Zero-sum free sets with small sum-set

Abstract

Abstract. Let A be a zero-sum free subset of Z n with |A| = k. We compute for k ≤ 7 the least possible size of the set of all subset-sums of A.

Long unsplittable zero-sum sequences over a finite cyclic group

“…As above we have S = (g l1 (a 2 g) 2 )T 1 T 2 T 3 T 4 T 5 where T i is square-free for all i ∈ [1,5], |T j | = 3 for j ∈ [0, s] and |T s+k | = 2 for k ∈ [0, 5 − s]. By Lemmas 2.3 and 2.8, we have…”

“…Since h(T ) = 5 and |T | ≥ 25, it follows from Lemma 2.9 that T = T 1 T 2 T 3 T 4 T 5 with T i square-free and |T i | ≥ 5 for all i ∈ [1,5]. By Lemma 2.8(3), [1,5].…”

“…Note that m − 2 = 4s + 3(5 − s) where s ∈ [1,4]. Since S = (g l1 (a 2 g) 2 )T with T = (a 2 g) l2−2 (a 3 g) l3 · · · (a r g) lr , by Lemma 2.9 we have T = T 1 T 2 T 3 T 4 T 5 with T i square-free for all i ∈ [1,5], |T j | = 4 for j ∈ [1, s] and |T s+k | = 3 for all k ∈ [1, 5 − s]. By Lemmas 2.3 and 2.8, we have n ≥ | (g l1 (a 2 g) 2 )| + 5 i=1 (| (T i )|) ≥ 3( n 3 − m + 8) + 2 + 10s + 6(5 − s) > n + s + 2 > n, yielding a contradiction.…”

The structure of maximal zero-sum free sequences

“…. , k}, this estimate is optimal up to the value of δ for odd k. This deficiency causes some trouble in our treatment of small primes, which motivated us to prove the following using computer calculations [5].…”

On the Davenport constant and on the structure of extremal zero-sum free sequences

“…Computational methods have already been used successfully for a variety of zero-sum problems (see recent work of G. Bhowmik, Y. Edel, C. Elsholtz, I. Halupczok, J.-C. Schlage-Puchta et al [6,4,5,1]). Inspired by former work in the groups C 2 2 ⊕ C 3 2n for n ≥ 3 odd, we found many examples of zero-sum free sequences S over G = C 2 ⊕ C 4 6 of length |S| = d * (G) + 1.…”