzo concerning Steinhaus triangles in finite cyclic groups

Let $X$ be a finite sequence of length $m\geq 1$ in $\mathbb{Z}/n\mathbb{Z}$. The \textit{derived sequence} $\partial X$ of $X$ is the sequence of length $m-1$ obtained by pairwise adding consecutive terms of $X$. The collection of iterated derived sequences of $X$, until length 1 is reached, determines a triangle, the \textit{Steinhaus triangle $ΔX$ generated by the sequence $X$}. We say that $X$ is \textit{balanced} if its Steinhaus triangle $ΔX$ contains each element of $\mathbb{Z}/n\mathbb{Z}$ with the same multiplicity. An obvious necessary condition for $m$ to be the length of a balanced sequence in $\mathbb{Z}/n\mathbb{Z}$ is that $n$ divides the binomial coefficient $\binom{m+1}{2}$. It is an open problem to determine whether this condition on $m$ is also sufficient. This problem was posed by Hugo Steinhaus in 1963 for $n=2$ and generalized by John C. Molluzzo in 1976 for $n\geq3$. So far, only the case $n=2$ has been solved, by Heiko Harborth in 1972. In this paper, we answer positively Molluzzo's problem in the case $n=3^k$ for all $k\geq1$. Moreover, for every odd integer $n\geq3$, we construct infinitely many balanced sequences in $\mathbb{Z}/n\mathbb{Z}$. This is achieved by analysing the Steinhaus triangles generated by arithmetic progressions. In contrast, for any $n$ even with $n\geq4$, it is not known whether there exist infinitely many balanced sequences in $\mathbb{Z}/n\mathbb{Z}$. As for arithmetic progressions, still for $n$ even, we show that they are never balanced, except for exactly 8 cases occurring at $n=2$ and $n=6$.

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