March 04, 2026
Let \(A\) be a nonempty subset of finite abelian group \(G\) of order \(n\). For an integer \(h \geq 2\), the restricted \(h\)-fold sumset \(h^\wedge A\) is the set of all sums of \(h\) distinct elements of \(A\). It is known that if \(G\) is a group of order \(n\) and \(A\) is a subset of \(G\) such that \(|A|\) is close to \(\frac{n}{2}\), then \(h^{\wedge}A = G\) under some conditions on \(h\) and \(n\). The constant \(\frac{1}{2}\) is optimal for groups of even order but not for groups of odd order. For an integer \(h \geq 4\), let \(\alpha_h\) be the unique positive root of the polynomial \(3^{h - 2} x^{h - 1} + x - 1\). In this paper, we show that for any \(\alpha > \alpha_h\), there exists a positive integer \(M_h(\alpha)\), which is determined precisely, such that for all \(n > M_h(\alpha)\) with \(n\) odd, if \(A\) is a subset of a finite abelian group \(G\) of order \(n\) and if \(|A| \geq \alpha n\), then \(h^{\wedge} A = G\). Moreover, \(\alpha_h > \alpha_{h + 1}\) for \(h \geq 4\) and \(\alpha_h\) approaches \(\frac{1}{3}\) as \(h\) increases, and the constant \(\frac{1}{3}\) is optimal when the smallest prime dividing \(n\) is \(3\). This result extends a theorem of Tang and Wei on \(4^{\wedge}A\) in the cyclic group \(\mathbb{Z}_n\) to \(h^{\wedge}A\) for every \(h \geq 4\), and to arbitrary finite abelian groups.
Let G be an additive abelian group of order \(n\), and let \(\mathbb{Z}_n\) be a cyclic group of order \(n\). Let \(h \geq 2\) be an integer, and let \(A\) be a subset of \(G\). The cardinality of a set \(A\) is denoted by \(|A|\). The \(h\)-fold sumset of \(A\), denoted by \(hA\), is defined as \[hA = \{a_1 + a_2 + \cdots + a_h : a_i \in A\}.\] The restricted \(h\)-fold sumset of \(A\), denoted by \(h^\wedge A\), is defined as \[h^\wedge A = \{a_1 + a_2 + \cdots + a_h : a_i \in A, a_i \neq a_j \;\text{for} \;i \neq j\}.\]
Understanding the growth and extremal behavior of these sumsets is central to problems in additive combinatorics. These kind of problems have been studied extensively in the litearture (see [1]–[4] and the references given therein). The study of sumsets can be traced back to Cauchy [5], who proved that if \(A\) and \(B\) are nonempty subsets of \(\Bbb Z_p\), then \(|A + B| \geq \min(p, |A| + |B| -1)\), where \(\Bbb Z_p\) is the group of prime order \(p\), and \(A + B = \{a + b: a \in A, b \in B\}\). This result, later rediscovered by Davenport in 1935, is now known as the Cauchy-Davenport Theorem [6], [7]. An immediate consequence of this theorem is that for any nonempty subset \(A \subseteq \mathbb{Z}_p\) and any positive integer \(h\), the \(h\)-fold sumset \(hA\) satisfies \(|hA| \ge \min(p, h|A| - h + 1)\). While the \(h\)-fold sumset \(hA\) has been extensively studied in the literature, relatively few results are known for the restricted \(h\)-fold sumset \(h^{\wedge}A\). The restricted \(h\)-fold sumset \(h^{\wedge}A\) often displays behavior that is noticeably different from the \(h\)-fold sumset \(hA\), revealing subtler additive structures. Because of these unique characteristics, analyzing restricted \(h\)-fold sumset \(h^{\wedge}A\) typically requires methods that are significantly different from those used for the \(h\)-fold sumset \(hA\). In 1964, Erdős and Heilbronn conjectured that for a subset \(A\) of the cyclic group \(\mathbb{Z}_p\), \[|2^\wedge A| \geq \min\{p, 2 |A| - 3\}.\]
For the restricted \(h\)-fold sumset \(h^\wedge A\) in \(\mathbb{Z}_p\), a corresponding result was established by Dias da Silva and Hamidoune [8] using techniques from exterior algebra.
Theorem 1 ([8]). Let \(h\) and \(k\) be positive integers such that \(h \leq k\). Let \(A \subseteq {\mathbb{Z}_p}\) be a nonempty set with \(k\) elements. Then \[\left|h^\wedge A\right| \geq \min\{p, h k - h^{2} + 1\}.\]
This result was subsequently reproved by Alon, Nathanson, and Ruzsa [9], [10] using the polynomial method, a powerful tool for solving various problems in additive combinatorics. In the special case \(h = 2\), this theorem corresponds to the Erdős–Heilbronn conjecture, originally proposed by Erdős and Heilbronn [11] in 1964.
Another important problem associated with the restricted \(h\)-fold sumset \(h^{\wedge}A\) is the following: Given a finite abelian group \(G\) and an integer \(h \geq 2\), how large does a set \(A \subseteq G\) need to be to guarantee that the set of all sums of \(h\) distinct elements covers the whole group \(G\), that is, \(h^{\wedge}A = G\)? This problem has been investigated by various researchers.
In 1999, Gallardo, Grekos and Pihko proved the following result on \(2^\wedge A\) in [12].
Theorem 2 ([12]). Let \(A \subseteq \mathbb{Z}_n\) such that \(|A| > n/2 + 1\), then \(2^\wedge A = \mathbb{Z}_n\).
In 2002, Gallardo et al. obtained the following results (see [13]) in connection with possible extensions of a well-known theorem of Erdős, Ginzburg, and Ziv.
Theorem 3 ([13]). Let \(n\) be an odd positive integer. For any \[\alpha > \alpha_{0} = \frac{\left(\sqrt{13} - 1\right)}{6},\] there exists \[N = N(\alpha) = \frac{9}{3 \alpha^{2} + \alpha - 1}\] such that for all \(n > N\) and \(A \subseteq \mathbb{Z}_n\), if \(|A| \geq \alpha n\), then we have \[3 ^\wedge A = \mathbb{Z}_n.\]
Theorem 4 ([13]). For any integer \(n \geq 12\), except \(n = 15\), and for any subset \(A \subseteq \mathbb{Z}_n\) such that \(|A| > \frac{n}{2}\), one has \[3^\wedge A = \mathbb{Z}_n.\]
The following conjecture was stated in [13].
Conjecture 5 ([13]). There is a constant \(c\), such that for any odd integer \(n\) and for any subset \(A\) of \(\mathbb{Z}_n\) such that \(|A| > \frac{2}{5}n + c\), one has \(3^{\wedge}A = \mathbb{Z}_n\).
This conjecture was proved by Lev in \(2002\) by establishing the following result for \(3^\wedge A\) in arbitrary finite abelian group [14].
