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\title{\bf Modification of Griffiths' result
for even integers}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address 

\author{Eshita Mazumdar\\
\small Department of Mathematics\\[-0.8ex]
\small IIT Bombay\\[-0.8ex] 
\small Mumbai, India.\\
\small\tt eshita@math.iitb.ac.in\\
\and
Sneh Bala Sinha \qquad  \\
\small Department of Mathematics\\[-0.8ex]
\small ISI Delhi\\[-0.8ex]
\small Delhi, India\\
\small\tt sneh@isid.ac.in
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Jul 6, 2016}{Oct 13, 2016}{Oct 28, 2016}\\
\small Mathematics Subject Classifications: 05E15, 11B50}

\begin{document}

\maketitle


\begin{abstract} 
  For a finite abelian group $G$ with $\exp(G)=n$, the arithmetical invariant
$\mathsf s_A(G)$  is defined to be the  least integer $k$ such that
any sequence $S$ with length $k$ of  elements in $G$
has a $A$ weighted zero-sum subsequence of  length $n$. 
When  $A=\{1\}$, it is  {\it the Erd\H{o}s-Ginzburg-Ziv constant}
and is denoted by $\mathsf s (G)$. For certain class of
sets $A$, we already have some general
bounds for these weighted constants corresponding to the cyclic
group $\Z_n$, which was given by Griffiths. 
For odd integer $n$, Adhikari and Mazumdar generalized
the above mentioned results in the sense that they
hold for more sets $A$. In 
the present paper we modify Griffiths' method for even $n$
and obtain general bound
for the weighted constants for certain class of weighted sets
which include sets that 
were not covered by Griffiths for $n\equiv 0 \pmod{4}$.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} the zero-sum problem; Kneser's theorem.
  
\end{abstract}

\section{Introduction}

Let $G$ be a finite abelian group (written additively).
By a sequence over $G$ we mean a finite sequence of terms from $G$
which is unordered and repetition of terms is allowed and
we view sequences over $G$ as elements of the free abelian monoid
$\mathcal{F}(G)$ and use multiplicative notation. So, our notation is
consistent with  \cite{GG}, \cite{GH} and  \cite{Gry}.

For $S \in \mathcal{F}(G)$, if 
$$S=x_1x_2 \cdots x_t= \prod_{g\in G}g^ {{\sf v}_g(S)},$$ 
then ${\sf v}_g(S)\geq 0$ is the {\it multiplicity}  of $g$ in $S$, and 
$$|S|=t=\sum_{g\in G} {\sf v}_g(S)$$ is the {\it length} of $S$. 
The sequence $S$ {\it contains} some 
$g\in G$ if ${\sf v}_g(S)\geq 1$.
If $S$ and $T$ are sequences over $G$,  then $T$ is said to be a {\it subsequence} of $S$ if
${\sf v}_g(T)\leq {\sf v}_g(S)$ for every
$g\in G$.

For  a non-empty subset $A$ of $\{1,2, \ldots, n-1\}$, where $n$
is  the exponent of $G$ (denoted by $\exp(G)$),
a sequence $S=x_1x_2 \cdots x_t$ of length $t$ over $G$ is said to be an {\it $A$-weighted zero-sum sequence,}
if there exists $\bar{a} =(a_1,a_2,\ldots,a_t)\in A^t$ 
such that $\sum_{i=1}^ta_ix_i=0$.

For integers $m<n$, we shall use the notation $[m,n]$ to denote the set
$\{m, m+1, \ldots, n\}$.  For a finite set $A$, we denote its size by $|A|$,
which  is the number of elements of $A$.
If $G$ is a finite abelian group with $\exp(G)=n$, then
for a non-empty subset $A$ of $[1, n-1]$,
one defines $\mathsf s_A(G)$  to be the  least integer $k$ such that
any sequence $S$ with length $k$ of  elements in $G$
has an $A$-{\it weighted zero-sum subsequence} of  length 
$\exp(G)=n$.
Taking $A=\{1\}$, one recovers the classical 
Erd\H{o}s-Ginzburg-Ziv constant $\mathsf s (G)$.
The above weighted versions and some other invariants with
weights were introduced by
Adhikari, Chen, Friedlander, Konyagin and Pappalardi
\cite{ACFKP}, Adhikari and Chen \cite{AC}  and
Adhikari, Balasubramanian, Pappalardi and  Rath \cite{ABPR}.
For developments regarding bounds on the constant $\mathsf s_A(G)$
in the case of abelian groups $G$ with higher rank and related references,
we refer to the recent  paper of  Adhikari, Grynkiewicz and  Sun 
\cite{AGS}.

