Zero-Sum Constants Related to the Jacobi Symbol

Santanu Mondal, Krishnendu Paul, and Shameek Paul School of Mathematical Sciences

Ramakrishna Mission Vivekananda Educational and Research Institute West Bengal 711202 India

          santanu.mondal.math18@gm.rkmvu.ac.in
           krishnendu.p.math18@gm.rkmvu.ac.in
               shameek.paul@rkmvu.ac.in

Abstract

Let A ⊆ ℤ_{n} be a subset. A sequence S = (x_{1}, …, x_{k}) is said to be an A-weighted zero-sum sequence if there exist a_{1}, …, a_{k} ∈ A such that a_{1}x_{1} + ⋯ + a_{k}x_{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 ℤ_{n} has an Aweighted zero-sum subsequence. The constant C_{A} is defined to be the smallest natural number k such that every sequence of k elements in 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 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 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 ℤ_{n}), the corresponding A-weighted constants are the same as those for the weight-set U(n).

1 Introduction

For a, b ∈ ℤ, we denote the set {x ∈ ℤ : a ≤ x ≤ b} by [a, b]. Let U(n) denote the group of units in the ring ℤ_{n}, and U(n)^{2} = { x^{2} : x ∈ U(n) }. For an odd prime p, let Q_{p} denote the set U(p)^{2}. For n = p_{1}p_{2} ⋯ p_{k} where p_{i} is a prime for each i ∈ [1, k], we define Ω(n) = k.

Definition 1. Let A ⊆ ℤ_{n} be a subset. A sequence S = (x_{1}, …, x_{k}) is said to be an Aweighted zero-sum sequence if there exist a_{1}, …, a_{k} ∈ A such that a_{1}x_{1} + ⋯ + a_{k}x_{k} = 0. We refer to A as a weight-set.

Definition 2. For a weight-set A ⊆ ℤ_{n}, the A-weighted Davenport constant D_{A} is defined to be the least positive integer k, such that every sequence in ℤ_{n} of length k has an A-weighted zero-sum subsequence.

Adhikari and Rath [4] gave the previous definition. Chintamani and Moriya [5] showed that D_{U(n)^{2}} = 2 Ω(n) + 1 when every prime divisor of n is at least seven. Grynkiewicz and Hennecart [7] generalized this by showing that D_{U(n)^{2}} ≥ 2Ω(n) + min{v_{3}(n), v_{5}(n)} + 1 when n is odd, with equality if either 3 ∤ n or v_{3}(n) ≥ v_{5}(n). Mazumdar and Sinha [10] made suitable modifications in the method of Griffiths [6] 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. [1, Lem. 2.1] showed that D_{{1,−1}} = ⌊log_{2}n⌋ + 1 for every positive integer n.

Mondal, Paul, and Paul [11] gave the following definition.

Definition 3. For a weight-set A ⊆ ℤ_{n}, the A-weighted constant C_{A} is defined to be the least positive integer k, such that every sequence in ℤ_{n} of length k has an A-weighted zero-sum subsequence of consecutive terms.

Mondal, Paul, and Paul [11, Cor. 3, Cor. 6] showed that C_{U(n)^{2}} = 3^{Ω(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 [12] showed the next result.

Theorem 4. For every positive integer n we have D_{U(n)} = Ω(n) + 1 and C_{U(n)} = 2^{Ω(n)}.

When p is an odd prime such that p ≡ 2 (mod 3), we can show that U(p)^{3} = U(p). Mondal, Paul, and Paul [11, Thm. 7, Lem. 2] showed that when p ≠ 7 is a prime such that p ≡ 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 [4, Thm. 2], and Mondal, Paul, and Paul [11, Thm. 4] showed the next result.

Theorem 5. Let p be an odd prime. Then C_{Q_{p}} = D_{Q_{p}} = 3.

Let m be a divisor of n. We refer to the ring homomorphism f_{n,m} : ℤ_{n} → ℤ_{m} given by a + nℤ ↦ a + mℤ as the natural map. As this map sends units to units, we get a group homomorphism U(n) → U(m), which we also refer to as the natural map. When n is odd and x ∈ ℤ_{n}, the Jacobi symbol (\frac{x}{n}) is defined in Section 2.

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.

• Let S(n) = \{ x ∈ U(n) : (\frac{x}{n}) = 1 \}.

If n is prime, then D_{S(n)} = 3, and D_{S(n)} = Ω(n) + 1 otherwise.

If n is prime, then C_{S(n)} = 3, and C_{S(n)} = 2^{Ω(n)} otherwise.

• Let L(n; p) = \{ x ∈ U(n) : (\frac{x}{n}) = (\frac{x}{p}) \} where p is a prime divisor of n.