Theorem 6. Let \(G\) be a finite abelian group, and let \(G_0 = \{g \in G: 2g = 0\}\). Let \(A\) be a subset of \(G\) such that \[|A| > \max\left\{\frac{5}{13}|G|, 120\left|G_0\right| + 355\right\}.\] Then either \(3^{\wedge}A = G\), or \(A\) is contained in a coset of an index two subgroup of \(G\), or \(A\) is contained in a union of two cosets of an index five subgroup of \(G\).
In 2019, Tang and Wei [15] obtained the following result for \(4^\wedge A\) in the cyclic group \(\mathbb{Z}_n\) of odd order \(n\).
Theorem 7 ([15]). Let \(n \geq 11\) be an odd positive integer. For any \[\alpha > \alpha_{0} = \frac{3}{486} \left(27 + \sqrt{741} \right) + \frac{3}{486} \left(27 - \sqrt{741} \right),\] there exists \[N = N(\alpha) = \frac{54}{9 \alpha^{3} + \alpha - 1}\] such that for all \(n > N\) and \(A \subseteq \mathbb{Z}_n\), if \(|A| \geq \alpha n\), then we have \[4^\wedge A = \mathbb{Z}_n.\]
In this paper, we generalize Theorem 3 and Theorem 7 for the restricted sumset \(h^{\wedge}A\) for \(h \geq 4\) in arbitrary finite abelian group \(G\) of odd order \(n\). More precisely, we prove the following theorem.
Theorem 8. Let \(h \geq 4\) be an integer, and let \(p(h)\) denote the number of partitions of \(h\). Let \(\alpha_h\) be the unique positive root of the polynomial \(3^{h - 2} x^{h - 1} + x - 1\). Then for any \(\alpha > \alpha_h\) there exists \[M_h(\alpha) = \max \left\{\frac{3^{h - 2} (h^2 - h)}{2 \left(3^{h - 2} \alpha^{h - 1} + \alpha - 1 \right)}, \frac{12 (p(h) - 4)) (h - 4)! + (3 h - 7) (h - 4)}{6 \alpha}\right\}\] such that for all \(n > M_h(\alpha)\) with \(n\) odd, if \(A\) is a subset of a finite abelian group \(G\) of order \(n\) and if \(|A| \geq \alpha n\), then \[h^{\wedge} A = G.\]
Moreover, the roots \(\alpha_h\) satisfy the following properties:
\(\alpha_h \in (\frac{1}{3},\frac{1}{2})\) and \(\alpha_{h} > \alpha_{h + 1}\) for every \(h \geq 4\),
\(\lim\limits_{h \to \infty}\alpha_h = \frac{1}{3}\).
We give numerical values of \(\alpha_h\) and \(M_h(\alpha)\) corresponding to some values of \(h\) and \(\alpha > \alpha_{h}\) in the following table:
| \(h\) | \(\alpha_h\) (upto \(3\) decimal places) | \(\alpha\) | \(M_h(\alpha)\) |
|---|---|---|---|
| \(4\) | \(0.404\) | \(0.405\) | \(18807.96\) |
| \(5\) | \(0.388\) | \(0.389\) | \(37255.68\) |
| \(6\) | \(0.377\) | \(0.378\) | \(392935.41\) |
| \(7\) | \(0.370\) | \(0.371\) | \(1097319.46\) |
| \(8\) | \(0.365\) | \(0.366\) | \(2777127.76\) |
| \(9\) | \(0.361\) | \(0.362\) | \(11349436.56\) |
| \(10\) | \(0.358\) | \(0.359\) | \(33321849.20\) |
| \(11\) | \(0.356\) | \(0.357\) | \(57366134.14\) |
Remark 9. Theorem 6 implies that for all \(n \geq 1235\) and \(n\) odd, if \(|A| > 0.4n\) then \(3^\wedge A = G\). This result further implies that if \(|A| > 0.4n + 1\) then \(4^\wedge A = G\), for all \(n \geq 1235\) and \(n\) odd. For \(n \geq 1235\), we have \(0.4n + 1 < 0.404n\). Therefore, for \(n \geq 1235\), Theorem 6 implies the \(h = 4\) case of our Theorem 8.
It is worth mentioning a result for the restricted \(h\)-fold sumset \(h^{\wedge}A\) which has appeared recently b?, chen? while we were preparing the final draft of this paper. We briefly discuss this result here which was proved in connection with the \(h\)-critical number of a finite abelian group \(G\). For a positive integer \(h\), the restricted \(h\)-critical number \(\chi^{\wedge}(G, h)\) of the finite abelian group \(G\) is defined as \[\chi^{\wedge}(G, h) = \min\{m \in \mathbb{N}: h^{\wedge}A = G~\text{whenever}~ A \subseteq G~ \text{with}~ |A| \geq m\}.\] For the groups of even order, the precise values \(\chi^{\wedge}(G, h)\) was determined by Roth and Lempel [16] in 1992 (see also [17] for detailed discussion on crticial numbers). Very recently, Chen and Huang b?, chen? has proved the following theorem in case of finite abelian groups of odd order.
Theorem 10 (b?, chen?). Let \(G\) be a finite abelian group of odd order \(n\), let \(p(G)\) be the smallest prime divisor \(n\). Let \(h\) be an integer such that \(3 \leq h \leq \dfrac{n}{p(G)} - 2\). Let \[c(n) = \begin{cases} \frac{2}{5} & \text{if } 5 \mid n, \\ \frac{5}{13} & \text{if } 5 \nmid n. \end{cases}\]
Then the \(h\)-critical number satisfies the upper bound \[\chi^{\wedge}(G, h) \leq \begin{cases} \lfloor c(n) n \rfloor + 9 & \text{if } p(G) = 3 \text{ and } n \geq 3 \cdot 46319, \\ \lfloor c(n) n \rfloor + 21 & \text{if } p(G) = 5 \text{ and } n \geq 5 \cdot 1235, \\ \lfloor c(n) n \rfloor + 3 & \text{if } p(G) \geq 7 \text{ and } n \geq 1235. \end{cases}\]
The above theorem gives an upper bound on \(\chi^{\wedge}(G, h)\) which is close to \(\frac{2}{5}n\) or \(\frac{5}{13}n\) depending on whether \(n\) is divisible by \(5\). But Theorem 8 gives more precise information about \(\chi^{\wedge}(G, h)\) and implies an improved upper bound: \(\chi^{\wedge}(G, h) \leq \alpha n\) for all \(\alpha > \alpha_h\) under some conditions on the order of the group \(n\), where \(\alpha_h\) strictly decreases as \(h\) increases, and approaches the limit \(\frac{1}{3}\). In fact, for \(h \geq 6\), the value of \(\alpha_h\) becomes even less than \(\frac{5}{13}\). We remark that if a set \(A\) is contained in a coset of a subgroup of \(G\) of index three, then \(|A| \leq \frac{n}{3}\) but \(h^{\wedge} A \neq G\). Thus the constant \(\frac{1}{3}\) is optimal.
