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\begin{center}
\vskip 1cm{\LARGE\bf 
Zero-Sum Constants Related to the \\
\vskip 0.1in
Jacobi Symbol
}
\vskip 1cm
\large
Santanu Mondal, Krishnendu Paul, and Shameek Paul \\
School of Mathematical Sciences \\ 
Ramakrishna Mission Vivekananda Educational and Research Institute \\
West Bengal 711202 \\
India \\
\href{mailto:santanu.mondal.math18@gm.rkmvu.ac.in}{\tt santanu.mondal.math18@gm.rkmvu.ac.in} \\
\href{mailto:krishnendu.p.math18@gm.rkmvu.ac.in}{\tt krishnendu.p.math18@gm.rkmvu.ac.in} \\ 
\href{mailto:shameek.paul@rkmvu.ac.in}{\tt shameek.paul@rkmvu.ac.in} 
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\vskip .2 in

\begin{abstract}
Let $A\subseteq \mathbb Z_n$ be a subset. A sequence $S=(x_1,\ldots,x_k)$ is said to be an $A$-weighted zero-sum sequence if there exist $a_1,\ldots,a_k\in A$ such that $a_1x_1+\cdots+a_kx_k=0$. We refer to $A$ as a weight-set. The $A$-weighted Davenport constant $D_A$ is defined to be the smallest natural number $k$ such that every sequence of $k$ elements in $\mathbb Z_n$ has an $A$-weighted zero-sum subsequence. The constant $C_A$ is defined to be the smallest natural number $k$ such that every sequence of $k$ elements in $\mathbb Z_n$ has an $A$-weighted zero-sum subsequence having consecutive terms. 

When $n$ is odd, let $S(n)$ be the set of all units in $\mathbb Z_n$ whose Jacobi symbol with respect to $n$ is 1. We compute the constants $C_{S(n)}$ and $D_{S(n)}$. For a prime divisor $p$ of $n$, we also compute these constants for a related weight-set $L(n;p)$. This is the set of all units $x$ in $\mathbb Z_n$ such that the Jacobi symbol of $x$ with respect to $n$ is the same as the Legendre symbol of $x$ with respect to $p$. We show that even though these weight-sets $A$ may have half the size of $U(n)$ (which is the set of units of $\mathbb Z_n$), the corresponding $A$-weighted constants are the same as those for the weight-set $U(n)$. 
\end{abstract}

\section{Introduction}\label{0}

For $a,b\in \mathbb Z$, we denote the set $\{x\in\mathbb Z:a\leq x\leq b\}$ by $[a,b]$. Let $U(n)$ denote the group of units in the ring $\mathbb Z_n$, and $U(n)^2=\{\,x^2:x\in U(n)\,\}$. For an odd prime $p$, let $Q_p$ denote the set $U(p)^2$. For $n=p_1p_2\cdots p_k$ where $p_i$ is a prime for each $i\in [1,k]$, we define $\Omega(n)=k$. 

\begin{definition} 
Let $A\subseteq \mathbb Z_n$ be a subset. A sequence $S=(x_1,\ldots,x_k)$ is said to be an $A$-weighted zero-sum sequence if there exist $a_1,\ldots,a_k\in A$ such that $a_1x_1+\cdots+a_kx_k=0$. We refer to $A$ as a weight-set. 
\end{definition}

\begin{definition} 
For a weight-set $A \subseteq \mathbb Z_n$, the {\it $A$-weighted Davenport constant} $D_A$ is defined to be the least positive integer $k$, such that every sequence in $\mathbb Z_n$ of length $k$ has an $A$-weighted zero-sum subsequence. 
\end{definition}

Adhikari and Rath \cite{AR} gave the previous definition. Chintamani and Moriya \cite{CM} showed that $D_{U(n)^2}=2\,\Omega(n)+1$ when every prime divisor of $n$ is at least seven. Grynkiewicz and Hennecart \cite{GH} generalized this by showing that $D_{U(n)^2}\geq 2\Omega(n)+\text{min}\{v_3(n),v_5(n)\}+1$ when $n$ is odd, with equality if either $3\nmid n$ or $v_3(n)\geq v_5(n)$. Mazumdar and Sinha \cite{MS} made suitable modifications in the method of Griffiths \cite{G} to consider the case when $n$ is an even integer. (However, their result cannot be used to determine $D_{U(n)^2}$ when $n$ is even.) Adhikari et al.\ \cite[Lem.\ 2.1]{ACFKP} showed that $D_{\{1,-1\}}=\lfloor \text{log}_{\,2}n\rfloor+1$ for every positive integer $n$.

Mondal, Paul, and Paul \cite{SKS1} gave the following definition.
\begin{definition}
For a weight-set $A \subseteq \mathbb Z_n$, the {\it $A$-weighted constant $C_A$} is defined to be the least positive integer $k$, such that every sequence in $\mathbb Z_n$ of length $k$ has an $A$-weighted zero-sum subsequence of consecutive terms. 
\end{definition} 

Mondal, Paul, and Paul \cite[Cor.\ 3, Cor.\ 6]{SKS1} showed that $C_{U(n)^2}=3^{\Omega(n)}$ when every prime divisor of $n$ is at least seven and  $C_{\{1,-1\}}=n$ when $n$ is a power of two. Mondal, Paul, and Paul \cite{SKS2} showed the next result. 

\begin{theorem}\label{un}
For every positive integer $n$ we have $D_{U(n)}=\Omega(n)+1$ and $C_{U(n)}=2^{\Omega(n)}$. 
\end{theorem}

When $p$ is an odd prime such that $p\equiv 2$ (mod 3), we can show that $U(p)^3=U(p)$. Mondal, Paul, and Paul \cite[Thm.\ 7, Lem.\ 2]{SKS1} showed that when $p\neq 7$ is a prime such that $p \equiv 1$ (mod 3), we have $D_{U(p)^3}=C_{U(p)^3}=3$, and also that $D_{U(7)^3}=3$ and $C_{U(7)^3}=4$. Adhikari and Rath \cite[Thm.\ 2]{AR}, and Mondal, Paul, and Paul \cite[Thm.\ 4]{SKS1} showed the next result. 

\begin{theorem}\label{q}
Let $p$ be an odd prime. Then $C_{Q_p}=D_{Q_p}=3$. 
\end{theorem}

Let $m$ be a divisor of $n$. We refer to the ring homomorphism $f_{n,m}:\mathbb Z_n\to \mathbb Z_m$ given by $a+n\mathbb Z\mapsto a+m\mathbb Z$ as the {\it natural map}. As this map sends units to units, we get a group homomorphism $U(n)\to U(m)$, which we also refer to as the natural map. When $n$ is odd and $x\in \mathbb Z_n$, the Jacobi symbol $\big(\frac{x}{n}\big)$ is defined in Section \ref{s}. 

The following are some of the results in this paper. We assume that $n$ is an odd, squarefree number whose every prime divisor is at least seven. 

\begin{itemize}
\item 
Let $S(n)=\big\{\,x\in U(n):\big(\frac{x}{n}\big)=1\,\big\}$. 
 
If $n$ is prime, then $D_{S(n)}=3$, and $D_{S(n)}=\Omega(n)+1$ otherwise. 
 
If $n$ is prime, then $C_{S(n)}=3$, and $C_{S(n)}=2^{\Omega(n)}$ otherwise.

\item 
Let $L(n;p)=\big\{\,x\in U(n) : \big(\frac{x}{n}\big)=\big(\frac{x}{p}\big)\,\big\}$ where $p$ is a prime divisor of $n$. 
 
If $\Omega(n)=2$, then $D_{L(n;p)}=4$, and $D_{L(n;p)}=\Omega(n)+1$ otherwise.

