On a problem of Molluzzo concerning Steinhaus triangles in finite cyclic groups
Résumé
Let $X$ be a finite sequence of length $m\geqslant 1$ in $\mathbb{Z}/n\mathbb{Z}$. The 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 Steinhaus triangle $\Delta X$ generated by the sequence $X$. We say that $X$ is balanced if its Steinhaus triangle $\Delta X$ 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 ${m+1 \choose 2}$. 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 $n\geqslant 3$. 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 $k\geqslant 1$. Moreover, for every odd integer $n\geqslant 3$, 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 $n\geqslant 4$, 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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