The organization of the paper is as follows. We fix some general notations in the next section. In Section 3, we discuss some basic concepts from group algebra and character theory of finite abelian group which will be required for the proofs. In Section 4, we prove auxiliary lemmas which will be reqiured for the proof of the main theorem. Finally, in Section 5, we prove Theorem 8.
A partition of a positive integer \(h\) is a \(h\)-tuple \((\lambda_1, \lambda_2, \dots, \lambda_h)\) such that \(0 \leq \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_h\) and \(\lambda_1 + \lambda_2 +\cdots + \lambda_h = h\). Let \(\mathbb{P}(h)\) denote set of all partitions of \(h\), and let \(p(h)\) be the number of partitions of \(h\). Then \[\mathbb{P}(h) = \{(\lambda_1, \lambda_2, \dots, \lambda_h): 0\leq \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_h \;\text{and} \;\lambda_1 + \lambda_2 +\cdots + \lambda_h = h\}.\] The set \(\mathbb{P}(h)\) is equipped with colexicographic order “\(\prec\)" defined as follows: \[(d_1, d_2, \dots, d_h) \prec (e_1, e_2, \dots, e_h)\] if and only if there exist \(k \in \{1, 2, \dots, h\}\) such that \(d_i = e_i\) for all \(i > k\) and \(d_k < e_k\). Then \(|\mathbb{P}(h)| = p(h)\). Let \(\boldsymbol{\lambda_i}\) denote the \(i^{\mathrm{th}}\) element of \(\mathbb{P}(h)\) in the colexicographic order. Then \[\boldsymbol{\lambda_1} = (1, 1, \dots, 1) \prec \boldsymbol{\lambda_2} = (0, 1, \dots, 1, 2 ) \prec \cdots \prec \boldsymbol{\lambda_{p(h)}} = (0, 0, \dots, 0, h).\] For \(r \in \{1,2 \dots, h\}\) and \(\boldsymbol{\lambda} \in \mathbb{P}(h)\), let \(\mu_{\boldsymbol{\lambda}}(r)\) denotes the multiplicity of \(r\) in \(\boldsymbol{\lambda}\). If \(\boldsymbol{\lambda} = \boldsymbol{\lambda_i}\), then we simply write \(\mu_{i}(r)\) in place of \(\mu_{\boldsymbol{\lambda}}(r)\). Let \(\boldsymbol{\lambda_i} = (\lambda_{i1}, \lambda_{i2}, \dots, \lambda_{ih})\). For \(h\geq 4\), let \(\boldsymbol{\lambda_{i_0}} = (0, 0, 1, \dots, 1, 3)\).
Let \(G\) be a finite abelian group, and let \(A = \{ a_1, a_2,\dots,a_k\} \subseteq G\). Let \[A^h = \{(a_{j_1},a_{j_2}, \dots, a_{j_h}): a_{j_r} \in A \;\text{for}\;r = 1, 2, \dots, h\}.\] For \(m \in G\), we define \[\label{R40m41} R(m) = \left|\{(a_{j_1}, a_{j_2}, \dots, a_{j_h}) \in A^h : m = a_{j_1} + a_{j_2} + \cdots + a_{j_{h}}, a_{j_r} \neq a_{j_s} \;\text{for}\;j_r \neq j_s\}\right|\tag{1}\] Corresponding to each \(\boldsymbol{\lambda_i} \in \mathbb{P}(h)\), we define \[\begin{align} R_i(m) = |\{(\underbrace{a_{j_1}, \ldots, a_{j_1}}_{\lambda_{i1} \;\text{times}}, \ldots, \underbrace{a_{j_h}, \ldots, a_{j_h}}_{\lambda_{ih} \;\text{times}}) \in A^h: m = \lambda_{i1}a_{j_{1}} + \cdots + \lambda_{ih}a_{j_{h}}\}|. \end{align}\] Let \(i_{1} = \min\{j: 1\leq j \leq h \;\text{and} \;\lambda_{ij} > 0\}\). Then \(R_i(m)\) can be expressed as \[\label{R95i40m41} R_i(m) = |\{(\underbrace{a_{j_{i_1}}, \ldots, a_{j_{i_1}}}_{\lambda_{ii_1} \;\text{times}}, \ldots, \underbrace{a_{j_h}, \ldots, a_{j_h}}_{\lambda_{ih} \;\text{times}}) \in A^h: m= \lambda_{ii_1} a_{j_{i_1}} + \cdots + \lambda_{ih} a_{j_{h}}\}|\tag{2}\] for \(i = 1, \ldots, p(h)\).
For any real number \(v\), we denote \[e(v) = \exp(2 \pi i v).\]
In this section, we follow the definitions and notations from [18].
Let \(\mathbb{C}\) be the field of complex numbers, and let \(G\) be the finite abelian group. The group algebra \(\mathbb{C}[G]\) is the all polynomials in the variable \(x\) with coefficients from \(\mathbb{C}\) and exponents from \(G\). More precisely, \(\mathbb{C}[G]\) consists of all formal expression of the form \(f = \sum_{g \in G} c_g x^g\) with \(c_g \in \mathbb{C}\), and multiplication and addition obeying the following rules:
\[(ax^g)(bx^h) = abx^{g + h} ~\text{for}~ a, b \in \mathbb{C},\]
\[\sum_{g \in G} a_g x^g + \sum_{g \in G} b_g x^g = \sum_{g \in G} (a_g + b_g) x^g,\]
\[c\left(\sum_{g \in G} a_g x^g \right) = \sum_{g \in G} c a_g x^g,\]
\[\left(\sum_{g \in G} a_g x^g\right) \left(\sum_{g \in G} b_g x^g\right) = \sum_{g \in G}\left(\sum_{h \in G} a_h b_{g - h}\right) x^g.\]
Let \(G\) be a finite abelian group, written additively. Let \(\widehat{G} := \mathrm{Hom}(G, \mathbb{C^{\times}})\) be the group of characteres of the finite abelian group \(G\), where the group operation in \(\widehat{G}\) is written multiplicatively. Then \(G \cong \widehat{G}\). We write \(g \mapsto \chi_g\) to denote an isomorphism from \(G\) onto \(\widehat{G}\). Then \(\widehat{G} = \{ \chi_g : g \in G \}\). This isomorphism depends on the choice of basis for \(G\). Given a subset \(A\) of \(G\), we fix a basis \(\{e_1, \dots, e_s\}\) of \(G\), so that each element \(a \in G\) can be expressed uniquely as \(a = \sum_{j = 1}^{s} y_je_j\), with \(y_j \in [0, \mathrm{ord}(e_j) - 1]\), where \(\mathrm{ord}(e_j)\) denote the order of \(e_j\). We use this basis to define the function \(S_A : G \to \mathbb{C}\) by \[S_A = \sum_{g \in A} \chi_g,\] where the \(\chi_g \in \widehat{G}\). If \[g = \sum_{j = 1}^{s} r_j e_j \in G\] and if \(u_j = \mathrm{ord}(e_j)\), then \(\chi_g\) is defined by \[\chi_g \left(\sum_{j = 1}^{s} y_j e_j \right) = \prod_{j = 1}^{s} e\left(\frac{r_j y_j}{u_j} \right) = e \left(\frac{1}{u} \sum_{j = 1}^{s} r_j y_j \frac{u}{u_j}\right),\] where \(y_j, r_j \in \mathbb{Z}\) and \(u = \mathrm{exp}(G)\), the exponent of the group \(G\). The following identities will be useful:
\(\chi_g(a) \chi_h(a) = \chi_{g + h}(a)\),
\(\chi_g(a + b) = \chi_g(a) \chi_g(b)\),
\(\chi_g(h) = \chi_h(g)\),
The image of \(G\) under \(\chi_g\) is a cyclic group of the \(d\)-th roots of unity, where \(d\) is order of \(g\).