If $\Omega(n)=2$, then $C_{L(n;p)}=6$, and $C_{L(n;p)}=2^{\Omega(n)}$ otherwise.
\end{itemize}

\begin{remark}
Adhikari and Hegde \cite{AH} showed that if $A=\mathbb Z_n\setminus\{0\}$ and $B=\{1,2,\ldots,\lceil n/2\rceil\}$, we have $D_A=D_B$. We make a similar observation in this paper. In Proposition \ref{indexs}, we show that $S(n)$ is a subgroup of $U(n)$ having index two when $n$ is not a square. Theorem \ref{un} shows that, when $n$ is odd, we have $D_{U(n)}=\Omega(n)+1$ and $C_{U(n)}=2^{\Omega(n)}$. In addition, if $n$ is not a prime, Theorems \ref{dsn} and \ref{csn} show that $D_{S(n)}=D_{U(n)}$ and $C_{S(n)}=C_{U(n)}$. Thus, even though these weight-sets may have different sizes, they can have the same constants. If $\Omega(n)\neq 2$, Theorems \ref{ld} and \ref{lc} show that $D_{L(n;\,p)}=D_{U(n)}$ and $C_{L(n;\,p)}=C_{U(n)}$.
\end{remark}

If $p$ is a prime divisor of $n$, we use the notation $v_p(n)=r$ to mean that $p^r\mid n$ and $p^{r+1}\nmid n$. Let $p$ be a prime divisor of $n$ and $v_p(n)=r$. We denote the image in $U(p^r)$ of $x\in U(n)$ under $f_{n,\,p^r}$ by $x^{(p)}$. For a sequence $S=(x_1,\,\ldots,\,x_l)$ in $\mathbb Z_n$, let $S^{(p)}$ denote the sequence $(x_1^{(p)},\,\ldots,\,x_l^{(p)})$ in $\mathbb Z_{p^r}$, which is the image of $S$ under $f_{n,\,p^r}$. Griffiths \cite[Obs.\ 2.2]{G} made the following observation. 
\begin{observation}\label{obs2}
Let $n=p_1^{r_1}\cdots p_k^{r_k}$ where the $p_i$'s are distinct primes and $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$. Suppose for every $i\in [1,k]$ there exist $c_{i,1},\ldots,c_{i,j},\ldots,c_{i,l}\in U(p_i^{r_i})$ such that $c_{i,1}x_1^{(p_i)}+\cdots+c_{i,j}x_j^{(p_i)}+\cdots+c_{i,l}x_l^{(p_i)}=0$. Then there exist $a_1,\ldots,a_j,\ldots,a_l\in U(n)$ such that for every $(i,j)\in [1,k]\times [1,l]$ we have $a_j^{(p_i)}=c_{i,j}$ and $a_1x_1+\cdots+a_jx_j+\cdots+a_lx_l=0$. 
\end{observation} 

\begin{proof}
Let $j\in [1,l]$. By the Chinese remainder theorem, there exists $a_j\in U(n)$ such that for every $i\in [1,k]$ we have that $a_j^{(p_i)}=c_{i,j}$. Let $x=a_1x_1+\cdots+a_jx_j+\cdots+a_lx_l$. For each $i\in [1,k]$ we see that $f_{n,\,p_i^{r_i}}(x)=x^{(p_i)}=c_{i,1}x_1^{(p_i)}+\cdots+c_{i,j}x_j^{(p_i)}+\cdots+c_{i,l}x_l^{(p_i)}=0$. So by using the Chinese remainder theorem once again, we see that $x=0$. 
\end{proof}

Mondal, Paul, and Paul \cite[Lem.\ 3]{SKS1} showed the next result, which will be used in Theorem \ref{lc2}. In the next two results, for a subset $A$ of $\mathbb Z_n$, we use the notation $C_A(n)$ and $D_A(n)$ for the constants $C_A$ and $D_A$ respectively. 

\begin{lemma}\label{clb}
Let $n=mq$. Let $A,B,C$ be subsets of $\mathbb Z_n,\mathbb Z_m,\mathbb Z_q$ respectively. Suppose $f_{n,m}(A)\subseteq B$ and $f_{n,q}(A)\subseteq C$. Then we have $C_A(n)\geq C_B(m)\,C_C(q)$. 
\end{lemma} 

We now prove a similar result for the weighted Davenport constant, which we will use in Theorem \ref{ld2}. Grynkiewicz, Marchan, and Ordaz \cite[Lem.\ 3.1]{GMO} proved a generalization of this result for abelian groups.

\begin{lemma}\label{dadd}
Let $n=mq$. Let $A,B,C$ be subsets of $\mathbb Z_n,\mathbb Z_m,\mathbb Z_q$ respectively. Suppose $f_{n,m}(A)\subseteq B$ and $f_{n,q}(A)\subseteq C$. Then we have $D_A(n)\geq D_B(m)+D_C(q)-1$. 
\end{lemma} 

\begin{proof}
Let $D_B(m)=k$ and $D_C(q)=l$. If $k=1$, we let $S_1'$ be the empty sequence, and if $l=1$, we let $S_2'$ be the empty sequence. Otherwise, there exists a sequence $S_1'=(u_1,\ldots,u_{k-1})$ of length $k-1$ in $\mathbb Z_m$, which has no $B$-weighted zero-sum subsequence, and there exists a sequence $S_2'=(v_1,\ldots,v_{l-1})$ of length $l-1$ in $\mathbb Z_q$, which has no $C$-weighted zero-sum subsequence. 

As $f_{n,m}$ is onto, for every $i\in [1,k-1]$ there exists $x_i\in \mathbb Z_n$ such that $f_{n,m}(x_i)=u_i$. As $f_{n,q}$ is onto, for every $j\in [1,l-1]$ there exists $y_j\in\mathbb Z_n$ such that $f_{n,q}(y_j)=v_j$. Consider the following sequence of length $k+l-2$ in $\mathbb Z_n$:  
\[S=(qx_1,\,\ldots,\,qx_{k-1},\,y_1,\,\ldots,\,y_{\,l-1}).\]

Let $S_1=(qx_1,\ldots,qx_{k-1})$ and $S_2=(y_1,\ldots,y_{\,l-1})$. Suppose $S$ has an $A$-weighted zero-sum subsequence $T$. If the sequence $T$ contains some term of $S_2$, by taking the image of $T$ under $f_{n,q}$ we get the contradiction that $S_2'$ has a $C$-weighted zero-sum subsequence, as $f_{n,q}(qx_i)=0$ and as $f_{n,q}(A)\subseteq C$. 

Thus, no term of $S_2$ is a term of $T$,   and so $T$ is a subsequence of $S_1$. Let $T'$ be the subsequence of $S_1'$, such that $u_i$ is a term of $T'$ if and only if $qx_i$ is a term of $T$. As $f_{n,m}(A)\subseteq B$, by dividing the $A$-weighted zero-sum which is obtained from $T$ by $q$ and by taking the image under $f_{n,m}$ we get the contradiction that $T'$ is a $B$-weighted zero-sum subsequence of $S_1'$. 

Hence, we see that $S$ does not have a $A$-weighted zero-sum subsequence. As $S$ has length $k+l-2$, it follows that $D_A(n)\geq k+l-1$. 
\end{proof}

\section{Some results about the weight-set \texorpdfstring{$S(n)$}{}}\label{s}

From this point onwards, we will assume that $n$ is odd. 

\begin{definition}\label{jacobi}
For an odd prime $p$ and $a\in U(p)$, the symbol $\Big(\dfrac{a}{p}\Big)$ is the {\it Legendre symbol with respect to $p$}, which is defined as follows:
$$\Big(\dfrac{a}{p}\Big)=\begin{cases}
   1, &amp; \text{if $ a\in Q_p $;}\\
  -1, &amp; \text{if $ a\notin Q_p $.}
\end{cases}$$

For a prime divisor $p$ of $n$, we use the notation $\Big(\dfrac{a}{p}\Big)$ to denote $\bigg(\dfrac{f_{n,\,p}(a)}{p}\bigg)$ where $a\in U(n)$. Let $n=p_1^{r_1}\cdots p_k^{r_k}$ where the $p_i$'s are distinct primes. 