When $A = \Z_n^* = \{ a \in [1,n-1]| (a,n)=1\} $,
the set of units of $\Z_n=\Z/n\Z$, 
Luca \cite{Luc} and Griffiths \cite{Gri}  proved 
independently the
following result which had been conjectured in \cite{ACFKP}:
\begin{equation} \label{1}
\mathsf s_A(\Z_n) \leq n + \Omega(n),
\end{equation}
where $\Omega(n)$ denotes the number of prime
factors of $n$, counted with multiplicity.
An example in \cite{ACFKP} had already established the inequality 
in the other direction:
$$\mathsf s_A(\Z_n) \geq n + \Omega(n).$$

Now we state the following results of
Griffiths \cite{Gri} which
generalizes result (\ref{1}) for integer $n$:

\begin{theorem}\label{thm1}
 Let $n=p_1^{a_1}\cdots p_k^{a_k}$ be an odd integer
and let $a= \sum_{s}a_s$. For each s,
let $A_s\subset \Z_{p_s^{a_s}}$ be a subset with 
its size $|A_s|> p_s^{a_s}/2$,
 and let $A=A_1\times\cdots\times A_k$. Then
for $m>a$, every sequence $x_1 \cdots x_{m+a}$ over $\Z_{n}$ has 0 as
an A-weighted m-sum.
\end{theorem}

\begin{theorem}\label{thm2}
 Let $n=2^{a_1}\cdots p_k^{a_k}$ be an even integer
and let $a= \sum_{s}a_s$. Let $A_1\subset \Z_{2^{a_1}}$ be 
such that $|A_1|> 2^{a_1-1}$ or $|A_1|> 2^{a_1-2}$ and 
$A_1\subset \Z_{2^{a_1}}^*$. For each $s\geq 2$,
let $A_s\subset \Z_{p_s^{a_s}}$ be a subset with 
$|A_s|>(1/2)p_s^{a_s}$, and let $A=A_1\times\cdots\times A_k$. Then 
for any even $m$, every sequence $x_1 \cdots x_{m+a}$ 
over $\Z_n$ has 0 as an A-weighted $m$-sum. 
\end{theorem}

For odd integer $n$ with suitable modifications in the method of 
Griffiths \cite{Gri}, Adhikari and Mazumdar established the following 
result in \cite{AM}:

\begin{theorem}\label{thm3}
Let $n=p_1^{a_1}\cdots p_k^{a_k}$ be an odd integer
and let $a= \sum_{s}a_s$. For each s,
let $A_s\subset \Z_{p_s^{a_s}}$ be a subset with $|A_s|> (4/9)p_s^{a_s}$,
 and let $A=A_1\times\cdots\times A_k$. Then
for $m>2a$, every sequence $x_1 \cdots x_{m+2a}$ over $\Z_n$ has 0 as
an A-weighted m-sum.
\end{theorem}

For general $n$, consider the set $A$ of squares in the group of units 
in the cyclic group $\Z_n$, it was proved by Adhikari, Chantal David and
Urroz \cite{ADU}
that if $n$ is a square-free
integer, coprime to $6$, then
\begin{equation} \label{3}
\mathsf s_A(\Z_n) = n +2\Omega(n).
\end{equation}

Later, removing the requirement that $n$ is square-free,
Chintamani and Moriya \cite{CM} showed that if $n$ is a power of $3$ or
$n$ is coprime to $30=2 \times 3 \times 5$, then the result (\ref{3}) holds,
where $A$ is again the set of squares in the group of 
units in $\Z_n$.

