k-barycentric Olson constant

Abstract

Let G be a finite Abelian group of order n. A k-sequence in G is said to be barycentric if it contains an element which is the "average" of its terms. The k-barycentric Olson constant BO(k, G) is introduced as the minimal positive integer t such that any t-set in G contains a k-barycentric set. Conditions for the existence of BO(k, G) are established and some values or bounds are given. Moreover, new values for the k-barycentric Davenport constant BD(k, G) and barycentric Ramsey numbers for stars BR(K 1,k , G) are given. These constants, defined in some of our previous works, use sequences instead of sets.

Full text

k-BARYCENTRIC OLSON CONSTANT
OSCAR ORDAZ, MAR´
IA TERESA VARELA and FELICIA VILLARROEL
Let Gbe a finite Abelian group of order n. A k-sequence in Gis said to be
barycentric if it contains an element which is the “average” of its terms. The k-
barycentric Olson constant BO(k, G) is introduced as the minimal positive integer
tsuch that any t-set in Gcontains a k-barycentric set. Conditions for the existence
of BO(k , G) are established and some values or bounds are given. Moreover,
new values for the k-barycentric Davenport constant BD(k , G) and barycentric
Ramsey numbers for stars BR(K1,k , G) are given. These constants, defined in
some of our previous works, use sequences instead of sets.
AMS 2000 Subject Classification: 05C55, 05C65, 05D10.
Key words: k-barycentric sequence, k-barycentric, k-barycentric Olson constant,
barycentric Ramsey number, zero-sum.
1. INTRODUCTION
Let Gbe an Abelian group of order n. Weighted sequences, i.e., sequences
constituted by terms of the form wiai, where aiare elements of Gand the coef-
ficients or weights are positive integers, appear initially in the Caro conjecture
[2] formulated in 1996.
Conjecture 1.1.Let w1, w2, . . . , wkbe positive integers such that w1+
w2+· · ·+wk= 0 (mod n). Let a1, a2, . . . , an+k−1in Gnot necessarily distinct.
Then there exist kdistinct indices i1, . . . , iksuch that w1ai1+w2ai2+· · · +
wkaik= 0.
Hamidoune [11] gave a solution to Caro conjecture under the additional
assumption (wi, n) = 1 ∀i. Recently, Grynkiewicz [10] gave a complete solution
to this conjecture in
Theorem 1.2 ([10]).Let m, n and k≥2be positive integers. If fis a
sequence of n+k−1elements from a nontrivial Abelian group Gof order n
and exponent m, and if W={wi}k
i=1 is a sequence of integers whose sum is
zero modulo m, then there exists a rearranged subsequence {bi}k
i=1 of fsuch
that
k
P
i=1
wibi= 0.Furthermore, if fhas a k-set partition A=A1, . . . , Aksuch
MATH. REPORTS 11(61),1(2009), 33–45

34 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 2
that |wiAi|=|Ai|for all i, then there exists a nontrivial subgroup Hof Gand
ak-set partition A1=A1
1, . . . , A1
kof fwith H⊆
k
P
i=1
wiA1
iand |wiA1
i|=|A1
i|
for all i.
A weighted k-sequence, k≥2,with terms wiai,wi= 1,1≤i≤k,
excepting some jfor which wj= 1 −k,ai∈Gand zero-sum, is called a
k-barycentric sequence. That is to say, a k-sequence ai, 1 ≤i≤k, with k≥2,
where there exists an ajsuch that a1+a2+· · · +aj+· · · +ak=kaj.
The study of barycentric sequences started in [6] and [7]. In [6] the
Davenport barycentric constant BD(G) and in [7] the Davenport k-barycentric
constant BD(k, G) are introduced. They are defined as the smallest positive
integer tsuch that every sequence of length tcontains a barycentric sequence
or a k-barycentric sequence, respectively.
Let H= (V(H), E(H)) be a graph with e(H) edges. In [7] the barycen-
tric Ramsey number BR(H, G) is defined as the minimum positive integer t
such that any coloring c:E(Kt)→Gof the edges of Ktby elements of G
yields a copy of H, say H0, with an edge e0such that
(1) X
e∈E(H0)
c(e) = kc(e0)
In this case His called a barycentric graph. This constant is a generalization of
the Ramsey zero-sum number R(H, G) defined when e(H) = 0 (mod n), as the
minimal positive integer ssuch that any coloring c:E(Ks)→Gof the edges of
Ksby elements of Gcontains a copy of H, say H, with P
e∈E(H)
c(e) = 0, where 0
is the zero element of G. The necessity of the condition e(H) = 0 (mod n) for
the existence of R(H, G) is clear. It comes from the monochromatic coloration
of the edges of H.
The main goals of this paper are:
– To define of the k-barycentric Olson constant BO(k, G) and study its
existence. The Abelian groups that will be studied are G=Znfor some prime
or composite n, in particular for 3 ≤n≤12 and 3 ≤k≤n. For those kand
nfor which BO(k, Zn) exists, this value or bounds of it are given.
– To establish new values for BD(k, G) from BO(k, G).
– To give new values for BR(K1,k, G) from BD(k, G).
Besides this introduction and the conclusion, the paper contains three
main sections. Section 2 presents the tools that are used in Sections 3 and 4.
Section 3 is devoted to a study of the k-barycentric Olson constant BO(k, G)
for some kand G. In Section 4, new values of BD(k, G) using the values of