For $a\in U(n)$, we define the Jacobi symbol $\Big(\dfrac{a}{n}\Big)$ to be $\Big(\dfrac{a}{p_1}\Big)^{r_1}\cdots \Big(\dfrac{a}{p_k}\Big)^{r_k}$. Observe that we have 
$\Big(\dfrac{a}{n}\Big)=\bigg(\dfrac{a^{(p_1)}}{p_1^{r_1}}\bigg)\cdots \bigg(\dfrac{a^{(p_k)}}{p_k^{r_k}}\bigg)$. 

Let $S(n)$ denote the kernel of the homomorphism $U(n)\to\{1,-1\}$ given by $a\mapsto \Big(\dfrac{a}{n}\Big)$. 
\end{definition}

Adhikari, David, and Urroz \cite[Sec.\ 3]{ADU} considered the set $S(n)$ as a weight-set. 

\begin{proposition}\label{indexs}
$S(n)$ is a subgroup having index two in $U(n)$ when $n$ is a non-square, and $S(n)=U(n)$ when $n$ is a square.  
\end{proposition}

\begin{proof}
Let $n=p_1^{r_1}\cdots p_k^{r_k}$ where the $p_i$'s are distinct primes. If $n$ is a square, then all the $r_i$ are even, and so $S(n)=U(n)$. If $n$ is not a square, there exists $j$ such that $r_j$ is odd. As for every $i\in [1,k]$ the map $f_{p_i^{r_i},\,p_i}$ is onto, by the Chinese Remainder theorem we see that there is a unit $b\in U(n)$ such that $\Big(\dfrac{b}{p_i}\Big)=1$ when $i\neq j$, and $\Big(\dfrac{b}{p_j}\Big)=-1$. It follows that $\Big(\dfrac{b}{n}\Big)=-1$ and so the homomorphism $U(n)\to\{1,-1\}$ given by $a\mapsto \Big(\dfrac{a}{n}\Big)$ is onto. Hence, we see that $S(n)$ has index two in $U(n)$. 
\end{proof}

\begin{remark}
In particular, if $n$ is squarefree, then $S(n)$ has index two in $U(n)$. It follows that when $p$ is an odd prime we have $S(p)=Q_p$. 
\end{remark}


\begin{observation}\label{observation}
Let $n=p_1\cdots p_k$ where the $p_i$'s are distinct prime numbers. For $a\in U(n)$, let $\mu(a)$ denote the cardinality of $\{\,j\in [1,k]:f_{n,\,p_j}(a)=a^{(p_j)}\notin Q_{p_j}\,\}$. As we have that 
\[\bigg(\dfrac{a}{n}\bigg)=\bigg(\dfrac{a^{(p_1)}}{p_1}\bigg)\cdots \bigg(\dfrac{a^{(p_j)}}{p_j}\bigg)\cdots \bigg(\dfrac{a^{(p_k)}}{p_k}\bigg),\]
it follows that $a\in S(n)$ if and only if $\mu(a)$ is even.  
\end{observation}

\begin{lemma}\label{u2s}
Let $d$ be a proper divisor of $n$ such that $d$ is not a square. Suppose $d$ is coprime with $n'$ where $n'=n/d$. Then we have that  $U(n')\subseteq f_{n,n'}(S(n))$.
\end{lemma}

\begin{proof}
Let $a'\in U(n')$. By the Chinese remainder theorem, there is an isomorphism $\psi:U(n)\to U(n')\times U(d)$. As $d$ is not a square, by Proposition \ref{indexs} there exists $b\in U(d)$ such that $b\notin S(d)$. If $a'\in S(n')$, let $a\in U(n)$ be a unit such that $\psi(a)=(a',1)$. If $a'\notin S(n')$, let $a\in U(n)$ be a unit such that $\psi(a)=(a',b)$. Then we have $a\in S(n)$ and $f_{n,n'}(a)=a'$. 
\end{proof}

\begin{lemma}\label{lifts'}
Let $S$ be a sequence in $\mathbb Z_n$ and $d$ be a proper divisor of $n$ which divides every term of $S$. Let $n'=n/d$ and $d$ be coprime with $n'$. Let $S'$ be the sequence in $\mathbb Z_{n'}$ which is the image of the sequence $S$ under $f_{n,n'}$. Let $A\subseteq\mathbb Z_n$ and $A'\subseteq \mathbb Z_{n'}$ be subsets such that $A'\subseteq f_{n,n'}(A)$. Suppose $S'$ is an $A'$-weighted zero-sum sequence. Then $S$ is an $A$-weighted zero-sum sequence.  
\end{lemma}

\begin{proof}
Let $S=(x_1,\ldots,x_k)$ be a sequence in $\mathbb Z_n$ and $S'=(x_1',\ldots,x_k')$ where $x_i'=f_{n,n'}(x_i)$ for every $i\in [1,k]$. Suppose $S'$ is an $A'$-weighted zero-sum sequence. Then for every $i\in [1,k]$ there exist $a_i'\in A'$ such that $a_1'x_1'+\cdots+a_k'x_k'=0$. Since  $A'\subseteq f_{n,n'}(A)$, for every $i\in [1,k]$ there exist $a_i\in A$ such that $f_{n,n'}(a_i)=a_i'$. As $a_1'x_1'+\cdots+a_k'x_k'=0$ in $\mathbb Z_{n'}$, it follows that $f_{n,n'}(a_1x_1+\cdots+a_kx_k)=0$. Let $x=a_1x_1+\cdots+a_kx_k\in \mathbb Z_n$. As $f_{n,n'}(x)=0$, we see that $n'\mid x$, and as every term of $S$ is divisible by $d$, we see that $d\mid x$. As $d$ is coprime with $n'$, it follows that $x$ is divisible by $n=n'd$, and so $x=0$. Thus, we see that $S$ is an $A$-weighted zero-sum sequence. 
\end{proof}

Griffiths \cite[Lem.\ 2.1]{G} proved the next result, which we restate here using our terminology.

\begin{lemma}\label{gri}
Let $p$ be an odd prime. If a sequence $S$ in $\mathbb Z_{p^r}$ has at least two terms coprime to $p$, then $S$ is a $U(p^r)$-weighted zero-sum sequence. 
\end{lemma} 

Chintamani and Moriya \cite[Lem.\ 1]{CM} proved the next result. 

\begin{lemma}\label{cm'}
Let $A=U(n)^2$ where $n=p^r$ and $p$ is a prime which is at least seven. Suppose we have elements $x_1,x_2,x_3\in U(n)$. Then we get that $Ax_1+Ax_2+Ax_3=\mathbb Z_n$. 
\end{lemma}

We will use the next result in Lemma \ref{gs'}.  

\begin{lemma}\label{kneser}
Let $n=p^r$ where $p$ is a prime which is at least seven. Let $A_1=U(n)^2$ and $A_2=U(n)\setminus U(n)^2$. Suppose $x_1,x_2,x_3\in U(n)$ and $f:\{1,2,3\}\to\{1,2\}$ is a function. Then $A_{_{f(1)}}x_1+A_{_{f(2)}}x_2+A_{_{f(3)}}x_3=\mathbb Z_n$. 
\end{lemma} 

\begin{proof}
From \cite[Thm.\ 2, p.\ 43]{IR} we see that when $n$ is a power of an odd prime, the group $U(n)$ is cyclic. So it follows that $-1$ is the unique element in $U(n)$ of order 2. Thus, the map $U(n)\to U(n)$ given by $x\mapsto x^2$ has kernel $\{1,-1\}$. Hence, the image of this map is a subgroup of $U(n)$ having index 2 and so there exists $c\in U(n)$ such that $A_2=cA_1$. 