Let \(x_1, \dots, x_k\) be variables. For \(t \geq 1\), let \[\label{power-sum-poly} p_t:= p_t(x_1,\dots, x_k) = x_1^t + \cdots + x_k^t.\tag{3}\] For \(h \geq 0\), the elementary symmetric polynomial \(e_h(x_1, \dots, x_k)\) is defined as \[\label{sym-poly} e_h:= e_h(x_1, \dots, x_k) = \begin{cases} 1, & ifh = 0; \\ \sum\limits_{1 \leq j_1 < \cdots < j_h \leq k} x_{j_1}\cdots x_{j_h} , & if1 \leq h \leq k; \\ 0, & ifh > k. \end{cases}\tag{4}\]
The following lemma for symmetric polynomials is crucial for the proof of main theorem.
Lemma 1 ([19]). For integers \(t \geq 1\) and \(h \geq 0\), let \(p_t\) and \(e_h\) be defined as in 3 and 4 . Then \[e_h = (-1)^h \sum_{\substack{m_1 + 2 m_2 + \cdots + h m_h = h \\ m_1 \geq 0, \dots, m_h \geq 0} } \prod_{r = 1}^{h} \frac{(-p_r)^{m_r}}{m_r!r^{m_r}}.\]
We prove the following lemma using Lemma 1.
Lemma 2. Let \(A = \{a_1, a_2, \dots,a_k\}\) be a nonempty subset of \(G\). Let \(R(m)\) and \(R_i(m)\) be defined as in 1 and 2 . Then for \(m \in G\), we have the following identity: \[R(m) = \sum_{i = 1}^{p(h)} (-1)^h h! \prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m).\]
Let \(x\) be a variable, and let \(x_j = x^{a_j}\) for \(j = 1, 2, \dots, k\). Consider the polynomial \(p_1:= x_1 + \cdots + x_k\) in the group algebra \(\mathbb{C}[G]\). Then \[\begin{align} p_1^h = (x_1 + x_2 + \cdots + x_k )^h & = (x^{a_1} + x^{a_2} + \cdots + x^{a_k} )^h \\ & = \sum_{1 \leq j_1,\dots, j_h \leq k} x^{a_{j_1} + a_{j_2} + \cdots + a_{j_h}}\\ & = \sum_{ m \in hA} \Biggl(\sum_{\substack{1 \leq j_1,\dots, j_h \leq k \\ a_{j_1} + a_{j_2} + \cdots + a_{j_h} = m } } x^m\Biggr)\\ & = \sum_{ m \in hA} x^m \Biggl(\sum_{\substack{1 \leq j_1,\dots, j_h \leq k \\ a_{j_1} + a_{j_2} + \cdots + a_{j_h} = m } }1\Biggr)\\ & = \sum_{ m \in hA} R_1(m) x^m. \end{align}\] Thus \[p_1^h = \sum_{m \in hA} R_1(m) x^m,\] and so it follows from Lemma 1 that \[\label{representation-lem-eq1} \sum_{m \in hA} R_1(m) x^m = h!e_h + (-1)^{h + 1} h!\sum_{\substack{m_1 + 2 m_2 + \cdots+ h m_h = h \\ m_1 \geq 0, \dots, m_h \geq 0 \\ m_1 \neq h} } \prod_{r = 1}^{h} \frac{(-p_r)^{m_r}}{m_r! r^{m_r}}.\tag{5}\] Any \(\boldsymbol{\lambda_i} \in \mathbb{P}(h)\) can be expressed as \[\boldsymbol{\lambda_i} = (\underbrace{0, \ldots, 0}_{m_0 \;\text{times}}, \underbrace{1, \ldots, 1}_{m_1 \;\text{times}}, \ldots, \underbrace{h, \ldots, h}_{m_h \;\text{times}}),\] where \(m_r \geq 0\) for \(r = 1, \dots, h\), and \(m_0 = h - \sum_{r = 1}^{h}m_r\). Then \(m_1 + 2 m_2 + \cdots + h m_h = h\). Clearly, \(m_r = \mu_i(r)\) for \(r \in \{1,2,\dots, h\}\). Conversely, if \(m_1 + 2 m_2 + \cdots + h m_h = h\), then this corresponds to the partition \(\boldsymbol{\lambda_i} \in \mathbb{P}(h)\) given by \[\boldsymbol{\lambda_i} = (\underbrace{0, \ldots, 0}_{m_0 \;\text{times}}, \underbrace{1, \ldots, 1}_{m_1 \;\text{times}}, \ldots, \underbrace{h, \ldots, h}_{m_h \;\text{times}}),\] where \(m_0 = h - \displaystyle\sum_{r = 1}^{h} m_r\). Thus there is a one-to-one correspondence between the set of partitions \(\mathbb{P}(h)\) and the set of \(h\)-tuples \((m_1, \ldots, m_h)\) of nonnegative integers satisfying \(m_1 + 2 m_2 + \cdots + h m_h = h\). Also \(m_1 = h\) corresponds to \(\boldsymbol{\lambda_1} = (1, 1, \dots, 1)\). Therefore, the identity 5 can be expressed as follows:
\[p_1^h = h! e_h + (-1)^{h+1} h!\sum_{\substack{\boldsymbol{\lambda} \in \mathbb{P}(h) \\ \boldsymbol{\lambda} \neq \boldsymbol{\lambda_1}}} \Biggl(\prod_{r = 1}^{h} \frac{(-p_r)^{\mu_{\boldsymbol{\lambda}} (r)}}{(\mu_{\boldsymbol{\lambda}} (r))! r^{\mu_{\boldsymbol{\lambda}} (r)}}\Biggr).\] The above identity can be written as \[\label{equation323461} p_1^h = h! e_h + (-1)^{h + 1} h!\sum_{i = 2}^{p(h)} \Biggl(\prod_{r = 1}^{h} \frac{(-p_r)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}}\Biggr).\tag{6}\] For \(\boldsymbol{\lambda_i} = (\lambda_{i1}, \lambda_{i2}, \dots, \lambda_{ih}) \in \mathbb{P}(h)\), let \(i_{1} = \min\{j: 1 \leq j \leq h \;\text{and} \;\lambda_{i j} > 0\}\). That is, \(\lambda_{i i_1}\) is the smallest positive part of the partition \(\boldsymbol{\lambda_i}\). Then it is easy to see that \[\prod_{r = 1}^{h} (p_r)^{\mu_i (r)} = \prod_{j = i_{1}}^{h} p_{\lambda_{i j}}.\] Hence it follows from 6 that