3k-barycentric Olson constant 35
BO(k, G) are established. Moreover, new values of BR(K1,k, G) from BD(k, G)
are given.
2. TOOLS
The following definition is introduced in [6] and [7] and constitutes a
natural extension of zero-sum sequences.
Definition 2.1.Let Abe a finite set with |A| ≥ 2 and Ga finite Abelian
group. A sequence f:A→Gis said to be barycentric if there exists a∈Asuch
that P
x∈A
f(x) = |A|f(a). The element f(a) is called barycenter. The sequence
is said to be k-barycentric when |A|=k. Moreover, when fis injective we use
barycentric set instead of barycentric sequence.
Hamidoune [11] gave the following condition.
Hamidoune-condition. Let Gbe an Abelian group of order n≥2 and
f:A→Ga sequence with |A| ≥ n+k−1. Then there exists a k-barycentric
subsequence of f. Moreover, in the case where k≥ |G|, the condition |A| ≥
k+D(G)−1, where D(G) is the Davenport constant, is sufficient for the
existence of a k-barycentric subsequence of f.
Definition 2.2.Let Gbe an Abelian group of order n≥2. The k-
barycentric Davenport constant BD(k, G) is the minimal positive integer t
such that every t-sequence in Gcontains a k-barycentric subsequence.
By Hamidoune-condition, we have BD(k, G)≤n+k−1.
In what follows, we present some results about orbits. More information
is available in [14].
The set Gn={fa,b :Zn→Zn, fa,b(x) = ax +b, a, b ∈Zn,(a, n) = 1}
is a group of order nφ(n) where φ(n) = |{0< q < n : (q, n) = 1}| is the
Euler phi-function. Let Xk
n={{x1, x2, . . . , xk}:xi∈Zn}.An action of Gn
on Xk
nis defined as fa,b({x1, . . . , xk}) = {fa,b (x1), . . . , fa,b(xk)}.It is easy to
see that θ({0}) is the only orbit of X1
nand that the orbits of X2
nare of the
form θ({0, z}).
The Bezout theorem [5] and the Chinese remainder theorem allow to
formulate
Lemma 2.3.θ({0, z}) = θ({0, t}), with t= (z, n). Moreover, each orbit
in X2
ncontains one and only one {0, t}with t|nand t < n. There are as many
orbits in X2
nas there are divisors of nin {1,2, . . . , n −1}.
Remark 2.4.The orbits of Xk
ncan be obtained considering for each
orbit θ({x1, . . . , xk−1}) of Xk−1
nthe sets {x1, . . . , xk−1, xk}with xk∈Zn\
{x1, . . . , xk−1}. The action of Gnon these sets defines its orbits.