For every $i\in [1,3]$ let 
$$y_i=
\begin{cases}
x_i, &amp; \text{if } f(i)=1; \\
cx_i, &amp; \text{if } f(i)=2.
\end{cases}$$
Let $x\in \mathbb Z_n$. By Lemma \ref{cm'} there exist $b_1,b_2,b_3\in U(n)^2$ with  $x=b_1y_1+b_2y_2+b_3y_3$. 

For every $i\in [1,3]$ let 
$$a_i=
\begin{cases}
b_i, &amp; \text{if } f(i)=1; \\                                                                                                                                
b_ic, &amp; \text{if } f(i)=2.
\end{cases}$$
For every $i\in [1,3]$ it follows that $a_i\in A_{f(i)}$ and $b_iy_i=a_ix_i$. Thus, we see that $x=a_1x_1+a_2x_2+a_3x_3$. 
\end{proof}

The next result follows immediately from Lemma \ref{kneser}. 

\begin{corollary}\label{cm}
Let $n=p^r$ where $p$ is a prime which is at least seven. Suppose $S$ is a sequence in $\mathbb Z_n$ such that at least three terms of $S$ are in $U(n)$. Then $S$ is a $U(n)^2$-weighted zero-sum sequence. 
\end{corollary}

\begin{remark}
The conclusion of Corollary \ref{cm} may not hold when $p\leq 5$. One can check that the sequence $(1,1,1)$ in $\mathbb Z_n$ is not a $U(n)^2$-weighted zero-sum sequence when $n=2,5$. Also, the sequence $(1,2,1)$ in $\mathbb Z_3$ is not a $U(3)^2$-weighted zero-sum sequence. 
\end{remark}

\section{The constants \texorpdfstring{$D_{S(n)}$}{} and \texorpdfstring{$C_{S(n)}$}{}}

\begin{lemma}\label{gs}
Let $n$ be squarefree and $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$. Suppose for every prime divisor $p$ of $n$, at least two terms of $S$ are coprime to $p$. If at most one term of $S$ is a unit, then $S$ is an $S(n)$-weighted zero-sum sequence. 
\end{lemma}

\begin{proof}
As we have assumed that $n$ is odd and for every prime divisor $p$ of $n$ at least two terms of $S$ are coprime to $p$, by Lemma \ref{gri} we see that for every prime divisor $p$ of $n$ the sequence $S^{(p)}=(x_1^{(p)},\ldots,x_j^{(p)},\ldots,x_l^{(p)})$ is a $U(p)$-weighted zero-sum sequence. Let $n=p_1\cdots p_i\cdots p_k$ where the $p_i$'s are distinct primes. For every $i\in [1,k]$ there exist $c_{i,1},\ldots,c_{i,j},\ldots,c_{i,l}\in U(p_i)$ such that $c_{i,1}x_1^{(p_i)}+\cdots +c_{i,j}x_j^{(p_i)}+\cdots +c_{i,l}x_l^{(p_i)}=0$. We will refer to this $U(p_i)$-weighted zero-sum in $\mathbb Z_{p_i}$ as the $i^{th}$ sum. 

By Observation \ref{obs2} we see that for every $j\in [1,l]$ there exists $a_j\in U(n)$ such that 
\begin{equation}\label{mains}
a_1x_1+\cdots+a_jx_j+\cdots+a_lx_l=0,
\end{equation}
and for every $i\in [1,k]$ we have $(a_1^{(p_i)},\ldots,a_j^{(p_i)},\ldots,a_l^{(p_i)})=(c_{i,1},\ldots,c_{i,j},\ldots,c_{i,l})$. We observe that for some $i\in [1,k]$, a different choice for the $i^{th}$ sum will give us a different $l$-tuple $(a_1,\ldots,a_l)$ in \eqref{mains}. For example, if for some $i\in [1,k]$ there exists $j\in [1,l]$ such that $x_j^{(p_i)}$ is zero,  we can make an arbitrary choice for $c_{i,j}$ in the $i^{th}$ sum. For every $i\in [1,k]$ we want to choose the $i^{th}$ sum so that all the $a_j$'s in \eqref{mains} are in $S(n)$. Consider the following matrices:  
\[
\begin{matrix}
 
{C=
\begin{pmatrix}
c_{1,1} &amp; \cdots &amp; c_{1,j} &amp; \cdots &amp; c_{1,l} \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 
c_{i,1} &amp; \cdots &amp; c_{i,j} &amp; \cdots &amp; c_{i,l} \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 
c_{k,1} &amp; \cdots &amp; c_{k,j} &amp; \cdots &amp; c_{k,l} \\ 
\end{pmatrix}} 

&amp; ~ &amp; \text{and} &amp; ~ &amp; 

{X=
\begin{pmatrix}
x_1^{(p_1)} &amp; \cdots &amp; x_j^{(p_1)} &amp; \cdots &amp; x_l^{(p_1)} \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 
x_1^{(p_i)} &amp; \cdots &amp; x_j^{(p_i)} &amp; \cdots &amp; x_l^{(p_i)} \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 
x_1^{(p_k)} &amp; \cdots &amp; x_j^{(p_k)} &amp; \cdots &amp; x_l^{(p_k)} \\ 
\end{pmatrix}}. 

\end{matrix}
\]

Suppose some entry $x_j^{(p_i)}$ of $X$ is $0$. From Proposition \ref{indexs} and Observation \ref{observation} we see that by making a suitable choice for $c_{i,j}$ we can ensure that in \eqref{mains} we have $a_j\in S(n)$. Thus, if the $j^{th}$ column of $X$ has a zero, we can get a $U(n)$-weighted zero-sum \eqref{mains} in which $a_j\in S(n)$. 

We observe that a term $x_j$ of $S$ is a unit if and only if the $j^{th}$ column of $X$ does not have a zero. Hence, if no term of $S$ is a unit, then every column of $X$ has a zero. So in this case $S$ is an $S(n)$-weighted zero-sum sequence. 

Suppose exactly one term of $S$ is a unit, say $x_{j_0}$. Then the $j_0^{th}$ column of $X$ does not have a zero and there is a zero in all the other columns of $X$. By multiplying the $1^{st}$ row of $C$ by a suitable element of $U(p_1)$, we can modify $c_{1,j_0}$ so that $a_{j_0}\in S(n)$. As the other columns of $X$ have a zero, we can modify those columns of $C$ suitably so that $a_j\in S(n)$ for $j\neq j_0$. Thus, it follows that $S$ is an $S(n)$-weighted zero-sum sequence. 
\end{proof}

\begin{lemma}\label{gs'}
Let $n$ be a squarefree integer with every prime divisor of $n$ at least seven. Let  $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$ such that, for every prime divisor of $n$, at least two terms of $S$ are coprime to it. Suppose there is a prime divisor $p$ of $n$ such that at least three terms of $S$ are coprime to $p$. Then $S$ is an $S(n)$-weighted zero-sum sequence.
\end{lemma}

\begin{proof}
If $\Omega(n)=1$, then $n$ is a prime say $p$. As at least three terms of $S$ are coprime to $p$, Corollary \ref{cm} implies $S$ is a $Q_p$-weighted zero-sum sequence with $Q_p=S(p)$.

Suppose $\Omega(n)\geq 2$. Let $n=p_1\cdots p_k$ where the $p_i$'s are distinct primes. By Lemma \ref{gri} for every $i\in [1,k]$ there exist $c_{i,1},\ldots,c_{i,l}\in U(p_i)$ such that $c_{i,1}x_1^{(p_i)}+\cdots +c_{i,l}x_l^{(p_i)}=0$. By Observation \ref{obs2} there exist $a_1,\ldots,a_l\in U(n)$ such that 
\begin{equation}\label{pre}
 a_1x_1+\cdots+a_lx_l=0. 
\end{equation}

Assume that $p=p_1$ and that $x_1^{(p)}$, $x_2^{(p)}$, and $x_3^{(p)}$ are units. A similar argument will work in the general case. Let us denote $c_{1,1},\ldots,c_{1,l}\in U(p_1)$ by $b_1,\ldots,b_l$. We want to choose the $b_i$'s in $U(p)$ so that the corresponding $a_i$'s in \eqref{pre} are in $S(n)$. 