\[\begin{align} \label{equation323461a} \sum_{m \in hA} R_1(m) x^m = h! e_h + (-1)^{h + 1} h! \sum_{i = 2}^{p(h)} \Biggl(\biggl(\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))!r^{\mu_i (r)}}\biggr)\biggl(\prod_{\substack{j = i_{1}}}^{h} p_{\lambda_{ij}}\biggr)\Biggr). \end{align}\tag{7}\]
Now, \[\begin{align} h! e_h = h!\sum\limits_{1 \leq j_1 < \cdots < j_h \leq k} x_{j_1}\cdots x_{j_h} &= h! \sum_{1 \leq j_1 < j_2 < \cdots < j_h \leq k} x^{a_{j_1}} x^{a_{j_2}} \cdots x^{a_{j_h}}\\ &= \sum_{\substack{1 \leq j_1, \dots, j_h \leq k \\ a_{j_r} \neq a_{j_s} \;\text{for} \;j_r \neq j_s }} x^{a_{j_1} + a_{j_2} + \cdots + a_{j_h}}\\ &= \sum_{m \in hA} \Biggl(\sum_{\substack{1 \leq j_{i_1}, \dots, j_h \leq k \\ a_{j_1} + a_{j_2} + \cdots + a_{j_h} = m \\ a_{j_r} \neq a_{j_s} \;\text{for} \;j_r \neq j_s }} x^{m}\Biggr)\\ &= \sum_{m \in hA} \Biggl(\sum_{\substack{1 \leq j_1, \dots, j_h \leq k \\a_{j_1} + a_{j_2} + \cdots + a_{j_h} = m \\a_{j_r} \neq a_{j_s} \;\text{for}\;j_r \neq j_s\\ }}1\Biggr)x^m, \end{align}\] and so it follows from the definition of \(R(m)\) that \[\label{equation323461b} h! e_h = \sum_{m \in hA} R(m) x^m.\tag{8}\] Next, \[\begin{align} \prod_{j = i_1}^{h} p_{\lambda_{ij}} = \prod_{j = i_1}^{h} (x_1^{\lambda_{ij}} + \cdots + x_k^{\lambda_{ij}}) &= \prod_{j = i_1}^{h} (x^{\lambda_{ij}a_1} + \cdots + x^{\lambda_{ij}a_k})\\ &= \sum_{1 \leq j_{i_{1}}, \dots, j_h \leq k} x^{\lambda_{i i_1} a_{j_{i_{1}}}}\cdots x^{\lambda_{i h} a_{j_h}}\\ &= \sum_{1 \leq j_{i_1}, \dots, j_h \leq k} x^{\lambda_{i i_1} a_{j_{i_1}} + \cdots + \lambda_{i h} a_{j_h}}\\ &= \sum_{m \in hA} \biggl(\sum_{\substack{1 \leq j_{i_1}, \dots, j_h \leq k \\ \lambda_{i i_1} a_{j_{i_1}} + \cdots + \lambda_{i h} a_{j_{h}} = m } } x^{m}\biggr)\\ &= \sum_{m \in hA} \biggl(\sum_{\substack{1 \leq j_{i_1}, \dots, j_h \leq k \\ \lambda_{i i_1} a_{j_{i_1}} + \cdots + \lambda_{i h} a_{j_{h}} = m }} 1\biggr) x^{m}, \end{align}\] and so it follows from the definition of \(R_i(m)\) that
\[\label{equation323461c} \prod_{j = i_1}^{h} p_{\lambda_{ij}} = \sum_{m \in hA} R_i(m) x^{m}.\tag{9}\] Therefore, it follows from 7 , 8 and 9 that
\[\begin{align} \sum_{m \in hA} R_1(m) x^m = &\sum_{m \in hA} R(m) x^m + (-1)^{h + 1} h! \sum_{i = 2}^{p(h)} \Biggl(\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} \biggl(\sum_{m \in hA} R_i(m) x^{m}\biggr)\Biggr), \end{align}\] which implies that \[\begin{align} \sum_{m \in hA} R_1(m) x^m = &\sum_{m \in hA} R(m) x^m + (-1)^{h + 1} h! \sum_{m \in hA}\Biggl(\sum_{i = 2}^{p(h)} \biggl(\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m)\biggr)\Biggr) x^{m}, \end{align}\] and so \[\label{equation323462} \begin{align} \sum_{m \in hA} R_1(m) x^m = \sum_{m \in hA} \Biggl(R(m) + (-1)^{h + 1} h! \sum_{i = 2}^{p(h)} \biggl(\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m)\biggr)\Biggr) x^{m} \end{align}\tag{10}\] By comparing the coefficients of \(x^m\) in 10 , we get \[R_1(m) = R(m) + (-1)^{h + 1} h!\sum_{i = 2}^{p(h)} \Biggl(\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m)\Biggr).\] This implies that \[R(m) = \sum_{i = 1}^{p(h)} (-1)^h h! \prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m).\]
Lemma 3 ([13]). Let \(d \geq 3\) be an odd integer, and let \(Y\) be a positive real number. For any \(\mathbf{y} = (y_1, \dots, y_d) \in \mathbb{R}^d\), let \[T(\mathbf{y}) = \sum_{j = 1}^{d} y_j e{\left(j/d\right)}.\] Then \[\max_{\mathbf{y} \in [0, Y]^d} |T(\mathbf{y})| = \frac{Y}{2\sin(\pi/2d)}.\]
We prove the following lemma using Lemma 3. The argument of the proof is similar to the argument in the proof of a similar result for a cyclic group \(\mathbb{Z}_n\) in [15].