If Ω(n) = 2, then D_{L(n;p)} = 4, and D_{L(n;p)} = Ω(n) + 1 otherwise.

If Ω(n) = 2, then C_{L(n;p)} = 6, and C_{L(n;p)} = 2^{Ω(n)} otherwise.

Remark 6. Adhikari and Hegde [3] showed that if A = ℤ_{n} \ {0} and B = {1, 2, …, ⌈n/2⌉}, we have D_{A} = D_{B}. We make a similar observation in this paper. In Proposition 11, we show that S(n) is a subgroup of U(n) having index two when n is not a square. Theorem 4 shows that, when n is odd, we have D_{U(n)} = Ω(n) + 1 and C_{U(n)} = 2^{Ω(n)}. In addition, if n is not a prime, Theorems 23 and 24 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 Ω(n) ≠ 2, Theorems 33 and 35 show that D_{L(n; p)} = D_{U(n)} and C_{L(n; p)} = C_{U(n)}.

If p is a prime divisor of n, we use the notation v_{p}(n) = r to mean that p^{r} | n and p^{r+1} ∤ n. Let p be a prime divisor of n and v_{p}(n) = r. We denote the image in U(p^{r}) of x ∈ U(n) under f_{n, p^{r}} by x^{(p)}. For a sequence S = (x_{1}, …, x_{l}) in ℤ_{n}, let S^{(p)} denote the sequence (x_{1}^{(p)}, …, x_{l}^{(p)}) in ℤ_{p^{r}}, which is the image of S under f_{n, p^{r}}. Griffiths [6, Obs. 2.2] made the following observation.

Observation 7. Let n = p_{1}^{r_{1}} ⋯ p_{k}^{r_{k}} where the p_{i}’s are distinct primes and S = (x_{1}, …, x_{l}) be a sequence in ℤ_{n}. Suppose for every i ∈ [1, k] there exist c_{i,1}, …, c_{i,j}, …, c_{i,l} ∈ U(p_{i}^{r_{i}}) such that c_{i,1}x_{1}^{(p_{i})}+⋯+c_{i,j}x_{j}^{(p_{i})}+⋯+c_{i,l}x_{l}^{(p_{i})} = 0. Then there exist a_{1}, …, a_{j}, …, a_{l} ∈ U(n) such that for every (i, j) ∈ [1, k] × [1, l] we have a_{j}^{(p_{i})} = c_{i,j} and a_{1}x_{1} + ⋯ + a_{j}x_{j} + ⋯ + a_{l}x_{l} = 0.

Proof. Let j ∈ [1, l]. By the Chinese remainder theorem, there exists a_{j} ∈ U(n) such that for every i ∈ [1, k] we have that a_{j}^{(p_{i})} = c_{i,j}. Let x = a_{1}x_{1} + ⋯ + a_{j}x_{j} + ⋯ + a_{l}x_{l}. For each i ∈ [1, k] we see that f_{n, p_{i}^{r_{i}}}(x) = x^{(p_{i})} = c_{i,1}x_{1}^{(p_{i})} + ⋯ + c_{i,j}x_{j}^{(p_{i})} + ⋯ + c_{i,l}x_{l}^{(p_{i})} = 0. So by using the Chinese remainder theorem once again, we see that x = 0.

Mondal, Paul, and Paul [11, Lem. 3] showed the next result, which will be used in Theorem 36. In the next two results, for a subset A of ℤ_{n}, we use the notation C_{A}(n) and D_{A}(n) for the constants C_{A} and D_{A} respectively.

Lemma 8. Let n = mq. Let A, B, C be subsets of ℤ_{n}, ℤ_{m}, ℤ_{q} respectively. Suppose f_{n,m}(A) ⊆ B and f_{n,q}(A) ⊆ C. Then we have C_{A}(n) ≥ C_{B}(m) C_{C}(q).

We now prove a similar result for the weighted Davenport constant, which we will use in Theorem 34. Grynkiewicz, Marchan, and Ordaz [8, Lem. 3.1] proved a generalization of this result for abelian groups.

Lemma 9. Let n = mq. Let A, B, C be subsets of ℤ_{n}, ℤ_{m}, ℤ_{q} respectively. Suppose f_{n,m}(A) ⊆ B and f_{n,q}(A) ⊆ C. Then we have D_{A}(n) ≥ D_{B}(m) + D_{C}(q) − 1.

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}, …, u_{k−1}) of length k − 1 in ℤ_{m}, which has no B-weighted zero-sum subsequence, and there exists a sequence S′_{2} = (v_{1}, …, v_{l−1}) of length l − 1 in ℤ_{q}, which has no C-weighted zero-sum subsequence.