36 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 4
We have the easy lemmas below.
Lemma 2.5.If {x1, . . . , xk}is a k-barycentric set, then all the elements
of θ({x1, . . . , xk})are k-barycentric. Such an orbit is called barycentric. More-
over, if {x1, . . . , xk}contains a t-barycentric set, then every element of θ({x1,
. . . , xk})also contains a t-barycentric set.
Lemma 2.6.The number of orbits of Xk
nis equal to the number of orbits
of Xn−k
n.
Proof. If x, y are in the same orbit of Xk
n, then Zn\xand Zn\yalso
are in the same orbit of Xn−k
n.Consequently, the bijection between the sets
Xk
nand Xn−k
noriginates a bijection between the orbits of Xk
nand Xn−k
n.
We also need the Dias da Silva-Hamidoune theorem below.
Theorem 2.7 ([8]).Let Hbe a subset of Zp. Let dbe an integer such
that 2≤d≤ |H|. Set ∧dH=nP
x∈S
x:S⊂H, |S|=do. Then | ∧dH| ≥
min{p, d(|H| − d)+1}.
We shall also use
Lemma 2.8 (Hamidoune [12]).Let Abe a subset of Znsuch that |A| ≥
n+3
2. Then ∧2A=Zn.
In what follows we present some results about decomposing a complete
graph into edge-disjoint subgraphs.
Theorem 2.9 (Harary [15]).Let Knbe a complete graph with nvertices.
Then Kn, with nodd, is the edge-disjoint union of n−1
2Hamiltonian cycles
while Kn, with neven, is the edge-disjoint union of n−2
2Hamiltonian cycles
and one perfect matching. Hence Kncan be decomposed into n−1perfect
matchings.
Corollary 2.10.Let Knbe a complete graph with nvertices, with n
odd. Then Kncan be decomposed into two complete graphs Kn+1
2
sharing a
vertex and a bipartite complete graph Kn−1
2,n−1
2
.
Corollary 2.11.Let Knbe a complete graph with nvertices, with n
even. Then Kncan be decomposed into two vertex-disjoint complete graphs
Kn
2and the remaining Kn
2,n
2into one perfect matching and one (n
2−1)-
regular graph.
Moreover, the following result is used to give the value of BR(K1,m, Zn)
with m= 0 (mod n).

5k-barycentric Olson constant 37
Theorem 2.12 (Bialostocki [1], Caro [4]).Let K1,m be the stars on m
edges with m= 0 (mod n).Then
BR(K1,m , Zn) = R(K1,m, Zn) = m+n−1if m=n= 0 (mod 2)
m+notherwise.
From now on, pwill be a prime number.
3. k-BARYCENTRIC OLSON CONSTANT
Let Gbe an Abelian group of order n≥3. We establish the existence
of BO(k, G) any derive some values or bounds. The existence of BO(k, Zp) is
established in Corollary 3.4 and Remark 3.2.
Definition 3.1.Let Gbe an Abelian group of order n≥3. The k-bary-
centric Olson constant BO(k , G) is the minimal positive integer tsuch that
every t-set in Gcontains a k-barycentric set, provided such an integer exists.
Remark 3.2.It is clear that when nis odd, BO(n, Zn) = nwhile if n
is even, BO(n, Zn) does not exist. Moreover, for nodd, the barycenter of
every Q⊆Znwith |Q|=n−1 is Zn\Q. Therefore, BO(n−1, Zn) does
not exist. For neven, BO(n−1, Zn) = n−1. Let S⊆Znwith |S|=n−1
and {b}=Zn\S. It is easy to see that Sis an (n−1)-barycentric set with
n
2−1
n−1bas barycenter.
Theorem 3.3 ([6]).Let s≥2,d≥2,p≥d+2+ 1
d−1. Let Abe a set
with s+delements, and f:A→Zpa sequence with |f(A)| ≥ p−1
d+d+ 1.
Then fcontains an s-barycentric subsequence.
Corollary 3.4.BO(k, Zp)≤pfor 3≤k≤p−2with p≥5.
Theorem 3.5.BO(k, Zn)≤nfor n≥6and n+1
2≤k≤n−2.
Proof. Let A⊆Znsuch that |A|=k+1. Then |A| ≥ n+3
2. By Lemma 2.8
there exists {u, v} ⊆ Asuch that u+v=P
A
x−ka +a, for some a∈Zn\A.
Then (A\ {u, v})∪ {a}is a k-barycentric set. 
Lemma 3.6 ([7]).If a 3-sequence in Zn,nodd, is barycentric, then its
elements are equal or pairwise different. Moreover, a 3-set in Znis barycentric
if and only if its elements are in arithmetic progression.
From this lemma we derive
Remark 3.7.BO(3, Zn)≤nfor n≥3.
The following result follows from the Dias da Silva-Hamidoune theorem.