Lemma 4. Let \(G\) be a finite abelian group of odd order \(n\), where \(n \geq 3\). Let \(A\) be a nonempty subset of \(G\), and let \(g \in G\) with \(g \neq 0\). Then \[|S_A(g)| \leq \frac{n}{3}.\]
Let \(\chi_g \in \widehat{G}\), where \(\chi_g \neq \chi_0\). The image of \(G\) under \(\chi_g\) is a cyclic group of the \(d\)-th roots of unity, where \(d\) is order of \(\chi_g\) in \(\widehat{G}\). Since \(n\) is odd and \(d\) divides \(n\), it follows that \(d\) must be an odd integer with \(d \ge 3\). We partition \(G\) into \(d\) cosets \(C_1, \ldots, C_d\) of the kernel of \(\chi_g\). These cosets are defined as \[C_j = \{ a \in G : \chi_g(a) = e{\left(j/d\right)}\} ~\text{for}~ j = 1, \dots, d.\] Let \[y_j = \left|\{ a \in A : \chi_g(a) = e{\left(j/d\right)} \}\right| ~\text{for}~ j = 1, 2, \dots, d.\] Then \[\sum_{a \in A} \chi_g(a) = \sum_{j = 1}^{d} y_j e{\left(j/d\right)}.\] Since \(|C_j| = \frac{n}{d}\), it follows that \(y_j = |A \cap C_j| \leq \frac{n}{d}\). Therefore, it follows from Lemma 3 that \[\left| \sum_{a \in A} \chi_g(a) \right| \leq \frac{n/d}{2 \sin(\pi/2d)}.\] We know that if \(\theta \in (0, \pi/6]\), then \(\sin \theta \geq \frac{3 \theta}{\pi}\). Since \(\pi/2d \in (0, \pi/6]\), it follows that \[\sin \left(\frac{\pi}{2 d}\right) \geq \frac{3 (\pi/2 d)}{\pi} = \frac{3}{2d}.\] Hence \[\left|\sum_{a \in A} \chi_g(a) \right| \leq \frac{n}{3}.\] Since \(\chi_g(a) = \chi_a(g)\), it follows that \[\left|\sum_{a \in A} \chi_a(g) \right| \leq \frac{n}{3},\] and so \[|S_A(g)| \leq \frac{n}{3}.\] This completes the proof.
Lemma 5 ([18]). Let \(G\) be a finite abelian group. Let \(\chi \in \widehat{G}\). Then \[\frac{1}{|G|} \sum_{g \in G} \chi(g) = \begin{cases} 1, & \chi = \chi_0 \\ 0, & \chi \neq \chi_0. \end{cases}\]
Lemma 6 ([18]). Let \(G\) be a finite abelian group and let \(A\) be a nonempty subset of \(G\). Then \[\sum_{g \in G \setminus \{0\}} |S_A(g)|^2 = |A| |G| - |A|^2\]
Lemma 7. Let \(A = \{a_1, a_2, \dots,a_k\}\) be a nonempty subset of a finite abelian group \(G\) of odd order. Let \(h\) be an integer with \(h \geq 4\), and let \(i_0\) be the index such that \(\boldsymbol{\lambda_{i_0}} = (0, 0, 1, \dots, 1, 3)\). Let \(R(m)\) and \(R_i(m)\) be defined as in 1 and 2 . Then for \(m \in G\), the following inequalities hold:
\(R_2(m) \leq k(k-1)\cdots (k - h + 3) + \binom{h - 2}{2} R_3(m)\),
\(R_3(m) \leq k^{h - 3}\),
\(R_i(m) \leq k^{h - 4}\) for \(i \geq 4\) with \(i \neq i_0\).
For \(i \geq 4\) with \(i \neq i_0\), let \(\boldsymbol{\lambda_i} = (\lambda_{i1}, \lambda_{i2}, \dots, \lambda_{ih}) \in \mathbb{P}(h)\). Then \[\prod_{r = 1}^{h} (\mu_i(r))! r^{\mu_i(r)} \geq 4.\]
Note that \[\begin{align} R_2(m) = \left|\{(a_{j_2}, \dots, a_{j_{h-1}}, a_{j_h}, a_{j_h}) \in A^h : m= a_{j_2} + \cdots + a_{j_{h - 1}} + 2 a_{j_{h}}\}\right|, \end{align}\] and \[R_3(m) = \left|\{(a_{j_3}, \dots, a_{j_{h - 2}}, a_{j_{h - 1}}, a_{j_{h - 1}}, a_{j_h}, a_{j_h}) \in A^h: m= a_{j_3} + \cdots + 2 a_{j_{h-1}} + 2 a_{j_{h}}\}\right|.\]
To get the upper bound for \(R_2(m)\), we consider two possiblities:
If \(a_{j_2}, \dots, a_{j_{h - 2}}\) all are distinct, then the number of choices for \(a_{j_2}, \dots, a_{j_{h - 1}}\) is \(k (k - 1) \cdots (k - h + 3)\), and any choice of \(a_{j_2}, \dots, a_{j_{h - 1}}\) fixes \(2 a_{j_{h}}\), and so it fixes the value of \(a_{j_{h}}\), since \(n\) is odd. Therefore, \[R_2(m) \leq k (k - 1) \cdots (k - h + 3).\]
If at least two of \(a_{j_2}, \dots, a_{j_{h - 2}}\) are same, then clearly \(R_2(m) \leq \binom{h - 2}{2} R_3(m)\).
Therefore, \[R_2(m) \leq k(k-1)\cdots (k - h + 3) + \binom{h - 2}{2} R_3(m).\]
Next, since the number of choices for \(a_{j_3}, \dots, a_{j_{h - 1}}\) is \(k^{h - 3}\), any of these choices fixes the value of \(a_{j_{h}}\), it follows that \[R_3(m) \leq k^{h - 3}.\]
A similar combinatorial argument proves that if \(i \geq 4\) and \(i \neq i_0\), then \[R_i(m) \leq k^{h - 4}.\]
Finally, to prove the last inequality, we consider three cases for \(\lambda_{i h}\):
If \(\lambda_{i h} < 3\), then \(\lambda_{i h} = \lambda_{i(h - 1)} = \lambda_{i(h - 2)} = 2\);
If \(\lambda_{i h} = 3\), then \(\lambda_{i(h - 1)}\geq 2\);
\(\lambda_{i h} > 3\),
In each of the three cases, it is easy to verify that \(\prod_{r = 1}^{h}(\mu_i(r))! r^{\mu_i(r)} \geq 4\). This completes the proof.