As f_{n,m} is onto, for every i ∈ [1, k − 1] there exists x_{i} ∈ ℤ_{n} such that f_{n,m}(x_{i}) = u_{i}. As f_{n,q} is onto, for every j ∈ [1, l − 1] there exists y_{j} ∈ ℤ_{n} such that f_{n,q}(y_{j}) = v_{j}. Consider the following sequence of length k + l − 2 in ℤ_{n}:

S = (qx_{1}, …, qx_{k−1}, y_{1}, …, y_{l−1}).

Let S_{1} = (qx_{1}, …, qx_{k−1}) and S_{2} = (y_{1}, …, y_{l−1}). Suppose S has an A-weighted zerosum 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) ⊆ 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) ⊆ 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) ≥ k + l − 1.

2 Some results about the weight-set S(n)

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

Definition 10. For an odd prime p and a ∈ U(p), the symbol (\frac{a}{p}) is the Legendre symbol

with respect to p, which is defined as follows:

(\frac{a}{p}) = \begin{cases}1, & if a ∈ Q_{p}; \\ −1, & if a ∉ Q_{p}. \end{cases}

For a prime divisor p of n, we use the notation (\frac{a}{p}) to denote (\frac{f_{n, p}(a)}{p}) where a ∈ U(n).

Let n = p_{1}^{r_{1}} ⋯ p_{k}^{r_{k}} where the p_{i}’s are distinct primes.

For a ∈ U(n), we define the Jacobi symbol (\frac{a}{n}) to be (\frac{a}{p_{1}} )^{r_{1}} ⋯ (\frac{a}{p_{k}} )^{r_{k}}. Observe that

we have (\frac{a}{n}) = (\frac{a^{(p_{1})}}{p_{1}^{r_{1}}} ) ⋯ (\frac{a^{(p_{k})}}{p_{k}^{r_{k}}} ).

Let S(n) denote the kernel of the homomorphism U(n) → \{1, −1\} given by a ↦ (\frac{a}{n}).

Adhikari, David, and Urroz [2, Sec. 3] considered the set S(n) as a weight-set.

Proposition 11. 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.

Proof. Let n = p_{1}^{r_{1}} ⋯ 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 ∈ [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 ∈ U(n) such that (\frac{b}{p_{i}} ) = 1 when i ≠ j, and (\frac{b}{p_{j}} ) = −1. It follows that

(\frac{b}{n}) = −1 and so the homomorphism U(n) → \{1, −1\} given by a ↦ (\frac{a}{n}) is onto. Hence, we see that S(n) has index two in U(n).

Remark 12. 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}.

Observation 13. Let n = p_{1} ⋯ p_{k} where the p_{i}’s are distinct prime numbers. For a ∈ U(n), let μ(a) denote the cardinality of { j ∈ [1, k] : f_{n, p_{j}}(a) = a^{(p_{j})} ∉ Q_{p_{j}} }. As we have that

(\frac{a}{n}) = (\frac{a^{(p_{1})}}{p_{1}} ) ⋯ (\frac{a^{(p_{j})}}{p_{j}} ) ⋯ (\frac{a^{(p_{k})}}{p_{k}} ),

it follows that a ∈ S(n) if and only if μ(a) is even.

Lemma 14. 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′) ⊆ f_{n,n′}(S(n)).

Proof. Let a′ ∈ U(n′). By the Chinese remainder theorem, there is an isomorphism ψ : U(n) → U(n′) × U(d). As d is not a square, by Proposition 11 there exists b ∈ U(d) such that b ∉ S(d). If a′ ∈ S(n′), let a ∈ U(n) be a unit such that ψ(a) = (a′, 1). If a′ ∉ S(n′), let a ∈ U(n) be a unit such that ψ(a) = (a′, b). Then we have a ∈ S(n) and f_{n,n′}(a) = a′.

Lemma 15. Let S be a sequence in 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 ℤ_{n′} which is the image of the sequence S under f_{n,n′}. Let A ⊆ ℤ_{n} and A′ ⊆ ℤ_{n′} be subsets such that A′ ⊆ f_{n,n′}(A). Suppose S′ is an A′-weighted zero-sum sequence. Then S is an A-weighted zero-sum sequence.

