A subset of an abelian group is sequenceable if there is an ordering $(x_1, \ldots, x_k)$ of its elements such that the partial sums $(y_0, y_1, \ldots, y_k)$, given by $y_0 = 0$ and $y_i = \sum_{j=1}^i x_j$ for $1 \leq i \leq k$, are distinct, with the possible exception that we may have $y_k = y_0 = 0$. We demonstrate the sequenceability of subsets of size $k$ of $\mathbb{Z}_n \setminus \{ 0 \}$ when $n = mt$ in many cases, including when $m$ is either prime or has all prime factors larger than $k! /2$ for $k \leq 11$ and $t \leq 5$ and for $k=12$ and $t \leq 4$. We obtain similar, but partial, results for $13 \leq k \leq 15$. This represents progress on a variety of questions and conjectures in the literature concerning the sequenceability of subsets of abelian groups, which we combine and summarize into the conjecture that if a subset of an abelian group does not contain $0$ then it is sequenceable.
Let {a 1 , ..., an} be a set of positive integers with a 1 < • • • < an such that all 2 n subset sums are distinct. A famous conjecture by Erdős states that an > c • 2 n for some constant c, while the best result known to date is of the form an > c • 2 n / √ n. In this paper, we weaken the condition by requiring that only sums corresponding to subsets of size smaller than or equal to λn be distinct. For this case, we derive lower and upper bounds on the smallest possible value of an.
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