Lemma 8. Let \(h\) and \(k\) be integers such that \(h \geq 4\) and \(k \geq \frac{h (h - 1)}{2}\). Then the following inequality holds: \[\begin{gather} k (k - 1)\cdots (k - h + 3) \leq k^{h - 2} - \frac{(h - 2)(h - 3)}{2} k^{h - 3} + \frac{(h - 2)(h - 3)(h - 4)(3 h - 7)}{24} k^{h - 4}. \end{gather}\]
The proof is trivial for \(h = 4\) and \(h = 5\). Now assume that \(h \geq 6\). Note that \[\begin{align} k (k - 1)\cdots (k - h + 3) &= k^{h - 2} + \sum_{r = 1}^{h - 3}(-1)^{r} \left(\sum_{0 \leq j_1 < \cdots < j_r \leq h - 3} j_1 \cdots j_r \right) k^{h - 2 - r}, \end{align}\] which implies that
\[\begin{gather} \label{bound32k-eq1} k (k - 1)\cdots (k - h + 3) \\ = k^{h - 2} - \frac{(h - 2)(h - 3)}{2} k^{h - 3} + \frac{(h - 2)(h - 3)(h - 4)(3h - 7)}{24} k^{h - 4}\\ + \sum_{r = 3}^{h - 3}(-1)^{r} \left(\sum_{0 \leq j_1 < \cdots < j_r \leq h - 3} j_1 \cdots j_r \right) k^{h - 2 - r}. \end{gather}\tag{11}\] Now, since \(k \geq \frac{h (h - 1)}{2}\), it follows that if \(r \geq 3\), then
\[\begin{align} \sum_{0 \leq j_1 < \cdots < j_{r + 1} \leq h - 3} j_1 \cdots j_{r + 1} &\leq \sum_{0 \leq j_1 < \cdots < j_{r} \leq h - 3} j_1 \cdots j_{r} \left(\sum_{0 \leq j_{r + 1} \leq h - 3} j_{r + 1}\right) \notag \\ &= \frac{(h - 2)(h - 3)}{2} \sum_{0 \leq j_1 < \cdots < j_{r} \leq h - 3} j_1 \cdots j_{r} \notag \\ & \leq k \sum_{0 \leq j_1 < \cdots < j_{r} \leq h - 3} j_1 \cdots j_{r}, \end{align}\] and so \[\begin{align} \label{bound32k-eq2} \left(\sum_{0 \leq j_1 < \cdots < j_{r} \leq h - 3} j_1 \cdots j_{r}\right) k^{h-2-r} - \left(\sum_{0 \leq j_1 < \cdots < j_{r + 1} \leq h - 3} j_1 \cdots j_{r + 1}\right) k^{h-3-r} \geq 0. \end{align}\tag{12}\] Let \(f(r)\) denote the left hand side of 12 . Now if \(h\) is odd, then it follows from 11 that \[\begin{align} \sum_{r = 3}^{h - 3}(-1)^{r} \left(\sum_{0 \leq j_1 < \cdots < j_r \leq h - 3} j_1 \cdots j_r \right) k^{h - 2 - r} = - \left(\sum_{s = 3}^{\frac{h-3}{2}}f(2s-1)\right) \leq 0, \end{align}\] and so it follows from 11 that \[\begin{gather} k (k - 1)\cdots (k - h + 3) \leq k^{h - 2} - \frac{(h - 2)(h - 3)}{2} k^{h - 3} + \frac{(h - 2)(h - 3)(h - 4)(3 h - 7)}{24} k^{h - 4}. \end{gather}\] If \(h\) is even, then it follows from 11 that \[\begin{gather} \sum_{r = 3}^{h - 3}(-1)^{r} \left(\sum_{0 \leq j_1 < \cdots < j_r \leq h - 3} j_1 \cdots j_r \right) k^{h - 2 - r} \\ = - \left(\sum_{s = 3}^{\frac{h-4}{2}}f(2s-1)\right) - \left(\sum_{0 \leq j_1 < \cdots < j_{h-3} \leq h - 3} j_1 \cdots j_{h-3} \right) k \leq 0, \end{gather}\] and so it follows from 11 that \[\begin{gather} k (k - 1)\cdots (k - h + 3) \leq k^{h - 2} - \frac{(h - 2)(h - 3)}{2} k^{h - 3} + \frac{(h - 2)(h - 3)(h - 4)(3 h - 7)}{24} k^{h - 4}. \end{gather}\] This completes the proof.
Let \(A = \{a_1, a_2, \dots, a_k\} \subseteq G\). It follows from the Lemma 2 that \[\begin{gather} \label{equation324461} R(m) = R_1(m) - \frac{h^2 - h}{2} R_2(m)+ \frac{h!}{8(h - 4)!} R_3(m) \\ + \sum_{i = 4}^{p(h)} (-1)^h h!\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m). \end{gather}\tag{13}\] We write \[\label{equation324464} R(m) = R_1(m) + S_1 + S_2,\tag{14}\] where \[\label{equation324462} S_1 = -\frac{h^2 - h}{2} R_2(m)+ \frac{h!}{8(h - 4)!} R_3(m),\tag{15}\] and \[\label{equation324463} S_2 = \sum_{i = 4}^{p(h)} (-1)^h h!\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m).\tag{16}\] We derive the lower bounds for \(R_1(m), S_1\) and \(S_2\) separately. First we estimate \(R_1(m)\). If \(m \in G\), then using Lemma 5, we can write \[\begin{align} R_1(m) = \sum_{\substack{a_{j_1}, a_{j_2}, \dots, a_{j_h} \in A \\ m = a_{j_1} + a_{j_2} + \cdots + a_{j_h}}} 1 & = \sum_{a_{j_1} \in A} \sum_{a_{j_2} \in A} \dots \sum_{a_{j_h} \in A} \frac{1}{n} \sum_{g \in G} \chi_{-m + a_{j_1} + a_{j_2} + \cdots + a_{j_h}}(g) \\ & = \frac{1}{n} \sum_{g \in G} S_A(g)^h \chi_{-m}(g)\\ & \geq \frac{k^h}{n} - \frac{1}{n} \sum_{g \in G \setminus\{0\}} \left|S_A(g)\right|^h\\ & \geq \frac{k^h}{n} - \left(\max_{\substack{ g \in G \setminus\{0\}}} \left|S_A(g)\right|\right)^{h-2} \left(\frac{1}{n} \sum_{g \in G \setminus\{0\}} \left|S_A(g)\right|^2\right), \end{align}\] and so it follows from Lemma 6 that \[R_1(m) \geq \frac{k^h}{n} - \left(\max_{\substack{ g \in G \setminus\{0\}}} \left|S_A(g)\right|\right)^{h-2} \left( k - \frac{k^2}{n}\right).\] Now it follows from Lemma 4 that \[\label{equation324465} R_1(m) \geq \frac{k^h}{n} - \frac{n^{h - 2}}{3^{h - 2}} \left(k - \frac{k^2}{n}\right) = \frac{k}{3^{h - 2}n} \left(3^{h - 2}k^{h - 1} - n^{h - 1} + n^{h - 2} k\right) .\tag{17}\]
