2016
DOI: 10.1142/s1793042116501141
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Inverse results for weighted Harborth constants

Abstract: Abstract. For a finite abelian group (G, +) the Harborth constant is defined as the smallest integer ℓ such that each squarefree sequence over G of length ℓ has a subsequence of length equal to the exponent of G whose terms sum to 0. The plus-minus weighted Harborth constant is defined in the same way except that the existence of a plus-minus weighted subsum equaling 0 is required, that is, when forming the sum one can chose a sign for each term. The inverse problem associated to these constants is the problem… Show more

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Cited by 6 publications
(5 citation statements)
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“…Much less is known about the Harborth constant than about the EGZ constant despite their similarities. Marchan, Ordaz, Ramos, and Schmid [13], [14] computed the Harborth constants for some classes of abelian groups, in particular, g(C 2 ⊕ C 2n ) = 2n + 3, n odd 2n + 2, n even for all n ≥ 1. They also consider the plus-minus weighted analogue of the Harborth constant, in which one may add either an element or its inverse in the sums of length exp(G).…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…Much less is known about the Harborth constant than about the EGZ constant despite their similarities. Marchan, Ordaz, Ramos, and Schmid [13], [14] computed the Harborth constants for some classes of abelian groups, in particular, g(C 2 ⊕ C 2n ) = 2n + 3, n odd 2n + 2, n even for all n ≥ 1. They also consider the plus-minus weighted analogue of the Harborth constant, in which one may add either an element or its inverse in the sums of length exp(G).…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…We say that S ⊆ G of size exp(G) ≤ |S| < g(G) passes if it does not fail. (See [14] for a discussion of this concept for G = C 2 ⊕C 2n .) In this section, we characterize all failing subsets of H n,m .…”
Section: Characterizing Failing Subsets For N and M Evenmentioning
confidence: 99%
“…Furthermore, in [150] the same authors determined all subsets A of size 2k + 1 in Z 2 × Z 2k for which 0 ∈ (2k)±A.…”
Section: F41 Fixed Number Of Termsmentioning
confidence: 99%
“…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.…”
Section: Introductionmentioning
confidence: 99%
“…(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]);…”
Section: Introductionmentioning
confidence: 99%