38 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 6
Theorem 3.8.BO(3, Zp)≤ dp
3e+ 1 for p≥5.
Proof. Let Abe a set in Zpwith |A|=dp
3e+1. By the theorem of Dias da
Silva-Hamidoune we have |∧2A| ≥ min{p, 2(|A| − 2)+1}= min{p, 2dp
3e−1}=
2dp
3e − 1. Set D={2x:x∈A}. Then | ∧2A|+|D| ≥ 3dp
3e> p. Therefore,
there exist 2x∈Dand y+z∈ ∧2Asuch that 2x=y+z, hence x+y+z= 3x,
i.e., {x, y, z}is a 3-barycentric set of A.
In particular, we deduce that
–BO(3, Z5) = 3;
– the set {0,4,6}shows that BO(3, Z7)≥4 so, by Theorem 3.8,
BO(3, Z7) = 4;
– the set {1,2,4,5}shows that BO(3, Z11 )≥5, hence BO(3, Z11) = 5;
– the set {0,1,3,9}shows that BO(3, Z13 )≥5 and, by Theorem 3.8 we
have BO(3, Z13)≤6,upper bound improved in the next result.
Lemma 3.9.BO(3, Z13) = 5.
Proof. The action of group G13 on X3
13 partitions it into three orbits:
θ({0,1,2}) (barycentric) and θ({0,1,3}), θ({0,1,4)}) (non barycentric). Now,
we apply Remark 2.4 to the orbits θ({0,1,3}), θ({0,1,4)}) in order to obtain 3-
barycentric-free orbits in X4
13. These orbits are: θ({0,1,3,4}), θ({0,1,3,9}),
θ({0,1,3,11}), θ({0,1,4,5}), θ({0,1,4,6}), θ({0,1,4,10}). Finally, none of
these orbits can be extended by a fifth element without forming a 3-barycentric
subset. 
Lemma 3.10.BO(3, Zn) = 5 for n= 6,8,9,10.
Proof. n= 6. Under the action of group G6, the set X3
6has the follow-
ing partition by orbits: θ({0,1,2}), θ({0,2,4}) (barycentric) and θ({0,1,3})
(non barycentric). The only 3-barycentric-free orbit in X4
6obtained from
θ({0,1,3}), is θ({0,1,3,4}), thus BO(3, Z6)>4. The only orbit in X5
6ob-
tained from θ({0,1,3,4}) contains a 3-barycentric set. Then BO(3, Z6)≤5.
n= 8. Under the action of group G8, the set X3
8has the following parti-
tion by orbits: θ({0,1,2}), θ({0,2,4}) (barycentric) and θ({0,1,3}), θ({0,1,4})
(non barycentric). The only 3-barycentric-free orbits in X4
8obtained from
the non barycentric orbits in X3
8, are θ({0,1,3,4}) and θ({0,1,4,5}), thus
BO(3, Z8)>4. None of these orbits can be extended by a fifth element in
order to form 3-barycentric-free orbits in X5
8. Then BO(3, Z8)≤5.
n= 9. Under the action of group G9, the set X3
9has the following
partition by orbits: θ({0,1,2}), θ({0,3,6}) (barycentric) and θ({0,1,6}) (non
barycentric). The only 3-barycentric-free orbits in X4
9obtained from θ({0,1,6}),
are θ({0,1,3,4}) and θ({0,1,3,7}). Thus BO(3, Z9)>4. None of these orbits
can be extended by a fifth element in order to form 3-barycentric-free orbits
in X5
9. Then BO(3, Z9)≤5.