Now we estimate the value of \(S_1\). It follows from Lemma 7 that \[\begin{align} S_1 &= - \frac{h^2 - h}{2} R_2(m) + \frac{h!}{8(h - 4)!} R_3(m) \\ & \geq - \frac{h^2 - h}{2} \left(k (k - 1) \cdots (k - h + 3) + \binom{h - 2}{2} R_3(m)\right) + \frac{h!}{8 (h - 4)!} R_3(m)\\ &= - \frac{h^2 - h}{2}\left(k (k - 1) \cdots (k - h + 3)\right) - \frac{h!}{8 (h - 4)!} R_3(m). \end{align}\] Since \(\alpha_h < \alpha \leq 1\), it follows that \[k \geq \alpha n > \alpha\frac{3^{h-2}(h^2-h)}{2 \left(3^{h - 2} \alpha^{h - 1} + \alpha - 1\right)} \geq \frac{(h^2 - h)}{2} = \frac{h(h - 1)}{2}.\] Therefore, an application of Lemma 8 gives \[S_1 \geq -\frac{h^2 - h}{2} k^{h - 2} + \frac{h!}{4 (h - 4)!} k^{h - 3} - \frac{h!(h - 4)(3 h - 7)}{48 (h - 4)!} k^{h - 4} - \frac{h!}{8 (h - 4)!} k^{h - 3}.\] By simplifying the above expression, we get \[\label{equation324466} S_1 \geq - \frac{h^2 - h}{2} k^{h - 2} + \frac{h!}{8 (h - 4)!} k^{h - 3} - \frac{h! (h - 4)(3 h - 7)}{48 (h - 4)!} k^{h - 4}.\tag{18}\] Note that in 16 , coefficient of \(R_{i_0}(m)\) is positive, and by Lemma 7, we have \(\prod_{r = 1}^{h}(\mu_i (r))! r^{\mu_i (r)} \geq 4\) for \(i \geq 4\) and \(i \neq i_0.\) Therefore, we get \[\begin{align} S_2 = \sum_{i = 4}^{p(h)} (-1)^h h! \prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m) \geq \sum_{\substack{ i = 4 \\ i \neq i_0}}^{p(h)}(-1)^h h!\prod_{r = 1}^{h} \frac{(-1)^{\mu_i (r)}}{(\mu_i (r))! r^{\mu_i (r)}} R_i(m), \end{align}\] and so \[\label{equation324467} S_2 \geq - \frac{h!(p(h) - 4)}{4} k^{h - 4}.\tag{19}\] Now using the inequalities 17 –19 in 14 , we get \[\begin{gather} \label{equation324467a} R(m) \geq \frac{k^h}{n} -\frac{n^{h - 2}}{3^{h - 2}} \left(k -\frac{k^2}{n}\right) - \left(\frac{h^2 - h}{2}\right) k^{h - 2} \\ + \left(\frac{h!}{8 (h - 4)!} \right)k^{h - 3} - \frac{h!(h - 4) (3 h - 7)}{48 (h - 4)!} k^{h - 4} - \frac{(p(h) - 4) h!}{4} k^{h - 4}. \end{gather}\tag{20}\] Since \[k \geq \alpha n\] and \[n > \max \left\{\frac{3^{h - 2} (h^2 - h)}{2 \left(3^{h - 2} \alpha^{h - 1} + \alpha - 1 \right)}, \frac{ 12 (p(h) - 4) ((h - 4)!) + (3 h - 7) (h - 4)}{6 \alpha}\right\},\] it follows that \[k > \max \left\{\alpha \frac{3^{h - 2} (h^2 - h)}{2 \left(3^{h - 2}\alpha^{h - 1} + \alpha - 1 \right)}, \frac{12 (p(h) - 4) (h - 4)! + (3 h - 7) (h - 4)}{6} \right\}.\] Therefore, \[\begin{align} \frac{k}{3^{h - 2} n} \left(3^{h - 2} k^{h - 1} - n^{h - 1} + n^{h - 2} k\right) & \geq \frac{\alpha n}{3^{h - 2} n} \left(3^{h - 2} \alpha^{h - 1} n^{h - 1} - n^{h - 1} + n^{h - 2} \alpha n\right)\\ & \geq \frac{\alpha n^{h - 1}}{3^{h - 2}} \left(3^{h - 2} \alpha^{h - 1} - 1 + \alpha\right)\\ & > n^{h - 2} \left(\frac{h^2 - h}{2}\right)\\ & \geq k^{h - 2}\left(\frac{h^2 - h}{2}\right). \end{align}\] Thus \[\label{equation324468} \frac{k^h}{n} - \frac{n^{h - 2}}{3^{h - 2}} \left(k - \frac{k^2}{n}\right) - \left(\frac{h^2 - h}{2}\right) k^{h - 2} > 0.\tag{21}\] Also, since \[k \geq \alpha n \geq \frac{12 (p(h) - 4) (h - 4)! + (3 h - 7) (h - 4)}{6},\] it follows that \[\label{equation324469} \left(\frac{h!}{8 (h - 4)!} \right) k^{h - 3} - \frac{h! (h - 4) (3 h - 7)}{48 (h - 4)!} k^{h - 4} - \frac{(p(h)- 4) h!}{4} k^{h - 4} \geq 0.\tag{22}\] Now using the inequalities 21 and 22 in 20 , we get \[\begin{align} R(m) > 0, \end{align}\] and so \[h^{\wedge} A = G.\]
Next we prove the properties of \(\alpha_h\). Let \(f_h(x) = 3^{h-2} x^{\,h-1} + x - 1\). The function \(f_h(x)\) is strictly increasing for \(x > 0\). Since \(f_h\left(\frac{1}{3}\right) < 0\) and \(f_h\left(\frac{1}{2}\right) > 0\), it follows that \(f_h(x)\) has a unique positive root \(\alpha_h\) that lies in the interval \(\left(\frac{1}{3}, \frac{1}{2}\right)\). Since \(f_h(\alpha_h) = 0\), we have \(\alpha_h - 1 = -3^{h-2} \alpha_h^{h-1}\). Since \(\alpha_h > \frac{1}{3}\), it follows that \(f_{h+1}(\alpha_h) > 0\). Thus \[f_{h+1}(\alpha_{h+1}) = 0 < f_{h+1}(\alpha_h).\] Since \(f_{h+1}(x)\) is strictly increasing for \(x > 0\), it follows that \(\alpha_{h+1} < \alpha_h\) for \(h \geq 4\). It is easy to show that \[\lim_{h \to \infty} \alpha_h = \frac{1}{3}.\] This completes the proof.
The research of the first named author is supported by the UGC Fellowship (NTA Ref. No.: 231610040283).
\(^{*}\)Corresponding author↩︎