Proof. Let S = (x_{1}, …, x_{k}) be a sequence in ℤ_{n} and S′ = (x′_{1}, …, x′_{k}) where x′_{i} = f_{n,n′}(x_{i}) for every i ∈ [1, k]. Suppose S′ is an A′-weighted zero-sum sequence. Then for every i ∈ [1, k] there exist a′_{i} ∈ A′ such that a′_{1}x′_{1} + ⋯ + a′_{k}x′_{k} = 0. Since A′ ⊆ f_{n,n′}(A), for every i ∈ [1, k] there exist a_{i} ∈ A such that f_{n,n′}(a_{i}) = a′_{i}. As a′_{1}x′_{1} + ⋯ + a′_{k}x′_{k} = 0 in ℤ_{n′}, it follows that f_{n,n′}(a_{1}x_{1} + ⋯ + a_{k}x_{k}) = 0. Let x = a_{1}x_{1} + ⋯ + a_{k}x_{k} ∈ ℤ_{n}. As f_{n,n′}(x) = 0, we see that n′ | x, and as every term of S is divisible by d, we see that d | 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.

Griffiths [6, Lem. 2.1] proved the next result, which we restate here using our terminology.

Lemma 16. Let p be an odd prime. If a sequence S in ℤ_{p^{r}} has at least two terms coprime to p, then S is a U(p^{r})-weighted zero-sum sequence.

Chintamani and Moriya [5, Lem. 1] proved the next result.

Lemma 17. 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} ∈ U(n). Then we get that Ax_{1} + Ax_{2} + Ax_{3} = ℤ_{n}.

We will use the next result in Lemma 22.

Lemma 18. 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) \ U(n)^{2}. Suppose x_{1}, x_{2}, x_{3} ∈ U(n) and f : {1, 2, 3} → {1, 2} is a function. Then A_{f(1)}x_{1} + A_{f(2)}x_{2} + A_{f(3)}x_{3} = ℤ_{n}.

Proof. From [9, Thm. 2, p. 43] 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) → U(n) given by x ↦ 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 ∈ U(n) such that A_{2} = cA_{1}.

For every i ∈ [1, 3] let

y_{i} = \begin{cases}x_{i}, & if f(i) = 1; \\ cx_{i}, & if f(i) = 2. \end{cases}

Let x ∈ ℤ_{n}. By Lemma 17 there exist b_{1}, b_{2}, b_{3} ∈ U(n)^{2} with x = b_{1}y_{1} + b_{2}y_{2} + b_{3}y_{3}. For every i ∈ [1, 3] let

a_{i} = \begin{cases}b_{i}, & if f(i) = 1; \\ b_{i}c, & if f(i) = 2. \end{cases}

For every i ∈ [1, 3] it follows that a_{i} ∈ A_{f(i)} and b_{i}y_{i} = a_{i}x_{i}. Thus, we see that x = a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}.

The next result follows immediately from Lemma 18.

Corollary 19. Let n = p^{r} where p is a prime which is at least seven. Suppose S is a sequence in 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.

Remark 20. The conclusion of Corollary 19 may not hold when p ≤ 5. One can check that the sequence (1, 1, 1) in ℤ_{n} is not a U(n)^{2}-weighted zero-sum sequence when n = 2, 5. Also, the sequence (1, 2, 1) in ℤ_{3} is not a U(3)^{2}-weighted zero-sum sequence.

3 The constants D_{S(n)} and C_{S(n)}

Lemma 21. Let n be squarefree and S = (x_{1}, …, x_{l}) be a sequence in 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.

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 16 we see that for every prime divisor p of n the sequence S^{(p)} = (x_{1}^{(p)}, …, x_{j}^{(p)}, …, x_{l}^{(p)}) is a U(p)-weighted zero-sum sequence. Let n = p_{1} ⋯ p_{i} ⋯ p_{k} where the p_{i}’s are distinct primes. For every i ∈ [1, k] there exist c_{i,1}, …, c_{i,j}, …, c_{i,l} ∈ U(p_{i}) such that c_{i,1}x_{1}^{(p_{i})} + ⋯ + c_{i,j}x_{j}^{(p_{i})} + ⋯ + c_{i,l}x_{l}^{(p_{i})} = 0. We will refer to this U(p_{i})-weighted zero-sum in ℤ_{p_{i}} as the i^{th} sum.

