Visible lattice points along curves

This paper concerns the number of lattice points in the plane which are visible along certain curves to all elements in some set S of lattice points simultaneously. By proposing the concept of level of visibility, we are able to analyze more carefully about both the “visible” points and the “invisible” points in the definition of previous research. We prove asymptotic formulas for the number of lattice points in different levels of visibility.

In 2018, Goins, Harris, Kubik and Mbirika [7] considered the integer lattice points in the plane which are visible to the origin (0, 0) along curves y = rx k with k ∈ N fixed and some r ∈ Q. They showed that the proportion of such integer lattice points is 1/ζ(k + 1). In the same year, Harris and Omar [8] futher considered the case of rational exponent k. Recently, Benedetti, Estupiñán and Harris [6] studied the proportion of visible lattice points to the origin along such curves in higher dimensional space.
All the above results are concerned about the lattice points visible to only one base point. It is natural to consider the distribution of lattice points which are visible to more base points simultaneously. For the case of visibility along straight lines, the earliest work originates from Rearick in 1960s. In his Ph.D. thesis, Rearick [13] first showed that the density of integer lattice points in the plane which are jointly visible along straight lines to N (N = 2 or 3) base points is p (1 − N/p 2 ), where the base points are mutually visible in pairs and the product is over all the primes. Then in [14], he generalized this result to lattice points in higher dimensional space and larger N .
The joint visibility of lattice points along curves has not been considered yet. In this paper, we focus on this topic and we also propose the concept of level of visibility. Level-1 visibility matches the definition of "k-visible" in [7]. We use higher level of visibility to analyze more carefully about the "invisible" points along certain curves. We give asymptotic formulas for the number of lattice points which are visible to a set of N base points along certain curves in different levels of visibility.

1.2.
Our results. For any positive integer k and integer lattice points (u, v), (m, n) ∈ N×N, let r ∈ Q be given by n − v = r(m − u) k and C be the curve y − v = r(x − u) k . If there is no integer lattice points lying on the segment of C between points (m, n) and (u, v), we say (m, n) is (Level-1) k-visible to (u, v). Further, if there is at most one integer lattice points lying on the segment of C between points (m, n) and (u, v), we say point (m, n) is Level-2 k-visible to (u, v).
One can see that k-visibility is mutual. Precisely, if a point (m, n) is Level-1 or Level-2 k-visible to the point (u, v) along the curve y − v = r(x − u) k , then (u, v) is also Level-1 or Level-2 k-visible to (m, n), respectively, along the curve y − n = (−1) k+1 r(x − m) k .
Throughout this paper, we always assume S is a given set of integer lattice points in the plane. We say an integer lattice point (m, n) is Level-1 k-visible to S if it belongs to the set is Level-2 k-visible to every point in S}. One may define higher Level k-visible points to S this way. But in this paper, we focus on Level-1 and Level-2 k-visible points.
For x ≥ 2, we consider visible lattice points along curves in the square [1, x]× [1, x]. Denote An important case is that the points of S are pairwise k-visible to each other. The cardinality of such S can't be too large. In fact, we have #S ≤ 2 k+1 by Proposition 2.1 in the next section. For such type of S, we obtain the following asymptotic formulas for N 1 k (S, x) and N 2 k (S, x). Theorem 1.1. Assume the elements of S are pairwise k-visible to each other and N = #S < 2 k+1 . For any k ≥ 2, we have where p runs over all primes, and Remark 1. If N = #S = 2 k+1 , by Proposition 2.1, there is no lattice point outside S which is (Level-1) k-visible to all elements of S.
By the above Theorem, the density of (Level-1) k-visible points to every elements of S is p 1 − N/p k+1 . For k = 1, it is done by the work in [14], where the author studied the visible points along straight lines. The special case N = 1 for k ≥ 2 in Theorem 1.1 covers the result in [7], where only the main term was given.
We also give asymptotic formulas for Level-2 k-visible points. Note that such set actually includes some "invisible" points in the definition of previous research. We are able to analyze more carefully about these "invisible" points. Theorem 1.2. Assume the elements of S are pairwise k-visible to each other and N = #S ≤ 2 k+1 . For any k ≥ 1, we have But there are still positive proportion of lattice points in the plane which are Level-2 k-visible to S.
For the special case N = 1 and k = 1, our problem is the same as the so-called "primitive lattice problem" inside a square. Nowak [12], Zhai [17] and Wu [16] have studied the number of primitive lattice points inside a circle. Primitive lattice points in general planar domains have also been studied by Hensley [9], Huxley and Nowak [10] and Baker [2] etc. Assuming the Riemann hypothesis(RH), they continuously improved the error term of the concerned asymptotic formulas by estimating certain exponential sums. One may wonder how much we can do to improve the estimates of E 1 (x) and E 2 (x) by similar argument under RH. However, we do not focus on pursuing the best possible error term in this paper.
Taking S = {(0, 0), (1, 1)}, we did numerical calculations for densities of Level-1 and Level-2 k-visible points for x = 10000 and k = 2, 3, . . . , 9 (See Table 1 Table 1. Densities of k-visible points to set of two elements We also calculate the case when S = {(0, 0), (1, 2), (2, 1)}, and we get the following data for densities of Level-1 and Level-2 k-visible points to S (See Table 2 and Figure 2).  Table 2. Densities of k-visible points to set of three elements Notations. We use Z to denote the set of integers; N to denote the set of positive integers; Q to denote the set of rational numbers; #S to denote the cardinality of a set S. As usual, we use the expressions f = O(g) or f g to mean |f | ≤ Cg for some constant C > 0. In the case when this constant C > 0 may depend on some parameters ρ, we write f = O ρ (g) or f ρ g.