Using Observation \ref{observation} we can choose $b_4,\ldots,b_l\in U(p)$ so that  $a_4,\ldots,a_l\in S(n)$.
Let $y=-(b_4x_4^{(p)}+\cdots+b_lx_l^{(p)})$.  By using Observation \ref{observation} and Lemma \ref{kneser} we can choose $b_1,b_2,b_3\in U(p)$ so that $a_1,a_2,a_3\in S(n)$ and $b_1x_1^{(p)}+b_2x_2^{(p)}+b_3x_3^{(p)}=y$. Thus, $S$ is an $S(n)$-weighted zero-sum sequence.
\end{proof}

\begin{theorem}\label{dsn}
Let $n$ be squarefree. If $n$ is prime we have $D_{S(n)}=3$. If $n$ is not a prime and every prime divisor of $n$ is at least seven, we have $D_{S(n)}=\Omega(n)+1$. 
\end{theorem}

\begin{proof}
From Theorem \ref{un} we have $D_{U(n)}=\Omega(n)+1$. As $S(n)\subseteq {U(n)}$ it follows that $D_{S(n)}\geq D_{U(n)}$ and so $D_{S(n)}\geq \Omega(n)+1$. If $\Omega(n)=1$, then $n$ is a prime and $S(n)=Q_n$. So by Theorem \ref{q}, we have $D_{S(n)}=3$. 

Suppose $\Omega(n)\geq 2$. We claim that $D_{S(n)}\leq \Omega(n)+1$. Let $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$ of length $l=k+1$ where $k=\Omega(n)$. We have to show that $S$ has an ${S(n)}$-weighted zero-sum subsequence. If at least one term of $S$ is zero, then that term will give us an ${S(n)}$-weighted zero-sum subsequence of length 1.  

\bigskip

\noindent\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

Let us assume without loss of generality that $x_i$ is divisible by $p$ for every $i\in [2,l]$. Let $T$ denote the subsequence $(x_2,\ldots,x_l)$ of $S$. Let $n'=n/p$ and let $T'$ be the sequence in $\mathbb Z_{n'}$ which is the image of $T$ under $f_{n,n'}$. From Theorem \ref{un}, we see that $D_{U(n')}=\Omega(n')+1$. As $T'$ has length $l-1=\Omega(n)=\Omega(n')+1$, it follows that $T'$ has a $U(n')$-weighted zero-sum subsequence. As $n$ is squarefree, $p$ is coprime to $n'$. Thus, by Lemmas \ref{u2s} and \ref{lifts'} we see that $S$ has an $S(n)$-weighted zero-sum subsequence. 

\bigskip
\noindent\textit{Case 2:} For every prime divisor $p$ of $n$, exactly two terms of $S$ are coprime to $p$.

Suppose $S$ has at most one unit. By Lemma \ref{gs}, we see that $S$ is an $S(n)$-weighted zero-sum sequence. So we can assume that $S$ has at least two units. By the assumption in this subcase, we see that $S$ will have exactly two units and the other terms of $S$ will be zero. As $S$ has length $k+1$ and as $k\geq 2$, some term of $S$ is zero. 

\bigskip\noindent
\textit{Case 3:} For every prime divisor $p$ of $n$ at least two terms of $S$ are coprime to $p$, and there is a prime divisor $p'$ of $n$ such that at least three terms of $S$ are coprime to $p'$.

In this case, we are done by Lemma \ref{gs'}.
\end{proof}
 

\begin{theorem}\label{csn}
Let $n$ be squarefree. If $n$ is a prime, then $C_{S(n)}=3$. If $n$ is not a prime and every prime divisor of $n$ is at least seven, then $C_{S(n)}=2^{\Omega(n)}$. 
\end{theorem}

\begin{proof}
If $n=p$ where $p$ is a prime, then $S(n)=Q_p$. As $p$ is odd, from Theorem \ref{q} we get that $C_{S(n)}=3$. Let $n=p_1\cdots p_k$ where $k\geq 2$. As $S(n)\subseteq U(n)$, it follows that $C_{S(n)}\geq C_{U(n)}$. As $n$ is odd, from Theorem \ref{un} we have $C_{S(n)}\geq 2^k$. 

Let $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$ of length $l=2^k$. If we show that $S$ has an $S(n)$-weighted zero-sum subsequence of consecutive terms, it will follow that $C_{S(n)}\leq 2^k$. If at least one term of $S$ is zero, we get an $S(n)$-weighted zero-sum subsequence of $S$ of length 1. 

\bigskip\noindent
\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

We will get a subsequence $T$ of consecutive terms of $S$ of length $l/2$ with all its terms divisible by $p$. Let $n'=n/p$ and let $T'$ be the image of $T$ under $f_{n,n'}$. From Theorem \ref{un}, we have $C_{U(n')}=2^{\Omega(n')}$. As the length of $T'$ is $2^{\Omega(n')}$, it follows that $T'$ has a $U(n')$-weighted zero-sum subsequence of consecutive terms. As $n'$ is coprime with $p$, by Lemmas \ref{u2s} and \ref{lifts'} we get that $T$ (and hence $S$) has an $S(n)$-weighted zero-sum subsequence of consecutive terms.

\bigskip
\noindent
\textit{Case 2:} For every prime divisor $p$ of $n$ exactly two terms of $S$ are coprime to $p$.

In this case, as $\Omega(n)=k$, there are at most $2k$ non-zero terms in $S$. Suppose $k\geq 3$. As $S$ has length $2^k$ and as $2^k&gt;2k$, some term of $S$ is zero and we are done. 
Now assume that $k=2$. Then $S$ has length four. If $S$ has at most one unit, by Lemma \ref{gs} this sequence $S$ is an $S(n)$-weighted zero-sum sequence. So we can assume that $S$ has at least two units. By the assumption in this subcase, we see that $S$ has exactly two units and so the other two terms of $S$ are zero.   

\bigskip
\noindent
\textit{Case 3:} For every prime divisor $p$ of $n$ at least two terms of $S$ are coprime to $p$, and there is a prime divisor $p'$ of $n$ such that at least three terms of $S$ are coprime to $p'$.

In this case, we are done by Lemma \ref{gs'}.
\end{proof}

\section{Some results about the weight-set \texorpdfstring{$L(n;p)$}{}}\label{l}

To determine the constant $D_{S(n)}$ for some non-squarefree $n$, we consider the following subset of $\mathbb Z_n$ as a weight-set. 

\begin{definition}
Let $p$ be a prime divisor of $n$ where $n$ is odd. We define 
\[L(n;p)=\Big\{\,a\in U(n)\,:\,\Big(\dfrac{a}{n}\Big) =\Big(\dfrac{a}{p}\Big)\,\Big\}.\] 
\end{definition}

Consider the homomorphism $\varphi:U(n)\to\{1,-1\}$ given by $\varphi(a)=\Big(\dfrac{a}{n}\Big)\Big(\dfrac{a}{p}\Big)$. Then the kernel of $\varphi$ is $L(n;p)$. It follows that $L(n;p)$ is a subgroup having an index at most two in $U(n)$. 

\begin{proposition}
Let $p$ be a prime divisor of $n$. Then $L(n;p)$ has index two in $U(n)$ unless $p$ is the unique prime divisor of $n$ such that $v_{p}(n)$ is odd.
\end{proposition}
 
\begin{proof}
Let $n=p^rm$ where $m$ is coprime to $p$. Let $\psi: U(n)\to U(p^r)\times U(m)$ be the isomorphism that is given by the Chinese remainder theorem. If we show that $-1$ is in the image of the homomorphism $\varphi: U(n)\to\{1,-1\}$ which was defined above, then the kernel of $\varphi$ will be a subgroup of index two in $U(n)$.