By Observation 7 we see that for every j ∈ [1, l] there exists a_{j} ∈ U(n) such that

a_{1}x_{1} + ⋯ + a_{j}x_{j} + ⋯ + a_{l}x_{l} = 0,  (1)

and for every i ∈ [1, k] we have (a_{1}^{(p_{i})}, …, a_{j}^{(p_{i})}, …, a_{l}^{(p_{i})}) = (c_{i,1}, …, c_{i,j}, …, c_{i,l}). We observe that for some i ∈ [1, k], a different choice for the i^{th} sum will give us a different l-tuple (a_{1}, …, a_{l}) in (1). For example, if for some i ∈ [1, k] there exists j ∈ [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 ∈ [1, k] we want to choose the i^{th} sum so that all the a_{j}’s in (1) are in S(n). Consider the following matrices:

C = (\begin{matrix}c_{1,1} & ⋯ & c_{1,j} & ⋯ & c_{1,l} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ c_{i,1} & ⋯ & c_{i,j} & ⋯ & c_{i,l} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ c_{k,1} & ⋯ & c_{k,j} & ⋯ & c_{k,l} \end{matrix}) and X = (\begin{matrix}x_{1}^{(p_{1})} & ⋯ & x_{j}^{(p_{1})} & ⋯ & x_{l}^{(p_{1})} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ x_{1}^{(p_{i})} & ⋯ & x_{j}^{(p_{i})} & ⋯ & x_{l}^{(p_{i})} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ x_{1}^{(p_{k})} & ⋯ & x_{j}^{(p_{k})} & ⋯ & x_{l}^{(p_{k})} \end{matrix}).

Suppose some entry x_{j}^{(p_{i})} of X is 0. From Proposition 11 and Observation 13 we see that by making a suitable choice for c_{i,j} we can ensure that in (1) we have a_{j} ∈ S(n). Thus, if the j^{th} column of X has a zero, we can get a U(n)-weighted zero-sum (1) in which a_{j} ∈ 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}} ∈ S(n). As the other columns of X have a zero, we can modify those columns of C suitably so that a_{j} ∈ S(n) for j ≠ j_{0}. Thus, it follows that S is an S(n)-weighted zero-sum sequence.

Lemma 22. Let n be a squarefree integer with every prime divisor of n at least seven. Let S = (x_{1}, …, x_{l}) be a sequence in 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.

Proof. If Ω(n) = 1, then n is a prime say p. As at least three terms of S are coprime to p, Corollary 19 implies S is a Q_{p}-weighted zero-sum sequence with Q_{p} = S(p).

Suppose Ω(n) ≥ 2. Let n = p_{1} ⋯ p_{k} where the p_{i}’s are distinct primes. By Lemma 16 for every i ∈ [1, k] there exist c_{i,1}, …, c_{i,l} ∈ U(p_{i}) such that c_{i,1}x_{1}^{(p_{i})} + ⋯ + c_{i,l}x_{l}^{(p_{i})} = 0. By Observation 7 there exist a_{1}, …, a_{l} ∈ U(n) such that

a_{1}x_{1} + ⋯ + a_{l}x_{l} = 0.  (2)

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}, …, c_{1,l} ∈ U(p_{1}) by b_{1}, …, b_{l}. We want to choose the b_{i}’s in U(p) so that the corresponding a_{i}’s in (2) are in S(n).

Using Observation 13 we can choose b_{4}, …, b_{l} ∈ U(p) so that a_{4}, …, a_{l} ∈ S(n). Let y = −(b_{4}x_{4}^{(p)}+⋯+b_{l}x_{l}^{(p)}). By using Observation 13 and Lemma 18 we can choose b_{1}, b_{2}, b_{3} ∈ U(p) so that a_{1}, a_{2}, a_{3} ∈ S(n) and b_{1}x_{1}^{(p)}+b_{2}x_{2}^{(p)}+b_{3}x_{3}^{(p)} = y. Thus, S is an S(n)-weighted zero-sum sequence.

Theorem 23. 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)} = Ω(n) + 1.

Proof. From Theorem 4 we have D_{U(n)} = Ω(n) + 1. As S(n) ⊆ U(n) it follows that D_{S(n)} ≥ D_{U(n)} and so D_{S(n)} ≥ Ω(n) + 1. If Ω(n) = 1, then n is a prime and S(n) = Q_{n}. So by Theorem 5, we have D_{S(n)} = 3.

Suppose Ω(n) ≥ 2. We claim that D_{S(n)} ≤ Ω(n) + 1. Let S = (x_{1}, …, x_{l}) be a sequence in ℤ_{n} of length l = k + 1 where k = Ω(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.

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 ∈ [2, l]. Let T denote the subsequence (x_{2}, …, x_{l}) of S. Let n′ = n/p and let T′ be the sequence in ℤ_{n′} which is the image of T under f_{n,n′}. From Theorem 4, we see that D_{U(n′)} = Ω(n′) + 1. As T′ has length l − 1 = Ω(n) = Ω(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 14 and 15 we see that S has an S(n)-weighted zero-sum subsequence.

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 21, 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 ≥ 2, some term of S is zero.

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 22.

Theorem 24. 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^{Ω(n)}.