7k-barycentric Olson constant 39
n= 10. Under the action of group G10, the set X3
10 has the follow-
ing partition by orbits: θ({0,1,2}), θ({0,2,4}) (barycentric) and θ({0,1,3}),
θ({0,1,5}) (non barycentric). The only 3-barycentric-free orbits in X4
10 ob-
tained from the non barycentric orbits in X3
10 are θ({0,1,3,4}), θ({0,1,3,8}),
θ({0,1,4,5}) and θ({0,1,5,6}). Thus BO(3, Z10 )>4. None of these orbits
can be extended by a fifth element in order to form 3-barycentric-free orbits
in X5
10. Then BO(3, Z10 )≤5.
Lemma 3.11.BO(3, Z4) = 3.
Proof. Lemma 2.6. 
Lemma 3.12.BO(4, Z7) = 5.
Proof. Under the action of G7, the only two orbits in X4
7are θ({0,1,2,3})
(non barycentric, thus 5 ≤BO(4, Z7)) and θ({0,1,2,4}) (barycentric). By
a simple inspection we can see that the only orbit in X5
7, obtained from
θ({0,1,2,3}), contains a 4-barycentric set. Then BO(4, Z7)≤5. By Lem-
mas 2.3 and 2.6, X5
7contains only one orbit, say θ({0,1,2,3,4}).Moreover,
this orbit contains a 4-barycentric set. Therefore, this is another way to obtain
the upper bound BO(4, Z7)≤5. 
The two theorems below and their corollaries give other bounds for
BO(k, Zp).
Theorem 3.13 ([6]).Let s, d be integers ≥2such that s > lp−1
dm.
Let Abe a set with |A|=s+d. Let f:A→Zpbe a sequence such that
|f(A)| ≥ lp−1
dm+d. Then there exists an s-barycentric subsequence of f.
Corollary 3.14.BOlp−1
dm+ 1, Zp≤lp−1
dm+ 1 + dfor d≥2and
p≥lp−1
dm+1+d.
Theorem 3.15 ([6]).Let s≥2,p≥7and Aa set with s+2 elements. If
f:A→Zpis a sequence with |f(A)|=p+3
2, then fcontains an s-barycentric
subsequence.
Corollary 3.16.BOlp−1
2m, Zp≤p+3
2for p≥7.
In particular, we have BO(5, Z11)≤7.
We have the following easy
Lemma 3.17.θ({0,1,2})is a barycentric orbit of X3
n,n≥3.
θ({0,1,2,5})is a barycentric orbit of X4
n,n≥6.
θ({0,1,2,3,4})is a barycentric orbit of X5
n,n≥5.

40 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 8
θ({0,1,2,3,5,7})is a barycentric orbit of X6
n,n≥8.
θ({0,1,2,3,4,5,6})is a barycentric orbit of X7
n,n≥7.
θ({0,1,2,3,4,5,6,7})is a barycentric orbit of X8
n,n= 10,12.
θ({0,1,2,3,4,7,8,9})is a barycentric orbit of X8
11.
θ({0,1,2,3,4,5,6,7,8})is a barycentric orbit of X9
n,n≥9.
θ({0,1,2,3,4,5,6,8,10,11})is a barycentric orbit of X10
12 .
θ({0,1,2,3,4,5,6,7,8,9,10})is a barycentric orbit of X11
n,n≥11.
Remark 3.18.The existence of BO(k, Zn) for 3 ≤n≤12 and 3 ≤k≤n
is ensured by Remark 3.2 and Lemma 3.17.
The exact values of the k-barycentric Olson constant are presented in
Table 1 and Table 2. The values that are not justified by lemmas are obtained
similarly using orbits technique as in Lemmas 3.9 through 3.12. In the lower
bound column, a longest set without a k-barycentric set is given.
TABLE 1
Exact values of BO(k, G)
Lower bound Justification k G BO(k , G)
{0,1}3Z33
{0,1}Lemma 3.11 3 Z43
{0,1}3Z53
{0,1,3,4}Lemma 3.10 3 Z65
{0,1,3}3Z74
{0,1,3,4}Lemma 3.10 3 Z85
{0,1,3,4}3Z95
{0,1,3,4}Lemma 3.10 3 Z10 5
{0,1,4,5}3Z11 5
{0,1,3,4}3Z12 5
Remark 3.2 4 Z4non defined
Remark 3.2 4 Z5non defined
{0,1,3,4}4Z65
{0,1,2,3}Lemma 3.12 4 Z75
{0,1,2,5,7}4Z86
{0,1,3,4,6,7}4Z97
{0,1,2,3,4}4Z10 6
{0,1,2,3,4}4Z11 6
{0,1,2,3,4}4Z12 6
Remark 3.2 5 Z55
{0,1,2,3}5Z65
{0,1,2,4}5Z75
{0,1,2,3,5}5Z86
{0,1,2,3,5}5Z96
{0,1,2,3,5,8}5Z10 7
{0,1,2,6,8,9}5Z11 7
{0,1,2,3,5,7}5Z12 7