\bigskip\noindent
\textit{Case 1:} $r$ is odd.

Suppose $m$ is a square. For every $a\in U(n)$, we have $\varphi(a)=\Big(\dfrac{a}{m}\Big)\Big(\dfrac{a}{p^{r+1}}\Big)=1$. Thus, $\varphi$ is the trivial map, and so $L(n;p)=U(n)$. 

Suppose $m$ is not a square. By Proposition \ref{indexs} we see that $S(m)$ has index two in $U(m)$. For $c\in U(m)\setminus S(m)$, there exists $a\in U(n)$ such that $\psi(a)=(1,c)$. Thus $\Big(\dfrac{a}{p}\Big)=\Big(\dfrac{1}{p}\Big)=1$ and so $\varphi(a)=\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{a}{m}\Big)=\Big(\dfrac{c}{m}\Big)=-1$.

\bigskip\noindent
\textit{Case 2:} $r$ is even.

Suppose $m=1$. Then $\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{a}{p}\Big)^r=1$ and so $\varphi(a)=\Big(\dfrac{a}{p}\Big)$. Let $b\in U(p)\setminus Q_p$. There exists $a\in U(n)$ such that $f_{n,p}(a)=b$. Thus $\varphi(a)=\Big(\dfrac{b}{p}\Big)=-1$. 

Suppose $m&gt;1$. Let $b\in U(p)\setminus Q_p$. There exists $b'\in U(p^r)$ such that $f_{p^r,\,p}(b')=b$. For $c\in S(m)$, there exists $a\in U(n)$ such that $\psi(a)=(b',c)$. Thus $\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{b}{p}\Big)^r\Big(\dfrac{c}{m}\Big)=1$ and so $\varphi(a)=\Big(\dfrac{a}{p}\Big)=\Big(\dfrac{b}{p}\Big)=-1$.
\end{proof}

\begin{remark}
In particular, if $n$ is a prime $p$, then $L(n;p)=U(p)$.
\end{remark}

The remaining results in this section are technical results, which will be used in the next section. 

\begin{lemma}\label{s2l}
Let $p$ and $p'$ be prime divisors of $n$ such that $p$ is coprime with $n'=n/p$. Then $S(n')\subseteq f_{n,\,n'}\big(L(n;p')\big)$. 
\end{lemma}

\begin{proof}
Let $b\in S(n')$ where $n'=n/p$. As $p$ is coprime with $n'$, by the Chinese remainder theorem we have an isomorphism $\psi:U(n)\to U(n')\times U(p)$. 

Suppose $p=p'$. Let $a\in U(n)$ be a unit such that $\psi(a)=(b,1)$. Thus $f_{n,\,n'}(a)=b$. We have $a\in L(n;p')$ as \[\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{b}{n'}\Big)\Big(\dfrac{1}{p}\Big)=\Big(\dfrac{1}{p}\Big)=\Big(\dfrac{a}{p}\Big)=\Big(\dfrac{a}{p'}\Big).\] 

Suppose $p\neq p'$. Then $p'$ divides $n'$. Let $c\in U(p)$ be a unit such that $\Big(\dfrac{c}{p}\Big)=\Big(\dfrac{b}{p'}\Big)$. Let $a\in U(n)$ be a unit such that $\psi(a)=(b,c)$. Thus $f_{n,\,n'}(a)=b$. We have $a\in L(n;p')$ as 
$$\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{b}{n'}\Big)\Big(\dfrac{c}{p}\Big)=\Big(\dfrac{c}{p}\Big)=\Big(\dfrac{b}{p'}\Big)=\Big(\dfrac{a}{p'}\Big).
\eqno\qed$$
\renewcommand{\qedsymbol}{}
\vspace{-\baselineskip}
\end{proof}

\begin{lemma}\label{u2l} 
Let $p'$ be a prime divisor of $n$ which is coprime to $n'=n/p'$. Then we have that $U(p')\subseteq f_{n,\,p'}\big(L(n;p')\big)$.  
\end{lemma}

\begin{proof}
Let $b\in U(p')$. As $n'=n/p'$ is coprime to $p'$, by the Chinese remainder theorem we have an isomorphism $\psi:U(n)\to U(n')\times U(p')$. There exists $a\in U(n)$ such that $\psi(a)=(1,b)$. Thus $f_{n,\,p'}(a)=b$. We have $a\in L(n;p')$ as $$\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{1}{n'}\Big)\Big(\dfrac{b}{p'}\Big)=\Big(\dfrac{b}{p'}\Big)=\Big(\dfrac{a}{p'}\Big).\eqno\qed$$
\renewcommand{\qedsymbol}{}
\vspace{-\baselineskip}
\end{proof}

The next result follows from a similar argument as in the proof of Observation \ref{obs2}. 

\begin{observation}\label{obs3}
Let $n=m_1m_2$ where $m_1$ and $m_2$ are coprime. Let $A\subseteq \mathbb Z_n$ be a subset and let $S$ be a sequence in $\mathbb Z_n$. For every $i\in [1,2]$ let $A_i\subseteq U(m_i)$ be given and $S_i$ denote the image of the sequence $S$ under $f_{n,\,m_i}$. Suppose $A_1\times A_2\subseteq \psi(A)$ where $\psi:U(n)\to U(m_1)\times U(m_2)$ is the isomorphism given by the Chinese remainder theorem. If $S_1$ is an $A_1$-weighted zero-sum sequence in $\mathbb Z_{m_1}$ and $S_2$ is an $A_2$-weighted zero-sum sequence in $\mathbb Z_{m_2}$, then $S$ is an $A$-weighted zero-sum sequence in $\mathbb Z_n$. 
\end{observation}

\begin{lemma}\label{gl'}
Let $n$ be a squarefree integer and let $n'=n/p'$, where $p'$ is a prime divisor of $n$. Suppose  $\psi:U(n)\to U(n')\times U(p')$ is the isomorphism given by the Chinese remainder theorem. Then we have that $S(n')\times U(p')\subseteq \psi\big(L(n;p')\big)$. 
\end{lemma}

\begin{proof}
Let $(b,c)\in S(n')\times U(p')$. There exists $a\in U(n)$ such that $\psi(a)=(b,c)$. Then we see that $a\in L(n;p')$ as 
\[\Big(\dfrac{a}{n}\Big)=\Big(\dfrac{b}{n'}\Big)\Big(\dfrac{c}{p'}\Big)=\Big(\dfrac{c}{p'}\Big)=\Big(\dfrac{a}{p'}\Big).\eqno\qed\]
\renewcommand{\qedsymbol}{}
\vspace{-\baselineskip}
\end{proof}

\section{The constants \texorpdfstring{$D_{L(n;\,p)}$}{} and \texorpdfstring{$C_{L(n;\,p)}$}{}}

\begin{lemma}\label{gl}
Let $n$ be a squarefree integer with every prime divisor of $n$ at least seven. Let  $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$ such that for every prime divisor $p$ of $n$ at least two terms of $S$ are coprime to $p$. Assume that $S'$ denotes the image of $S$ under $f_{n,n'}$, where $n'=n/p'$ with $p'$ a prime divisor of $n$. Suppose at most one term of $S'$ is a unit, or there is a prime divisor $p$ of $n/p'$ such that at least three terms of $S$ are coprime to $p$. Then $S$ is an $L(n;p')$-weighted zero-sum sequence. 
\end{lemma}

\begin{proof}
Let $n'=n/p'$ and let $S'$ denote the image of the sequence $S$ under $f_{n,n'}$.  As at least two terms of $S^{(p')}$ are coprime to $p'$, Lemma \ref{gri} implies that $S^{(p')}$ is a $U(p')$-weighted zero-sum sequence.