Proof. If n = p where p is a prime, then S(n) = Q_{p}. As p is odd, from Theorem 5 we get that C_{S(n)} = 3. Let n = p_{1} ⋯ p_{k} where k ≥ 2. As S(n) ⊆ U(n), it follows that C_{S(n)} ≥ C_{U(n)}. As n is odd, from Theorem 4 we have C_{S(n)} ≥ 2^{k}.

Let S = (x_{1}, …, x_{l}) be a sequence in 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)} ≤ 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.

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 4, we have C_{U(n′)} = 2^{Ω(n′)}. As the length of T′ is 2^{Ω(n′)}, it follows that T′ has a U(n′)-weighted zero-sum subsequence of consecutive terms. As n′ is coprime with p, by Lemmas 14 and 15 we get that T (and hence S) has an S(n)-weighted zero-sum subsequence of consecutive terms.

Case 2: For every prime divisor p of n exactly two terms of S are coprime to p.

In this case, as Ω(n) = k, there are at most 2k non-zero terms in S. Suppose k ≥ 3. As S has length 2^{k} and as 2^{k} > 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 21 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.

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 22.

4 Some results about the weight-set L(n; p)

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

Definition 25. Let p be a prime divisor of n where n is odd. We define

L(n; p) = \{ a ∈ U(n) : (\frac{a}{n}) = (\frac{a}{p}) \}.

Consider the homomorphism φ : U(n) → \{1, −1\} given by φ(a) = (\frac{a}{n})(\frac{a}{p}). Then the

kernel of φ is L(n; p). It follows that L(n; p) is a subgroup having an index at most two in U(n).

Proposition 26. 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.

Proof. Let n = p^{r}m where m is coprime to p. Let ψ : U(n) → U(p^{r}) × 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 φ : U(n) → {1, −1} which was defined above, then the kernel of φ will be a subgroup of index two in U(n).

Case 1: r is odd.

Suppose m is a square. For every a ∈ U(n), we have φ(a) = (\frac{a}{m})(\frac{a}{p^{r+1}} ) = 1. Thus, φ

is the trivial map, and so L(n; p) = U(n).

Suppose m is not a square. By Proposition 11 we see that S(m) has index two in U(m).

For c ∈ U(m) \ S(m), there exists a ∈ U(n) such that ψ(a) = (1, c). Thus (\frac{a}{p}) = (\frac{1}{p}) = 1

and so φ(a) = (\frac{a}{n}) = (\frac{a}{m}) = (\frac{c}{m}) = −1.

Case 2: r is even.

Suppose m = 1. Then (\frac{a}{n}) = (\frac{a}{p})^{r} = 1 and so φ(a) = (\frac{a}{p}). Let b ∈ U(p) \ Q_{p}. There

exists a ∈ U(n) such that f_{n,p}(a) = b. Thus φ(a) = (\frac{b}{p}) = −1.

Suppose m > 1. Let b ∈ U(p) \ Q_{p}. There exists b′ ∈ U(p^{r}) such that f_{p^{r}, p}(b′) = b. For

c ∈ S(m), there exists a ∈ U(n) such that ψ(a) = (b′, c). Thus (\frac{a}{n}) = (\frac{b}{p})^{r}(\frac{c}{m}) = 1 and

so φ(a) = (\frac{a}{p}) = (\frac{b}{p}) = −1.

Remark 27. In particular, if n is a prime p, then L(n; p) = U(p).

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

Lemma 28. Let p and p′ be prime divisors of n such that p is coprime with n′ = n/p. Then S(n′) ⊆ f_{n, n′}(L(n; p′)).

Proof. Let b ∈ S(n′) where n′ = n/p. As p is coprime with n′, by the Chinese remainder theorem we have an isomorphism ψ : U(n) → U(n′) × U(p).

Suppose p = p′. Let a ∈ U(n) be a unit such that ψ(a) = (b, 1). Thus f_{n, n′}(a) = b. We have a ∈ L(n; p′) as

(\frac{a}{n}) = (\frac{b}{n′} )(\frac{1}{p}) = (\frac{1}{p}) = (\frac{a}{p}) = (\frac{a}{p′} ).

Suppose p ≠ p′. Then p′ divides n′. Let c ∈ U(p) be a unit such that (\frac{c}{p}) = (\frac{b}{p′} ). Let

a ∈ U(n) be a unit such that ψ(a) = (b, c). Thus f_{n, n′}(a) = b. We have a ∈ L(n; p′) as

(\frac{a}{n}) = (\frac{b}{n′} )(\frac{c}{p}) = (\frac{c}{p}) = (\frac{b}{p′} ) = (\frac{a}{p′} ).

