Abstract
We obtain an asymptotic formula for the cube-full numbers in an arithmetic progression , where . By extending the construction derived from Dirichlet’s hyperbola method and relying on Kloosterman-type exponential sum method, we improve the very recent error term with .
1. Introduction and Main Results
Let be a fixed integer and be a positive integer. We call a powerful number (or -full number) if or for a prime dividing , also divides . Let denote the set of powerful numbers. Suppose , and this defines square-full numbers and cube-full numbers, respectively. Erds and Szerkeres [1] first introduced powerful numbers and gavewhere are effective constants and . From then on, many authors have studied the powerful numbers and got a lot of relevant conclusions (see [2–18] and references therein).
In 2013, Liu and Zhang [19] investigated the distribution of square-full numbers in arithmetic progressions and got an asymptotic formulaunder the condition of . By utilizing the method of exponent pairs, Srichan [20] then obtainedwhere the error terms had been corrected by Watt [MR3265055].
Recently, Chan [21] got a new asymptotic formulawhich improved his own result with Tsang [22]. As a critical step, he [21] mainly dealt with a sum in the form of by following closely Montgomery and Vaughan’s construction [23]. It is somewhat similar to Dirichlet’s hyperbola method shown in Figure 1.
Actually, they divided the above sum into four parts as shown in Figure 2 and then discussed them separately.
Motivated by this idea, we turn to discuss the following sum with three parameters:
By extending the construction from Montgomery and Vaughan [23], the summation (5) is divided into eight parts as shown in Figure 3.
Then, relying again on Kloosterman-type exponential sum method, an asymptotic formula of (5) is obtained. Finally, an asymptotic formula of cube-full numbers in an arithmetic progression is derived.
Before we formulate our result we need to give some definitions that will be used below. For modulus , we define
It is clear that , , and for any , and if .
Theorem 1. For ,wherein which , , and are defined below.
Note that compared with the result in [20], we improve the error term when .
The key in our proof of Theorem 1 is the following.
Theorem 2. Let , and be two parameters such thatThen, we havewhere
2. Some Lemmas
Before we start the proof, let us give a few lemmas which are needed later.
Lemma 1. For ,
Proof. The first result can be found in Lemma 3 of [21]. Note thatThe second and third one can be proved in the same way. The proofs of the last three are slightly different. For example, by orthogonal property of additive characters, we havewhere is the set of all characters such that , the principal character.
Lemma 2. For and ,
Proof. Let be the distance from to the nearest integer; then, we have
Lemma 3. For ,
Proof. Using the trivial estimation of the innermost sum, we haveThen, by orthogonal property of additive characters, we obtainInterchanging the order of summations and combining Lemma 2 and Eq. (12.48) on page 324 of [24], we haveFinally, we getBy definition of , (see Lemma 2.2 in [25]).
Lemma 4. If we definethen we have
Proof. Following much the same way as Lemma 4 of [21], we first suppose with . By the “reciprocity” formula where and , and the additive multiplicity of exponential functionwith , , , , and , it follows thatNow we just need to discuss the argument in the following cases:(I)Prime moduli case. Now Theorem 2 obtained by Moreno and Moreno [26], which is a special form of the Bombieri-Weil bound [27], implies provided that is not the form of with , where is the algebraic closure of . Let with and . Then, we have derived from This is impossible if , by comparing the degrees of both sides of the above. If , the validity of (26) can be easily checked.(II)Prime power moduli case with . Obviously, we only need to consider it with the assumption . Following the proofs of Lemma 12.2 and 12.3 in [24] with the equation of , we obtain where with .Note that and . Now we concentrate on the number of solutions of congruence equation with .
with . Then, the congruence equation isIf , then . Relying on the properties of indices, we deduce that (32) has no solution when and one solution when . Next, we assume . If and , then (32) has at most seven solutions. And if and , (32) also has at most seven solutions. Then, we havewhere with . Firstly, in the same way, if , by the analysis in the case , the sum in (30) is empty unless in which case one has . Now suppose ; according to Chapter 3 in [24], we know if , then we have . Therefore, if (for otherwise and imply , a contradiction), we have ; hence, as there are at most seven solutions to (32). If or 7, , and hence .
In any case, we haveCombining (25), (26), (33), and (34), we finally obtainwhich completes the proof of Lemma 4.
By applying Lemma 4, we have the following.
Lemma 5. For ,
Proof. Note that if , then and . So,The remaining part of the proof is similar to Theorem 4 in [21].
3. Proof of Theorem 2
Consider three positive parameters , , and . By extending the construction from Montgomery and Vaughan [23] as shown in Figure 3, we have
First, we estimate .
For , we haveby Lemma 3.
Then, we estimate and . For , we know
The first term in the above formula is
In order to simplify our final result, by using Euler’s summation formula which can be found in Theorem 3.2 in [28], the constant can be rewritten as
Thus, we obtain the asymptotic formula of as
Next, we deal with .
If we let , then the first term in the above formula is
And the second term is
So, we can get
And for , following closely Chan [21] as shown in Figure 2 instead of Dirichlet’s hyperbola method shown in Figure 1, we just need to divide the interval in the same way. Note that the sumcan be estimated with the help of asymptotic formula given at the end of page 101 in [21]; then, using Lemma 5, we can getwhere satisfies .
Picking to have the same size as and combining the previous results, we have
For , in the same way as , we have
And we can obtain
For , again in the same way as the proof of , we have
If we let , then the first term in the above formula isand the second term is
Then, we can obtain
If we pick to have the same size as , we can getwhere satisfies .
Similar to the proofs of and , we can get the following:
Finally, we discuss as follows: dividing interval into parts with the length of each one being . The th part is
Let be the intersection of the plane , , and curved surface , which isand then we obtain
Similarly, we have
We define the intersection of and as
Now we divide the interval in the th part and follow the construction in [23]; firstly, we have rectangles
In the remaining regionswe place additional rectangles . If we let , in the same way we place a further rectangles and so on, then we can get which is
The remaining regions are
In this part, expanding and to a cube with height , we can get , , , , and the remaining region correspondingly. Then, we have the asymptotic formula
If we let by Lemma 5, we know
Then, we further obtainin which the main term is
Now we deal with the error term of . Note that
Thus, the area of is at mostwhich implies that the estimation of the first error term is
Similarly, we have the same error bound for the second, third, and the fourth error term. The error terms in the fifth are
For the last error term, we extend into a ring sector which volume iswhere
Then, we can add and get
Therefore, we have
Now we come to simplify the above error terms. Firstly, we have
When and , we know that
If and , then
If we pick and , we getwith the condition .
Therefore, if we choose , and , then we get all error terms from to :where , , and satisfy the conditions
Further if we choose , then we obtainwhere , and . Continue to simplify the error term and getwhere and or . If , we can choose ; otherwise, we can choose .
Finally, we havewhere
4. Proof of Theorem 1
First recall that any cube-full number can be written uniquely as where , are square-free numbers and . Following the proof of Theorem 2.3 in [20], we have
In view of the identity , we get
From the identityand for , otherwise and , we get
Using again andwe have
From , we know that if , then we can pick of size , and if , we simply pick . Then, we obtain
From the result of Theorem 2, we have
For the final term of the -term, we have
In the same way, we have
The first two terms of the error term are
If we takeand assume , then we have and
Now we pick and get
Consequently, we obtainwhich is Theorem 2.
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research was funded by the National Natural Science Foundation of China (nos. 11871317, 11926325, and 11926321).