Preliminaries
We define the degree-k greatest common divisor of m, n ∈ Z as gcd k (m, n) := max{d ∈ N : d | m, d k | n}. The size of the image S is at most 2 k+1 . If S has more than 2 k+1 points, there must be two distinct elements which map to the same element in S, say and hence gcd k (u 2 − u 1 , v 2 − v 1 ) ≥ 2, which contradicts our assumption on S.
By the definition of k-visible points and elementary argument, we get the following lemma. One may refer to [7] (Proposition 3) for similar argument. Here we omit the proof. We also need the following well-known result for l-fold divisor function τ l (n) = d 1 ···d l =n 1.

Proof of Theorem 1.1
Given a set S, if we shift S such that it contains the origin, the error occurs to our counting function is O S (x). Thus, we may assume (0, 0) ∈ S. Denote the elements of S as (u j , v j ), 0 ≤ j ≤ N − 1 with (u 0 , v 0 ) = (0, 0). By Proposition 2.1, the contribution of points (m, n) with m = u j or n = v j for some j, j is O(|S|x) = O k (x). Hence, we only need to estimate the contribution of points (m, n) with m = u j and n = v j for all 0 ≤ j, j ≤ N − 1. Throughout all our proofs, we implicitly assume the input of gcd k ( * , * ) has no zero coordinates unless otherwise specified.

By Lemma 2.2 we have
Applying the formula d|n µ(d) = 1, if n = 1; 0, otherwise, (3.3) where µ is the Möbius function, we write Let D > 0 be a parameter to be chosen later. Divide the sum over d 0 , · · · , d N −1 into two parts: d 0 · · · d N −1 ≤ D and d 0 · · · d N −1 > D, and denote their contributions to N 1 k (S, x) by ≤ and > respectively. Then we have For ≤ , we change the order of the summation and obtain Note that (u 0 , v 0 ) = (0, 0), then for any given d 0 , · · · , d N −1 , the inner sum over m, n in the above formula actually equals Since the points in S are mutually k-visible, then by Lemma 2.2 we have This implies gcd(d j , d l ) = 1 for 0 ≤ j = l ≤ N − 1.
It then follows that

Then by Lemma 2.3 we obtain
Writing n = d 0 · · · d N −1 , we then have

Using Lemma 2.3, we obtain
Thus the second sum > in (3.5) is empty since we already exclude zero inputs of gcd k ( * , * ) in the beginning of the proof. Inserting (3.8) into (3.5) yields (1.1) in Theorem 1.1 with ii) If k < N < 2 k+1 , we need to make another choice for D, and deal with > more carefully. Taking absolute value of µ(d), we obtain Using the bounds τ (n) ε n ε for any ε > 0, N < 2 k+1 and gcd k (m − u j , n − v j ) ≤ x 1/k , we have By the definition of gcd k , we have > k,ε x ε It follows that > k,ε x 2+ε Collecting all the above gives
By the above argument, we write Without loss of generality, we may assume (u 0 , v 0 ) = (0, 0). We only need to estimate the inner sum of the second term in (4.10) for l = 0, other cases are similar. Denote 1≤j≤N −1

1.
We have By changing the order of summation and making the substitutions m = 2d 0 s and n = (2d 0 ) k t, we obtain In order to get estimates of I(x), we need to analyze the conditions in the inner sum. Fix d 0 , · · · , d N −1 , in order for those congruence equations having solutions, we need Since points (u j , v j ) are k-visible to point (u 0 , v 0 ), then Lemma 2.2 gives gcd k (u j , v j ) = 1. It follows that gcd(2d 0 , d j ) = 1 for 1 ≤ j ≤ N − 1. Moreover, in order for those congruence equations having solutions, we also need the following equations By the assumption of pairwise k-visibility of elements of S, we have gcd k (u j 2 − u j 1 , v j 2 − v j 1 ) = 1, and thus gcd(d j 1 , d j 2 ) = 1 for any 1 ≤ j 1 = j 2 ≤ N − 1.
As what we did in Section 3, we divide the sum over d 0 , · · · , d N −1 into two parts according to d 0 · · · d N −1 ≤ D or not. Denote them by I ≤ and I > respectively, then For I ≤ , we have and by Lemma 2.3, we get Making the substitution n = d 0 · · · d N −1 , we obtain where h(n) = n=d 0 ···d N −1 d 1 ,··· ,d N −1 odd 1.
Extending the sum over n and using the bound h(n) ≤ τ N (n), and by Lemma 2.3, we derive Note that h(n) is multiplicative with h(2) = 1 and h(p) = N for p > 2 prime. Thus i) If 1 ≤ N ≤ k, then we choose D = x. In this case, since each d j ≤ x 1/k , d 0 · · · d N −1 ≤ x N/k ≤ x. Thus the second term I > in (4.13) is empty. Inserting (4.14) into (4.13) yields (1.2) in Theorem 1.2 with ii) If k < N ≤ 2 k+1 , we need to make another choice for D and deal with I > . By similar argument as before, we obtain Using the bound τ (n) ε n ε for any ε > 0, by a similar argument as in the proof of Theorem 1.1, we obtain I > ε x ε m,n≤x Plugging this into (4.10), we obtain (1.2) in Theorem 1.2 with E 2 (x) k,ε x 2− 2k N +k +ε for k < N ≤ 2 k+1 .