Lemma 29. Let p′ be a prime divisor of n which is coprime to n′ = n/p′. Then we have that U(p′) ⊆ f_{n, p′}(L(n; p′)).

Proof. Let b ∈ U(p′). As n′ = n/p′ is coprime to p′, by the Chinese remainder theorem we have an isomorphism ψ : U(n) → U(n′) × U(p′). There exists a ∈ U(n) such that ψ(a) = (1, b). Thus f_{n, p′}(a) = b. We have a ∈ L(n; p′) as

(\frac{a}{n}) = (\frac{1}{n′} )(\frac{b}{p′} ) = (\frac{b}{p′} ) = (\frac{a}{p′} ).

The next result follows from a similar argument as in the proof of Observation 7.

Observation 30. Let n = m_{1}m_{2} where m_{1} and m_{2} are coprime. Let A ⊆ ℤ_{n} be a subset and let S be a sequence in ℤ_{n}. For every i ∈ [1, 2] let A_{i} ⊆ U(m_{i}) be given and S_{i} denote the image of the sequence S under f_{n, m_{i}}. Suppose A_{1}×A_{2} ⊆ ψ(A) where ψ : U(n) → U(m_{1})×U(m_{2}) is the isomorphism given by the Chinese remainder theorem. If S_{1} is an A_{1}-weighted zero-sum sequence in ℤ_{m_{1}} and S_{2} is an A_{2}-weighted zero-sum sequence in ℤ_{m_{2}}, then S is an A-weighted zero-sum sequence in ℤ_{n}.

Lemma 31. Let n be a squarefree integer and let n′ = n/p′, where p′ is a prime divisor of n. Suppose ψ : U(n) → U(n′) × U(p′) is the isomorphism given by the Chinese remainder theorem. Then we have that S(n′) × U(p′) ⊆ ψ(L(n; p′)).

Proof. Let (b, c) ∈ S(n′) × U(p′). There exists a ∈ U(n) such that ψ(a) = (b, c). Then we see that a ∈ L(n; p′) as

(\frac{a}{n}) = (\frac{b}{n′} )(\frac{c}{p′} ) = (\frac{c}{p′} ) = (\frac{a}{p′} ).

5 The constants D_{L(n; p)} and C_{L(n; p)}

Lemma 32. Let n be a squarefree integer with every prime divisor of n at least seven. Let S = (x_{1}, …, x_{l}) be a sequence in 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.

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 16 implies that S^{(p′)} is a U(p′)-weighted zero-sum sequence.

If at most one term of S′ is a unit, by Lemma 21 we see that S′ is an S(n′)-weighted zero-sum sequence in 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 22 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 ψ : U(n) → U(n′) × U(p′) be the isomorphism given by the Chinese remainder theorem. By Lemma 31 we see that S(n′) × U(p′) ⊆ ψ(L(n; p′)). Hence, by Observation 30 we see that S is an L(n; p′)-weighted zero-sum sequence.

Theorem 33. Let n be a squarefree number such that every prime divisor of n is at least seven and Ω(n) ≠ 2. Suppose p′ is a prime divisor of n. Then D_{L(n; p′)} = Ω(n) + 1.

Proof. Let p′ be a prime divisor of n. We have D_{U(n)} ≤ D_{L(n; p′)}, as L(n; p′) ⊆ U(n). From Theorem 4 we have D_{U(n)} = Ω(n) + 1 and so D_{L(n; p′)} ≥ Ω(n) + 1. If Ω(n) = 1, then L(n; p′) = U(n) and so by Theorem 4 we have D_{L(n; p′)} = 2.

Let n be a squarefree number such that every prime divisor is at least seven and Ω(n) ≥ 3. Suppose S = (x_{1}, …, x_{l}) is a sequence in ℤ_{n} of length Ω(n) + 1. It suffices to show that S has an L(n; p′)-weighted zero-sum subsequence.

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 > 1. Let T denote the subsequence (x_{2}, …, x_{l}) of S. Let n′ = n/p and let T′ denote the sequence in 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 Ω(n′) + 1.

So it follows from Theorem 23 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 15 and 28 we see that T has an L(n; p′)-weighted zero-sum subsequence.

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}, …, x′_{l}) be the image of S under f_{n,n′}. Suppose at most one term of S′ is a unit. By Lemma 32 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 ≠ 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 Ω(n) ≥ 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 16 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 15 and 29 we see that T is an L(n; p′)-weighted zero-sum subsequence of S.

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 32.

Theorem 34. Let n = p′q where p′ and q are distinct primes which are at least seven. Then D_{L(n;p′)} = 4.

