An Improved Asymptotic on the Representations of Integers as Sums of Products

In this paper, we study the number of representations of a natural number N as a sum of k terms, each being a product of l factors. Let ](N; k, l) denote this number. *is problem was studied by Estermann [1, 2] in the case k � 2 or 3 and l � 2 by using some properties of Dirichlet L-function. His method is not easy to be generalized. Later, Chace [3] generalized Estermann’s result to k≥ 3 and l≥ 2. For such k and l, he got


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(3) Here, μ(N; k, l) is defined in (72). e way he studied the problem is different from Estermann. e main tools he used are the Hardy-Littlewood method and some results from the divisor problem in arithmetic progressions. He got the main term μ(N; k, l) which is a sum of terms of the form SI, where S are the "singular series" and I are the "singular integrals." ey occur in the applications of the Hardy-Littlewood method. In this paper, we improve Chace's result in the cases k ≥ 3 and l � 2, 3. We get the following result. Theorem 1. Suppose k ≥ 3 and l � 2, 3. en, where μ(N; k, l) defined by (72) satisfies (2) and the error term satisfies We can compare the exponents in (3) with our results (5). For l � 2, 3, we see that 2 − 2/(2l ) when k ≥ 4. Our error terms are better than Chace's. e proof of the theorem is an application of the Hardy-Littlewood method (cf, Chapter 3 of [4]), the Voronoi summation formula, and some results from the Kloosterman sums. e estimates on the minor arc were studied by Chace, and his result is sufficient for us. Hence, we will not focus on the minor arc in this paper. e main difficulty arises in treating the error term of the major arcs. In Section 2, we make some preparations for our proof. We state the Voronoi summation formula and give some lemmas related to the Kloosterman sums in this section. In Sections 3.1 and 3.2, we obtain a bound for the contribution from the minor arcs. In I3.3, we prove our theorem.

Lemma 1 (Voronoi's summation formula). With the above notations, we have, for
To simplify the integrals in (15), we define where σ > 0. By (12), it is clear that To use Lemma 1 in practice, we need to estimate the residue in (15), G ± l (x) and A ± l (n, (h/q)). We first compute the residue. For 0 ≤ j ≤ l − 1, we define where a > 0 is an integer, and the coefficients c j (b, q) are sums of terms of the form for some function f. e coefficients c j (b, q) are given explicitly in equation (2.13) in [6]. It has been shown in [6] that Proof. By (9), we can rewrite E l (s, (a/q)) as where for Rs > 1. erefore, Chace ( eorem 1 and (2.1) in [6]) gave that where Substituting this into (23), we have e proof of the lemma is complete. Our next lemma, proved by Jiang and Lü (Lemma 2.7 in [7]), is to evaluate the integrals in (15). □ Lemma 3. Let G ± l (x) be defined as in (16). en, we have, for l ≥ 2, Proof. is can be proved by taking J � X 1− δ with 0 < δ < 1 in Lemma 2.7 of [7]. e following two lemmas give estimates for A ± l (n, (a/q)). is two lemmas play an important role in our proof. We use some results for the Kloosterman sum to obtain the power saving in the q aspect. □ Lemma 4. Suppose 1 ≤ a ≤ q, (a, q) � 1, and let A ± l (n, (a/q)) be defined as in (11). We have Proof. By (13), we obtain where for m � (m 1 , . . . , m l ) ∈ Z l . Here, the notation d|m means that d divides each component of m. en, we obtain that the left hand side of the desired equation in our lemma equals to (31) Now, we define for m � (m 1 , . . . , m l ) ∈ Z l and w ∈ Z. Here, x denotes the multiplicative inverse of x modulo w. In Section 6 in [8], Smith gave that where e[t] � (1, . . . , 1, t) ∈ Z l for all t ∈ Z. erefore, (31) can be written as Taking the change of variables that m � q 1 b, we get From eorem 6 in [9], we know that e lemma follows immediately. □ Lemma 5. Suppose 1 ≤ a ≤ q, (a, q) � 1, and l � 2, 3. Let A ± l (n, (a/q)) be defined as before. en, we have Proof. By (13), for l � 2, we have I n 1 , n 2 , a; q � For l � 3, one has I n 1 , n 2 , n 3 , a; q � q d| n 2 ,n 3 ,q ( ) Now, if we use Weil's classical bound then it follows from (39) that I n 1 , n 2 , n 3 , a; q ≪ ε q (3/2)+ε d| n 2 ,n 3 ,q ( ) For d|(n 2 , n 3 , q) and (a, q) � 1, noting that n � n 1 n 2 n 3 by (11), we have (n 1 d, n 2 n 3 a, q) ≤ (n 1 n 2 n 3 a, q) � (n, q). erefore, I n 1 , n 2 , n 3 , a; q ≪ ε q (3/2)+ε (n, q) (1/2) .
Hence, we have for l � 2, 3. e proof of the lemma is complete.

Proof of Theorem 1
In this section, we prove eorem 1. e main difficulty arises in treating the major arcs. e results from Kloosterman's sums will play a role in our proof. roughout this section, 1 ≤ X ≤ (N/2). We choose a smooth function f compactly supported in [X − X δ , 2X + X δ ] with 0 < δ < 1, satisfying f(x) � 1 in [X, 2X] and (6) and (7). To apply the circle method, we choose the parameters P and Q such that Journal of Mathematics PQ � N, By Dirichlet's lemma on rational approximations, each α ∈ I ≔ [(1/Q), 1 + (1/Q)] may be written in the form as follows: for some integers a and q with 1 ≤ a ≤ q ≤ Q and (a, q) � 1. We denote by M(q, a) the set of α satisfying (45) and define the major arc by M(a, q).
e upper bound of the integral of S k l (α) on m is given by Chace [3]. Lemma 6. Suppose k ≥ 3 and l ≥ 2 are integers. en, is estimate on the minor arc is sufficient for us. It remains to consider the integral of S l (α) on the major arc. Let Now, for α � a/q + β ∈ M, we have where f β (x) ≔ f(x)e(xβ) and It is clear that f β (x) is a smooth function compactly supported in [X − X δ , 2X + X δ ]. Moreover, it satisfies for any v ≥ 0, and for any v ≥ 1. Now, by Lemma 1, we get where is the Mellin transformation of f β (x) and It follows from Lemma 2 that From Equation (4.8) in [3], we know that where where μ(N; k, l) � We complete the proof of eorem 1.

Data Availability
e data used to support the findings of this study are included within the article.

Conflicts of Interest
e author declares that there are no conflicts of interest.