If at most one term of $S'$ is a unit, by Lemma \ref{gs} we see that $S'$ is an $S(n')$-weighted zero-sum sequence in $\mathbb Z_{n'}$. This is because $n'$ is squarefree and for every prime divisor $p$ of $n'$ at least two terms of $S'$ are coprime to $p$. 

If there is a prime divisor $p$ of $n/p'$ such that at least three terms of $S$ are coprime to $p$, by Lemma \ref{gs'} we see that $S'$ is an $S(n')$-weighted zero-sum sequence since at least three terms of $S'$ are coprime to $p$. 

As $n$ is squarefree, $n'$ is coprime to $p'$. Let $\psi:U(n)\to U(n')\times U(p')$ be the isomorphism given by the Chinese remainder theorem. By Lemma \ref{gl'} we see that $S(n')\times U(p')\subseteq \psi\big(L(n;p')\big)$. Hence, by Observation \ref{obs3} we see that $S$ is an $L(n;p')$-weighted zero-sum sequence. 
\end{proof}

\begin{theorem}\label{ld}
Let $n$ be a squarefree number such that every prime divisor of $n$ is at least seven and $\Omega(n)\neq 2$. Suppose $p'$ is a prime divisor of $n$. Then $D_{L(n;\,p')}=\Omega(n)+1$. 
\end{theorem}

\begin{proof}
Let $p'$ be a prime divisor of $n$. We have $D_{U(n)}\leq D_{L(n;\,p')}$, as $L(n;p')\subseteq U(n)$. From Theorem \ref{un} we have $D_{U(n)}=\Omega(n)+1$ and so $D_{L(n;\,p')}\geq \Omega(n)+1$. If $\Omega(n)=1$, then $L(n;p')=U(n)$ and so by Theorem \ref{un} we have $D_{L(n;\,p')}=2$. 

Let $n$ be a squarefree number such that every prime divisor is at least seven and $\Omega(n)\geq 3$. Suppose $S=(x_1,\ldots,x_l)$ is a sequence in $\mathbb Z_n$ of length $\Omega(n)+1$. It suffices to show that $S$ has an $L(n;p')$-weighted zero-sum subsequence. 

\bigskip\noindent
\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

Let us assume without loss of generality that $x_i$ is divisible by $p$ for $i&gt;1$. Let $T$ denote the subsequence $(x_2,\ldots,x_l)$ of $S$. Let $n'=n/p$ and let $T'$ denote the sequence in $\mathbb Z_{n'}$ which is the image of $T$ under $f_{n,n'}$. We see that $n'$ is a squarefree number, which is not a prime, every prime divisor of $n'$ is at least seven, and $T'$ has length $\Omega(n')+1$. 

So it follows from Theorem \ref{dsn} that $T'$ has an $S(n')$-weighted zero-sum subsequence. As $n$ is squarefree, it follows that $p$ is coprime to $n'$. So by Lemmas \ref{lifts'} and \ref{s2l} we see that $T$ has an $L(n;p')$-weighted zero-sum subsequence.

\bigskip
\noindent
\textit{Case 2:} For every prime divisor $p$ of $n/p'$,  there are exactly two terms of $S$ which are coprime to $p$, and at least two terms of $S$ are coprime to $p'$.

Let $n'=n/p'$ and let $S'=(x_1',\ldots,x_l')$ be the image of  $S$ under $f_{n,n'}$. Suppose at most one term of $S'$ is a unit. By Lemma \ref{gl} we see that $S$ is an $L(n;p')$-weighted zero-sum sequence. Suppose at least two terms of $S'$ are units. Under the assumptions in this case, two terms $x_{j_1}'$ and $x_{j_2}'$ of $S'$ are units, and the other terms of $S'$ are zero. It follows that all the terms of $S$ are divisible by $n'$ except $x_{j_1}$ and $x_{j_2}$.  

Hence, if some term $f_{n,p'}(x_j)$ of $S^{(p')}$ is zero for $j\neq j_1,j_2$, then $x_j=0$. So we can assume that all the terms of $S^{(p')}$ are non-zero except possibly two terms. As $\Omega(n)\geq 3$, the sequence $S$ has length at least four. Let $T$ be a subsequence of $S$ of length at least two which does not contain the terms $x_{j_1}$ and $x_{j_2}$. 

As all the terms of $T^{(p')}$ are non-zero and as $T^{(p')}$ has length at least 2, by Lemma \ref{gri} we see that $T^{(p')}$ is a $U(p')$-weighted zero-sum sequence. Also, all the terms of $T$ are divisible by $n'$.  Hence, by Lemmas \ref{lifts'} and \ref{u2l} we see that $T$ is an $L(n;p')$-weighted zero-sum subsequence of $S$.  

\bigskip\noindent
\textit{Case 3:} For every prime divisor $p$ of $n$, there are at least two terms of $S$ which are coprime to $p$, and there is a prime divisor $p$ of $n/p'$ such that at least three terms of $S$ are coprime to $p$.

In this case, we are done by Lemma \ref{gl}.
\end{proof}

\begin{theorem}\label{ld2}
Let $n=p'q$ where $p'$ and $q$ are distinct primes which are at least seven. Then $D_{L(n;p')}=4$.
\end{theorem}

\begin{proof} 
Let $n$ be as in the statement of the theorem. As $L(n;p')\subseteq U(n)$, we have that  $f_{n,\,p'}\big(L(n;p')\big)\subseteq U(p')$. Also observe that $f_{n,\,q}\big(L(n;p')\big)\subseteq Q_q$. As from Theorem \ref{un} we have $D_{U(p')}=2$ and from Theorem \ref{q} we have $D_{Q_q}=3$, by Lemma \ref{dadd} it follows that $D_{L(n;\,p')}\geq 4$. 

Let $S=(x_1,x_2,x_3,x_4)$ be a sequence in $\mathbb Z_n$. We will show that $S$ has an $L(n;p')$-weighted zero-sum subsequence. It will follow that $D_{L(n;\,p')}=4$. If some term of $S$ is zero, then we are done. So we can assume that all the terms of $S$ are non-zero. We continue with the notation and terminology that were used in the proof of Theorem \ref{ld}. 

\bigskip\noindent
\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

We can find a subsequence $T$ of $S$ of length three such that all the terms of $T$ are divisible by $p$. Let $n'=n/p$ and let $T'$ be the sequence in $\mathbb Z_{n'}$ which is the image of $T$ under $f_{n,n'}$. As all the terms of $S$ are non-zero, no term of $T$ can be divisible by $n'$. So $T'$ is a sequence of non-zero terms of length three. As $n'$ is a prime, we have  $S(n')=Q_{n'}$. By Corollary \ref{cm} we see that $T'$ is a $Q_{n'}$-weighted zero-sum subsequence. Thus, by Lemmas \ref{lifts'} and \ref{s2l} we see that $T$ is an $L(n;p')$-weighted zero-sum subsequence of $S$.

\bigskip\noindent
\textit{Case 2:} Exactly two terms of $S$ are coprime to $q$.

Let us assume that $x_1$ and $x_2$ are coprime to $q$. If $T=(x_3,x_4)$, the sequence $T^{(q)}$ has both terms zero. Hence, we get that $T^{(q)}$ is an $S(q)$-weighted zero-sum sequence. As $S$ has all terms non-zero, we see that both the terms of $T^{(p')}$ are non-zero. So by Lemma \ref{gri} we get that $T^{(p')}$ is a $U(p')$-weighted zero-sum sequence. Let $\psi:U(n)\to U(q)\times U(p')$ be the isomorphism given by the Chinese remainder theorem. By Lemma \ref{gl'} we have $S(q)\times U(p')\subseteq \psi\big(L(n;p')\big)$. Thus, by Observation \ref{obs3} we see that $T$ is an $L(n;p')$-weighed zero-sum subsequence of $S$. 