Proof. Let n be as in the statement of the theorem. As L(n; p′) ⊆ U(n), we have that f_{n, p′}(L(n; p′)) ⊆ U(p′). Also observe that f_{n, q}(L(n; p′)) ⊆ Q_{q}. As from Theorem 4 we have D_{U(p′)} = 2 and from Theorem 5 we have D_{Q_{q}} = 3, by Lemma 9 it follows that D_{L(n; p′)} ≥ 4.

Let S = (x_{1}, x_{2}, x_{3}, x_{4}) be a sequence in 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 33.

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 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 19 we see that T′ is a Q_{n′}-weighted zero-sum subsequence. Thus, by Lemmas 15 and 28 we see that T is an L(n; p′)-weighted zero-sum subsequence of S.

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 16 we get that T^{(p′)} is a U(p′)-weighted zero-sum sequence. Let ψ : U(n) → U(q)×U(p′) be the isomorphism given by the Chinese remainder theorem. By Lemma 31 we have S(q)×U(p′) ⊆ ψ(L(n; p′)). Thus, by Observation 30 we see that T is an L(n; p′)-weighed zero-sum subsequence of S.

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 32.

Theorem 35. Let n be a squarefree number such that every prime divisor of n is at least seven and Ω(n) ≠ 2. Suppose p′ is a prime divisor of n. Then C_{L(n; p′)} = 2^{Ω(n)}.

Proof. If n is a prime, then n = p′ and L(n; p′) = U(n). So from Theorem 4 we have C_{L(n; p′)} = 2. Let p′ = p_{k} and n = p_{1} ⋯ p_{k} where k ≥ 3. As L(n; p′) ⊆ U(n), we have C_{L(n; p′)} ≥ C_{U(n)}. So from Theorem 4, we have C_{L(n; p′)} ≥ 2^{Ω(n)}. Let S = (x_{1}, …, x_{l}) be a sequence in ℤ_{n} of length l = 2^{Ω(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′)} ≤ 2^{Ω(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.

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 Ω(n′) = Ω(n) − 1 ≥ 2 and T′ has length 2^{Ω(n′)}, by Theorem 24 we see that T′ has an S(n′)-weighted zero-sum subsequence of consecutive terms. By Lemma 28 we get S(n′) ⊆ f_{n,n′}(L(n; p′)). So by Lemma 15 we see that T (and hence S) has an L(n; p′)-weighted zero-sum subsequence of consecutive terms.

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 33. 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′.

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 32.

Theorem 36. Let n = p′q where p′ and q are distinct primes which are at least seven. Then C_{L(n; p′)} = 6.

Proof. Let n be as in the statement of the theorem. By Theorems 4 and 5 we see that C_{U(p′)} = 2 and C_{Q_{q}} = 3. As f_{n, p′}(L(n; p′)) ⊆ U(p′) and f_{n, q}(L(n; p′)) ⊆ Q_{q}, by Lemma 8 it follows that C_{L(n; p′)} ≥ 6.

Let S = (x_{1}, …, x_{6}) be a sequence in 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.

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 ℤ_{n′}. As n′ is a prime, it follows that S(n′) = Q_{n′}. By Corollary 19 we get that T′ is a Q_{n′}-weighted zero-sum sequence. By using Lemmas 15 and 28 it follows that T is an L(n; p′)-weighted zero-sum subsequence of S of consecutive terms.

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 ≠ 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 16 we get that T^{(p′)} is a U(p′)-weighted zero-sum sequence. So by Lemmas 15 and 29 it follows that T is an L(n; p′)-weighted zero-sum subsequence of consecutive terms of S.

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 32.

6 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)} ≥ 4 and hence D_{S(15)} ≥ Ω(15) + 2. This shows that the statement of Theorem 23 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. [1] proposed to characterize when two weight-sets A ⊆ ℤ_{n} have the same value of D_{A}. In this paper, we have seen that if A ⊆ ℤ_{n} is such that S(n) ⊆ A ⊆ U(n) and if n is not a prime, then D_{A} = D_{U(n)}. We have also seen that if A ⊆ ℤ_{n} is such that L(n; p) ⊆ A ⊆ U(n) and if Ω(n) ≠ 2, then again D_{A} = D_{U(n)}. We can investigate whether there are other weight-sets A ⊆ ℤ_{n} such that D_{A} = D_{U(n)}. We can also ask similar questions regarding the constant C_{A}.

7 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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2020 Mathematics Subject Classification: Primary 11B50; Secondary 11B75. Keywords: Davenport constant, zero-sum sequence, Jacobi symbol.

Received May 18 2023; revised versions received May 21 2023; November 29 2023. Published in Journal of Integer Sequences, January 14 2024.

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