9k-barycentric Olson constant 41
TABLE 2
Exact values of BO(k, G)
Lower bound Justification k G BO(k, G)
Remark 3.2 6 Z6non defined
Remark 3.2 6 Z7non defined
{0,1,2,3,4,5}6Z87
{0,1,2,3,4,6}6Z98
{0,1,2,3,4,5}6Z10 7
{0,1,2,3,5,10}6Z11 7
{0,1,2,3,5,8}6Z12 7
{0,1,2,3,4,5}Remark 3.2 7 Z77
{0,1,2,3,4,5}7Z87
{0,1,2,3,4,5}7Z97
{0,1,2,3,4,5,7}7Z10 8
{0,1,2,3,5,7,10}7Z11 8
{0,1,3,4,5,7,10}7Z12 8
Remark 3.2 8 Z8non defined
Remark 3.2 8 Z9non defined
{0,1,2,3,4,5,6,8}8Z10 9
{0,1,2,3,5,7,9,10}8Z11 9
{0,1,2,3,4,5,7,8}8Z12 9
{0,1,2,3,4,6,7,8}Remark 3.2 9 Z99
{0,1,2,3,5,6,7,8}9Z10 9
{0,1,2,3,4,5,6,7}9Z11 9
{0,1,2,3,4,5,6,7}9Z12 9
Remark 3.2 10 Z10 non defined
Remark 3.2 10 Z11 non defined
{0,1,2,3,4,5,6,7,8,9}10 Z12 11
{0,1,2,3,4,5,6,7,8,10}Remark 3.2 11 Z11 11
{0,1,2,3,4,5,6,7,8,9}11 Z12 11
Remark 3.2 12 Z12 non defined
4. k-BARYCENTRIC DAVENPORT CONSTANT
AND BARYCENTRIC RAMSEY NUMBERS FOR STARS
Let Gbe an Abelian group of order nand k≤n. If BO(k, G) exists,
then we have BO(k, G)≤BD(k, G), otherwise n+ 1 ≤BD(k, G).From the
Hamidoune-condition we have BD(k, G)≤n+k−1, i.e., so that its existence
is assured. When k=|G|=n, the k-barycentric sequences are zero-sum.
Gao [9] showed that ZS(G) = n+D(G)−1, where ZS(G) is the smallest
positive integer tsuch that every sequence of length tcontains a zero-sum
n-subsequence. Then B D(n, G) = Z S(G).Moreover, since D(Zn) = n(see
[17]), we have BD(n, G) = 2n−1.

42 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 10
New values of BD(k, G) different from those presented in [7] are given
here.
In this section, the values of BR(K1,k, Zn) for 3 ≤n≤12 and 3 ≤k≤n
are established. The upper bound is derived from the fact that BR(K1,k, Zn)≤
BD(k, Zn) + 1. The process of obtaining the lower bound is as follows. Let µ
be the sequence that yields the lower bound of BD(k, Zn).Select an adequate
decomposition of the complete graph KBD(k,Zn), using Theorem 2.9, Corol-
lary 2.10 and Corollary 2.11. The edges of each subgraph, part of this decom-
position, are colored; µis the sequence of edge colors of the K1,B D(k,Zn)−1stars.
In some cases, such as BR(K1,4, Z6) = 7, its upper bound is derived from
BO(4, Z6) = 5: if there exists a K1,6in K7colored with five or six colors, we
are done. Else, if all K1,6are colored with three or four colors, then each orbit
of X3
6,X4
6, respectively, is used to color the edges of K7. For example, let
θ({0,1,2}) be an orbit of X3
6. It is easy to see that coloring any K1,6in K7
with 0,1,2 we can identify a K1,4barycentric. If all K1,6are colored with only
two colors; then they must be colored by {0,1},{0,2}or {0,3}. Therefore,
assuming that K7is K1,4barycentric free, the graph induced by the edges
colored by 0 has odd order and each vertex has odd degree, and we reach
a contradiction. For the lower bound, Theorem 2.9 is used. The two-edge
disjoint Hamiltonian cycles of K6are colored by 1 and 3, respectively, and the
perfect matching by 0.
In what follows we give a few representative examples to illustrate the
methods used to compute BD(k, G).
Lemma 4.1.BD(3, Z4) = 5.
Proof. Since BO(3, Z4) = 3, the sequence that yields the lower bound
must have at most two different elements. Since the orbits of X2
4are θ({0,1})
and θ({0,2}), the length of 3-barycentric free sequence 0011 is a lower bound
for BD(3, Z4).Moreover, in every sequence of length 5, there exist at least
three different elements or three identical elements. Then it contains a 3-
barycentric sequence. The proof is complete. 
Lemma 4.2.BD(3, Z6) = 5.
Proof. Since BO(3, Z6) = 5, the lower bound is obtained for sequences
with four different elements. For example, the sequence 0134 is 3-barycentric
free. Moreover, if in a sequence of length 5 there exists an element repeated
three times, then we are done. Otherwise, it contains three different elements.
Since the only orbit in X3
6not barycentric is θ({0,1,3}), it is sufficient to con-
sider the sequence 00113, which contains the 3-barycentric sequence 003. 