But still we lack any information on bounds in case when $n$ is an even integer.
In this paper we mainly focus on the case when $n$ is an even integer and
get the following theorem:

\begin{theorem}\label{thm4}
Let $n=2^{a_1}\cdots p_k^{a_k}$ be an even integer
and let $a= \sum_{s}a_s$. Let $A_1\subset \Z_{2^{a_1}}$
such that $|A_1|\geq 2^{a_1-1}$. For each $s\geq 2$,
let $A_s\subset \Z_{p_s^{a_s}}$ be a subset with 
$|A_s|>(5/12)p_s^{a_s}$, and let $A=A_1\times\cdots\times A_k$. Then for any integer $m$ multiple of $4$,
every sequence $x_1 \cdots x_{m+3a}$ over $\Z_n$ has 0 as
an A-weighted m-sum.
\end{theorem}

\smallskip
For $a_1 \geq 2$ from Theorem \ref{thm4} it follows that
any sequence of length $n+3a$ of elements of $\Z_n$ has 0 as
an A-weighted $n$-sum. In other words, if $A$ is as in
Theorem \ref{thm4}, $$\mathsf s_A(\Z_n) \leq n + 3 \Omega(n),$$ for
$n\equiv0 \pmod{4}$.

It has been observed in \cite{ACFKP} that if we consider 
$n\equiv0 \pmod{4}$ such that $n$ is the product of
$\Omega(n)$ (not neccessarily distinct) primes
$n= q_1\cdots q_{\Omega(n)}$ and $A=\Z_n^*$, then 
the sequence $1, q_1,q_1q_2,\ldots,q_1q_2\cdots q_{\Omega(n)-1}$
has no non-empty subset with weighted sum $0$.
Thus adjoining $n-1$ zeros  we have a sequence of length
$n+\Omega(n)-1$ which does not have $0$ as a $A$-weighted
$n$-sum. So in this case we can conclude that
$$n+ \Omega(n)\leq \mathsf s_A(\Z_n) \leq n + 3 \Omega(n),$$
for $n\equiv0 \pmod{4}$.

Clearly, in the Theorem \ref{thm4} we include more sets 
which are not covered by the results of Griffiths.

In the next section we shall give the proof of Theorem \ref{thm4} and
discuss some results. But before that let us first state the Kneser's theorem 
(\cite{Kne}, also see \cite[Chapter 4]{Nat}), which is 
one of the most useful tools for Section 2.

Recall that the stabilizer of a subset $S$
of an abelian group $G$ is defined as 
$$ stab(S) = \{x \in G : x + S = S\}.$$

\begin{theorem}[Kneser's Theorem]\label{KT}
Let $G$ be an abelian group and $A,B$ be finite,
non-empty subsets of $G$. Then $$ |A+B|
\geq |A+H|+|B+H| -|H|,$$
where $H$ is the
stabilizer of $A+B.$
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem \ref{thm4}}

To prove Theorem \ref{thm4} we need the following 
lemmas.

\begin{lemma}\label{lem1}
Let $A\subset \Z_{2^a}$ be a subset such that 
$|A|\geq2^{a-1}$. If $x,y,z,w \in \Z_{2^a}^*$,
the group of units in $\Z_{2^a}$, then given
 any even $t\in \Z_{2^a}$, there exist $\alpha,\beta,\gamma,
 \delta \in A$ such that 
$$\alpha x+ \beta y+ \gamma z + \delta w =t.$$
\end{lemma}

\begin{proof}
There are two cases.
\begin{itemize}
\item When $|A| = 2^{a-1}$ and $A$ contains all odd elements or 
all even elements 
of $\mathbb{Z}_{2^a}$.
Given $x, y, z, w \in \Z_{2^a}^*$ and $A \subset \Z_{2^a}$ with 
$|A| = 2^{a-1}$, $Ax+ Ay+ Az+ Aw$ will contain only even 
elements from $\Z_{2^a}$ which will imply 
$$ |Ax+ Ay+ Az+ Aw| \leq 2^{a-1}.$$
Now we calculate the cardinality $|Ax+ Ay+ Az+ Aw|.$
As we observed that
\begin{align*}
|Ax+ Ay+ Az+ Aw| & \geq  |Ax|= |A|= 2^{a-1},
\end{align*}
it follows that $Ax+Ay+Az+Aw= Z_{2^{a-1}}.$
Thus we have the theorem in this case.

