Zero-sum problems with subgroup weights

Abstract

In this note, we generalize some theorems on zero-sums with weights from Mathematics Subject Classification 2001 : 11B13

On weighted zero-sum sequences

“…For any abelian p-group G, our upper bound for s A (G) in terms of |A| is essentially best possible, as illustrated by the following example (see also [4] for the particular case G = Z r p ).…”

“…and proved that f {±1} (n, 2) = 2n − 1 when n is odd. If p is a prime, A ⊆ [1, p − 1], and {a mod p : a ∈ A} is a subgroup of the multiplicative group Z * p = Z p \ {0}, then the authors in [4] showed that f A (p, r) ≤ r(p − 1) |A| + p for 1 ≤ r < p|A| p − 1 ;…”

Weighted EGZ-constant for p-groups of rank 2

“…In the case of G = Z r p , for p a prime number, Adhikari et al (see [2]) proved that s A (G) = p + r, for all p > r. In this direction Luca [19] (see also [16]) proved that s A (Z n ) = n + Ω(n) and he classified the extremal A-weighted zero-sum free sequences for n = p k , where Ω(n) denotes the total number of prime divisors of n (counted with multiplicity). This result was conjectured by Adhikari et al (see [3]).…”

“…We list here some of these contributions. (i) s A (Z r 2 ) = 2 r + 1 (see [11]); (ii) s A (Z 3 ) = 4 (see [3]), s A (Z 2 3 ) = 5 (see [4]), s A (Z 3 3 ) = 9 (see [9,13]), s A (Z 4 3 ) = 21, s A (Z 5 3 ) = 41 and s A (Z 6 3 ) = 113 (see [13]); (iii) s A (Z 4 ) = 6 and s A (Z 6 ) = 8 (see [3]); s A (Z 2 4 ) = 8 (see [1]); (iv) s A (Z 2 ⊕Z 4 ) = 7 (see [14,15]), s A (Z 2 2 ⊕Z 4 ) = 8 (see [15]) and s A (Z 2 ⊕Z 6 ) = 9 (see [14]); The following results relate these two constants.…”

“…(ii) s A (Z 3 ) = η A (Z 3 ) + 3 − 1 (see [3]), s A (Z 2 3 ) = η A (Z 2 3 ) + 3 − 1 (see [4,15]); (iii) s A (Z 4 ) = η A (Z 4 ) + 4 − 1 and s A (Z 6 ) = η A (Z 6 ) + 6 − 1 (see [3]); (iv) s A (Z 2 4 ) = η A (Z 2 4 ) + 4 − 1 (see [1,15]); (v) s A (Z 2 ⊕ Z 4 ) = η A (Z 2 ⊕ Z 4 ) + 4 − 1 (see [14,15]), s A (Z 2 2 ⊕ Z 4 ) = η A (Z 2 2 ⊕ Z 4 ) + 4 − 1 (see [15]) and s A (Z 2 ⊕ Z 6 ) = η A (Z 2 ⊕ Z 6 ) + 6 − 1 (see [14]); (vi) s A (Z p s ⊕ Z p r ) = η A (Z p s ⊕ Z p r ) + p r − 1, where p is an odd prime number and s ∈ {1, 2} (see [5,6]);…”