11 k-barycentric Olson constant 43
In Table 3 and Table 4 we give the exact values of BD(k, Zn) and
BR(K1,k , Zn) for 3 ≤n≤12 and 3 ≤k≤n. In the lower bound column
is given a longest sequence without a k-barycentric subsequence.
TABLE 3
Exact values of BD(k, G) and BR(K1,k, G)
Lower bound k G BD(k, G)B R(K1,k , G)
0101 3 Z35 6
0101 3 Z45 6
0101 3 Z55 6
0113 3 Z65 6
013013 3 Z77 8
013013 3 Z87 8
01340134 3 Z99 10
01340134 3 Z10 9 10
01340134 3 Z11 9 10
01340134 3 Z12 9 10
000111 4 Z47 7
000111 4 Z57 8
000111 4 Z67 7
0001222 4 Z78 9
01450011 4 Z88 9
0120022 4 Z98 9
01560011 4 Z10 9 10
01341133 4 Z11 9 10
01270022 4 Z12 9 10
00001111 5 Z59 10
00001111 5 Z69 10
00001111 5 Z79 10
00001111 5 Z89 10
00001111 5 Z99 10
00001111 5 Z10 9 10
00001111 5 Z11 9 10
000011115 5 Z12 10 11
5. CONCLUSION
The k-barycentric Olson constants BO(k, G) play an important role in
the computation of other constants such as the k-barycentric Davenport con-
stant BD(k, G) and the barycentric Ramsey numbers stars BR(K1,k , G). We
only discussed the existence of BO(k, G) for some group G. It is an open prob-
lem to prove the existence of and to compute BO(k, G) for a general Abelian
group Gby using results of additive group theory.

44 Oscar Ordaz, Mar´ıa Teresa Varela and Felicia Villarroel 12
TABLE 4
Exact values of BD(k, G) and BR(K1,k, G)
Lower bound k G BD(k , G)BR(K1,k, G)
0000011111 6 Z611 11
0000011111 6 Z711 12
0000011111 6 Z811 11
0000011111 6 Z911 12
0000011111 6 Z10 11 11
00000122222 6 Z11 12 13
000001111167 6 Z12 13 14
000000111111 7 Z713 14
000000111111 7 Z813 14
000000111111 7 Z913 14
000000111111 7 Z10 13 14
000000111111 7 Z11 13 14
000000111111 7 Z12 13 14
00000001111111 8 Z815 15
00000001111111 8 Z915 16
00000001111111 8 Z10 15 15
00000001111111 8 Z11 15 16
00000001111111 8 Z12 15 15
0000000011111111 9 Z917 18
0000000011111111 9 Z10 17 18
0000000011111111 9 Z11 17 18
0000000011111111 9 Z12 17 18
000000000111111111 10 Z10 19 19
000000000111111111 10 Z11 19 20
000000000111111111 10 Z12 19 19
00000000001111111111 11 Z11 21 22
00000000001111111111 11 Z12 21 22
0000000000011111111111 12 Z12 23 23
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Received 2 November 2007 Universidad Central de Venezuela
Facultad de Ciencias
Departamento de Matem´aticas y Centro ISYS
Ap. 47567, Caracas 1041-A, Venezuela
flosav@cantv.net
Universidad Sim´on Bolivar
Departamento de Matem´aticas Puras y Aplicadas
Ap. 89000, Caracas 1080-A, Venezuela
and
Universidad de Oriente
Departamento de Matem´aticas
Escuela de Ciencias, N´ucleo Sucre
Cuman´a, Venezuela