\item
When $A$ contains even as well as odd elements of $\Z_{2^a}$.
Using Theorem \ref{KT} we get
\begin{align*}
|Ax+ Ay| & \geq  |Ax+H_1|+ |Ay+H_1|-|H_1|,
\end{align*}
where $H_1$ is the stabilizer of $Ax+ Ay$.
For $|H_1|= 2^a$, $Ax+Ay= \Z_{2^a}.$
For $|H_1|=2^{a-1}$, $H_1$ contains all even elements of 
$\Z_{2^a}$ and the order of $\Z_{2^a}/H_1$ will be $2$.
For an even $u \in A$ we have $u+ H_1 =H_1$ and for an odd $v \in A$
we have $v + H_1 \neq H_1.$
Therefore we get $$ \Z_{2^a} = H_1 \bigcup (v+H_1) .$$
Thus by Theorem \ref{KT} we have 
$$ |Ax+ Ay| \geq  |Ax+H_1|+ |Ay+H_1|-|H_1| 
\geq |Ax+H| = 2^a.$$
and hence $Ax+ Ay= \Z_{2^a}$.
If $|H_1|\leq 2^{a-2}$, then $$|Ax+ Ay| \geq  |Ax+H_1|+ |Ay+H_1|-|H_1| 
\geq 3.2^{a-2}.$$
Similarly, using Theorem \ref{KT} we have 
$$|t-Az- Aw| \geq  |t-Az+H_2|+ |(-Aw)+H_2|-|H_2|,$$
where $H_2$ is the stabilizer of $t-Az- Aw.$
Now, for $|H_2|=2^a$ or $2^{a-1},$ $t-Az- Aw= \Z_{2^a}.$
For $|H_2|\leq2^{a-2},|t-Az- Aw| \geq 3.2^{a-2}$.
In all the cases we will find that $$|Ax+Ay|+ 
|t-Az-Aw|> 2^a.$$ So, by the pigeonhole principle,
we conclude that any even number $t$ can be written 
as $\alpha x+ \beta y+ \gamma z + \delta w$ for 
some $\alpha,\beta,\gamma,\delta \in A$.\qedhere
\end{itemize}
\end{proof}

\begin{lemma}\label{lem2}
Let $p^a$ be an odd prime power and 
$A\subset \Z_{p^a}$ be a subset such that 
$|A|>\frac{5}{12}p^a$. If $x,y,z,w \in \Z_{p^a}^*$,
the group of units in $\Z_{p^a}$, then given
 any $t\in \Z_{p^a}$, there exist $\alpha,\beta,\gamma,
 \delta \in A$ such that 
$$\alpha x+ \beta y+ \gamma z + \delta w =t.$$
\end{lemma}

\begin{proof}
Consider the following sets 
$$A_1=\{\alpha x: \alpha\in A\}, ~B_1=\{\beta y: \beta\in A\},$$
$$~C_1=\{-\gamma z: \gamma\in A\}, ~D_1=\{t-\delta w: \delta\in A\}.$$ 
Observe that $|A_1|=|B_1|=|C_1|=|D_1|=|A|$. Therefore
by using Theorem \ref{KT}, 
we have
\begin{equation}\label{4}
|A_1 +B_1|\geq |A_1| + |B_1| -|H_1|,
\end{equation}
where $H_1$ is the stabilizer of $A_1 +B_1$.
Now if  $H_1 =\Z_{p^a}$ then it would imply that $A_1 +B_1=\Z_{p^a}.$
Thus we get $A_1 +B_1 +\gamma z + \delta w=\Z_{p^a}$
and we are through.
Otherwise, we have
$$|H_1| \leq p^{a-1} =\frac{p^a}{p} \leq \frac{p^a}{3}.$$
Hence using (\ref{4}) we have
$$|A_1 +B_1|\geq |A_1|+ |B_1|- |H_1|  > 
\frac{5p^a}{6} - \frac{p^a}{3} = \frac{p^a}{2}.$$
Similarly, if the stabilizer of $C_1+D_1$ has cardinality $p^a$ then
we are through.
Otherwise $$|C_1+D_1| >\frac{p^a}{2},$$ and
therefore we get
$$ |A_1 +B_1| + |C_1+D_1| > \frac{p^a}{2} + \frac{p^a}{2} = p^a,$$
which implies that the sets $A_1 +B_1$ and $C_1+D_1$ intersect and 
we are done.
\end{proof}