\bigskip
\noindent
\textit{Case 3:} At least three terms of $S$ are coprime to $q$, and at least two terms of $S$ are coprime to $p'$.

In this case, we are done by Lemma \ref{gl}.
\end{proof}

\begin{theorem}\label{lc}
Let $n$ be a squarefree number such that
every prime divisor of $n$ is at least seven and $\Omega(n)\neq 2$. Suppose $p'$ is a prime divisor of $n$. Then $C_{L(n;\,p')}=2^{\Omega(n)}$. 
\end{theorem}

\begin{proof}
If $n$ is a prime, then $n=p'$ and $L(n;p')=U(n)$. So from Theorem \ref{un} we have $C_{L(n;\,p')}=2$. Let $p'=p_k$ and $n=p_1\cdots p_k$ where $k\geq 3$. As $L(n;p')\subseteq U(n)$,  we have $C_{L(n;\,p')}\geq C_{U(n)}$. So from Theorem \ref{un}, we have $C_{L(n;\,p')}\geq 2^{\Omega(n)}$. Let $S=(x_1,\ldots,x_l)$ be a sequence in $\mathbb Z_n$ of length $l=2^{\Omega(n)}$. If we show that $S$ has an $L(n;p')$-weighted zero-sum subsequence of consecutive terms, it will follow that $C_{L(n;\,p')}\leq 2^{\Omega(n)}$. If at least one term of $S$ is zero, we get an $L(n;p')$-weighted zero-sum subsequence of $S$ of length one. 

\bigskip\noindent
\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

We can find a subsequence $T$ of consecutive terms of $S$ of length $l/2$ such that all the terms of $T$ are divisible by $p$. Let $n'=n/p$ and let $T'$ be the image of $T$ under $f_{n,n'}$. As $\Omega(n')=\Omega(n)-1\geq 2$ and $T'$ has length $2^{\Omega(n')}$, by Theorem \ref{csn} we see that $T'$ has an $S(n')$-weighted zero-sum subsequence of consecutive terms. By Lemma \ref{s2l} we get $S(n')\subseteq f_{n,n'}\big(L(n;p')\big)$. So by Lemma \ref{lifts'} we see that $T$ (and hence $S$) has an $L(n;p')$-weighted zero-sum subsequence of consecutive terms. 

\bigskip
\noindent
\textit{Case 2:} For every prime divisor $p$ of $n/p'$, there are exactly two terms of $S$  which are coprime to $p$, and at least two terms of $S$ are coprime to $p'$.

In this case, we can use a slight modification of the argument which was used in the same case of the proof of Theorem \ref{ld}. We just observe that if $S$ is a sequence of length at least eight such that at most two terms of $S$ are not divisible by $n'$, then we can find a subsequence $T$ of consecutive terms of $S$ having length at least two such that all the terms of $T$ are divisible by $n'$. 

\bigskip
\noindent
\textit{Case 3:} For every prime divisor $p$ of $n$, there are at least two terms of $S$ which are coprime to $p$, and there is a prime divisor $p$ of $n/p'$ such that at least three terms of $S$ are coprime to $p$.

In this case, we are done by Lemma \ref{gl}.
\end{proof}

\begin{theorem}\label{lc2}
Let $n=p'q$ where $p'$ and $q$ are distinct primes which are at least seven. Then $C_{L(n;\,p')}=6$.
\end{theorem}

\begin{proof}
Let $n$ be as in the statement of the theorem. By Theorems \ref{un} and \ref{q} we see that $C_{U(p')}=2$ and $C_{Q_q}=3$. As $f_{n,\,p'}\big(L(n;p')\big)\subseteq U(p')$ and $f_{n,\,q}\big(L(n;p')\big)\subseteq Q_q$, by Lemma \ref{clb} it follows that $C_{L(n;\,p')}\geq 6$.

Let $S=(x_1,\ldots,x_6)$ be a sequence in $\mathbb Z_n$. It is enough to show that $S$ has an $L(n;p')$-weighted zero-sum subsequence of consecutive terms. We can assume that all the terms of $S$ are non-zero. 

\bigskip\noindent
\textit{Case 1:} There is a prime divisor $p$ of $n$ such that at most one term of $S$ is coprime to $p$.

In this case, we can find a subsequence $T$ of $S$ of consecutive terms of length three whose all terms are divisible by $p$. As all the terms of $S$ are non-zero, all the terms of $T$ are coprime to $n'$ where $n'=n/p$. If $T'$ is the image of $T$ under $f_{n,n'}$, then $T'$ is a sequence of non-zero terms of length three in $\mathbb Z_{n'}$. As $n'$ is a prime, it follows that $S(n')=Q_{n'}$. By Corollary \ref{cm} we get that $T'$ is a $Q_{n'}$-weighted zero-sum sequence. By using Lemmas \ref{lifts'} and \ref{s2l} it follows that $T$ is an $L(n;p')$-weighted zero-sum subsequence of $S$ of consecutive terms.

\bigskip
\noindent
\textit{Case 2:} Exactly two terms of $S$ are coprime to $q$.

Let the terms $x_{j_1}$ and $x_{j_2}$ be coprime to $q$. As $S$ has length six, we can find a subsequence $T$ of consecutive terms of $S$ of length two, such that neither $x_{j_1}$ nor $x_{j_2}$ is a term of $T$. As $x_j$ is divisible by $q$ when $j\neq j_1,j_2$, all the terms of $T$ are divisible by $q$. As $S$ has all terms non-zero, all the terms of $T$ are coprime to $p'$. 

By Lemma \ref{gri} we get that $T^{(p')}$ is a $U(p')$-weighted zero-sum sequence. So by Lemmas \ref{lifts'} and \ref{u2l} it follows that $T$ is an $L(n;p')$-weighted zero-sum subsequence of consecutive terms of $S$. 

\bigskip
\noindent
\textit{Case 3:} At least three terms of $S$ are coprime to $q$, and at least two terms of $S$ are coprime to $p'$.

In this case, we are done by Lemma \ref{gl}.
\end{proof}

\section{Concluding remarks}  

We have $S(15)=\{1,2,4,8\}$. We can check that the sequence $S=(1,1,1)$ does not have a $S(15)$-weighted zero-sum subsequence. So it follows that $D_{S(15)}\geq 4$ and hence $D_{S(15)}\geq\Omega(15)+2$. This shows that the statement of Theorem \ref{dsn} is not true in general if some prime divisor of $n$ is smaller than seven. It will be interesting to find the Davenport constant $D_{S(n)}$ for non-squarefree $n$. 

Adhikari et al.\ \cite{ACFKP} proposed to characterize when two weight-sets $A\subseteq\mathbb Z_n$ have the same value of $D_A$. In this paper, we have seen that if $A\subseteq \mathbb Z_n$ is such that $S(n)\subseteq A\subseteq U(n)$ and if $n$ is not a prime, then $D_A=D_{U(n)}$. We have also seen that if $A\subseteq\mathbb Z_n$ is such that $L(n;p)\subseteq A\subseteq U(n)$ and if $\Omega(n)\neq 2$, then again $D_A=D_{U(n)}$. We can investigate whether there are other weight-sets $A\subseteq \mathbb Z_n$ such that $D_A=D_{U(n)}$. We can also ask similar questions regarding the constant $C_A$.  

\section{Acknowledgments}
Santanu Mondal would like to acknowledge CSIR, Govt.\ of India for a research fellowship whose file number is 09/934(0013)/2019-EMR-I. We thank the referee for the corrections and suggestions. 

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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B50; Secondary 11B75.

\noindent \emph{Keywords:} Davenport constant, zero-sum sequence, Jacobi symbol. 

\bigskip
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\vspace*{+.1in}
\noindent
Received May 18 2023; 
revised versions received May 21 2023; November 29 2023.
Published in {\it Journal of Integer Sequences}, January 14 2024.

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