\begin{lemma}\label{lem3}
Let $A\subset \Z_{2^a}$ be such that $|A|\geq 2^{(a-1)}$. 
Let $x_1\cdots x_m$ be a sequence over $\Z_{2^a}$
such that for each $b \in [1,a]$, the cardinality of the set 
$\{i\mid x_i\neq0\pmod{2^b} \}$ is divisible by $4$. 
Then $x_1\cdots x_m$ is an $A$-weighted zero-sum 
sequence.
\end{lemma}

\begin{proof}
Let $c$ be minimal such that $\{i\mid  x_i\neq0\pmod{2^c}\}$ is 
non-empty. If no such $c$ exists then 
all terms are $0$ and we are done instantly.
Therefore, $\{i\mid x_i\neq0\pmod{2^c}\}$ has at least four elements; 
without loss of generality let 
$x_1,x_2,x_3, x_4\neq 0 \pmod{2^c}$.
Also there are only even number of elements in $x_5,\ldots,x_m$
which are $\neq0\pmod{2^c}$ and   $\equiv0\pmod {2^{c-1}}$. 
Others are congruent to $0$ modulo $2^{c}$.
Set $$x_i'= x_i/2^{c-1}\in \Z_{2^{a-(c-1)}},$$ 
for $i \in [1,m]$. 
Therefore $x_i'$ for $i= 1,2,3,4$ are odd elements and 
$-x_5'-\cdots-x_m'$ is even element in $\Z_{2^{a-(c-1)}}.$
Also if elements of $A$ meet less than $2^{a-(c-1)-1}$ congruence 
classes modulo $2^{a-(c-1)}$, then $|A|< 2^{a-(c-1)-1} \times 2^{(c-1)} 
=2^{a-1},$ which is a contradiction to our assumption.
Therefore, the elements of $A$ must meet at least $2^{a-(c-1)-1}$ 
congruence classes modulo $2^{a-(c-1)}$.
Let $\alpha \in A$ be any element.
Therefore, $-\alpha x_5'-\cdots- \alpha x_m'$ is an even element 
in $\Z_{2^{a-(c-1)}}.$
Now using Lemma \ref{lem1}, there exist  $\alpha_1,\alpha_2,
\alpha_3, \alpha_4\in A$ such that
$$\alpha_1 x_1'+\alpha_2 x_2' +\alpha_3 x_3' + \alpha_4 x_4'
=  -\alpha x_5'-\cdots- \alpha x_m'$$  in $\Z_{2^{a-(c-1)}},$
and we are through in $\Z_{2^a}$ taking $\alpha = 
\alpha_5=\cdots=\alpha_m$.
\end{proof}

\noindent{\bf Remark.} For an odd prime power $n=p^a$, the same condition 
on a sequence $x_1\cdots x_m$ in $\Z_{p^{a}}$ ( namely 
that $A\subset \Z_{p^a}$ be such that $|A|> \frac{5}{12}p^{a}$ and 
the cardinality of the set 
$\{i\mid x_i\neq0\pmod{p^b} \}$ is divisible by $4$ for all
$b=1,\ldots,a$) is sufficient to imply that $x_1\cdots x_m$
 is an $A$-weighted zero-sum sequence.
For a proof we should proceed as above. 
Only at the last part
we need to apply Lemma \ref{lem2} instead of Lemma \ref{lem1}.

\begin{lemma}\label{lem4}
Given disjoint subsets $X_1,\ldots,X_{a}$ of the set 
$V=[1,m+3a]$, where $m$ is a multiple of $4$, there exists a 
set $I\subset [1,m+3a]$ with $|I|=m$ 
such that $|I \cap X_s|$ is a multiple of $4$, for all $s=1,\ldots,a$.
\end{lemma}

\begin{proof}
Let $X_{a+1}=V\setminus \cup_{b=1}^{a} X_b$.
For $b=1,\ldots,a+1$, let $W_b\subset X_b$, be
a maximal sized subset whose cardinality is a multiple of 
$4$. Then, $|W_b|\in \{|X_b|, |X_b|-1, |X_b|-2, |X_b|-3 \}$.
So, $|\cup_{b=1}^{a+1} W_b| = \sum_{b=1}^{a+1}|W_b|\geq 
\sum_{b=1}^{a+1}(|X_b|-3) = (m+3a)-3(a+1) =m-3 $. For each $b\in [1,a+1]$, 
$|W_b|$ is divisible by $4$ and also
it is given that $m$ is divisible by $4$. Therefore, $|\cup_{b=1}^{a+1} W_b| \geq m$. Thus we can obtain an $
I$ with cardinality $m$ having the required property.
\end{proof}

\noindent{\bf Remark.} The same is true for a nested family 
$X_1\subset X_2\subset \cdots \subset X_a$. To see this, set $Y_1=X_1$,
and $Y_b= X_b\diagdown X_{b-1}$, for $b=2,\ldots,a$. These sets are
disjoint. Now by applying Lemma \ref{lem4} to $Y_1,\ldots,Y_a$,
we obtain $I\subset \{1,2,\ldots,m+3a\}$ with $|I|=m$ and 
$|I\cap Y_b|$ multiple of $4$, for $b=1,\ldots,a$. 
Hence we are done because this implies $|I\cap X_b|$ is 
multiple of $4$ for all $b=1,\ldots,a$. 

\medskip
\noindent{\bf Observation :} Let $n=p_1^{a_1}\cdots p_k^{a_k}$ be a positive
integer. Then, $\Z_n$ is isomorphic to
$\Z_{p_1^{a_1}} \times \cdots \times \Z_{p_k^{a_k}}$
and an element $x \in \Z_n$ can be written as
$x=(x^{(1)}, \ldots, x^{(k)})$, where $x^{(s)} \equiv x \pmod{p_s^{a_s}}$
for each $s$.
It has been observed in \cite{Gri} that if
$A=A_1\times A_2  \times \cdots \times A_k$ is a subset of $\Z_n$, 
where $A_s\subset \Z_{p_s^{a_s}}$ for each $s\in [1,k]$,
then a sequence $x_1 \cdots x_m$ over $\Z_n$ is
an $A$-weighted zero-sum sequence in $\Z_n$ if and only if
for each $s\in [1,k]$, the sequence
$x_1^{(s)}\cdots x_m^{(s)}$  is 
an $A_s$-weighted zero-sum sequence in $\Z_{p_s^{a_s}}$.

\begin{proof}[Proof of Theorem~\ref{thm4}]
Given a sequence $x_1\cdots x_{m+3a}$ over $\Z_n$, 
we 
define $X_b^{(s)}\subset [1,m+3a]$ 
for $s \in [1,k]$ and $b \in [1,a_s]$  by 
$$X_b^{(s)} =\{i:x_i\neq0\pmod{p_s^b}\}.$$

Using Lemma \ref{lem4} and the Remark just after that,
we can say that there exists $I\subset [1,m+3a]$ such that 
$|I|=m$ and $|I\cap X_b^{(s)}|$ is a multiple of $4$ 
for all $s$, $b$. 
Let $I=\{ i_1,\ldots,i_m\}$.
For $s=1,$ by Lemma \ref{lem3}, $x^{(1)}_{i_1}\cdots x^{(1)}_{i_m}$ is an $A_1$- weighted
zero-sum sequence.
For $s\geq 2,$ by remark just after Lemma \ref{lem3} 
$x^{(s)}_{i_1}\cdots x^{(s)}_{i_m}$ is a $A_s$-
weighted zero-sum sequence.
Now using the above observation  it follows that
$x_{i_1},\ldots,x_{i_m}$ is  an A-weighted zero-sum sequence.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
The authors would like to thank the reviewer for their useful comments.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{thebibliography}{99}

\bibitem{ABPR}
S. D. Adhikari, R. Balasubramanian, F. Pappalardi and P. Rath,\newblock Some zero-sum constants with weights,
\newblock{\em Proc. Indian Acad. Sci. (Math. Sci.)}, {\bf 118} (2008), 183--188.

\bibitem{AC}
S. D. Adhikari and  Y. G. Chen,\newblock Davenport constant with 
weights and some related questions - II,
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