k-FREE NUMBERS AND INTEGER PARTS OF αp

In this note, we obtain asymptotic results on integer parts of αp that are free of kth powers of primes, where p is a prime number and α is a positive real number.


Introduction and Statement of Results
Let α and β be real numbers such that α > 0. Let ⌊x⌋ denote the largest integer not greater than x. Sequences of the form {⌊αn + β⌋} ∞ n=1 are called Beatty sequences. A Beatty sequence is said to be homogeneous if β = 0. Beatty sequences have been attracting a lot of attention since they can be viewed as analogues of arithmetic progressions, therefore they show up in a broad context. The interested reader is referred to [1, 2, 4-6, 8-11, 14-16, 19, 24].
Let k ⩾ 2 be an integer. An integer is said to be k-free if it is not divisible by a kth power of a prime. Very recently in [3], an asymptotic formula with an explicit error term is obtained for k-free values of homogeneous Beatty sequences at prime arguments (i.e. sequences of the form {⌊αp⌋} ∞ p=2 ) provided that α is of finite type (see Definition 1). This result can be viewed as a natural analogue of the result of Mirsky [20]. In this article, we pursue this result and obtain two asymptotic formulas that are of the same flavour. The results we present here are well applicable to non-homogeneous Beatty sequences. Theorem 1. Let k ≥ 2 be an integer. Let {α i } ℓ i=1 be a finite type subset of irrational numbers each greater than one. Assume that {α i } ℓ i=1 satisfies (1) for some τ > 0. Let α = (α 1 , α 2 , . . . , α ℓ ) and π(x, k, α) = #{p ⩽ x : ⌊α i p⌋ is k-free for each i = 1, . . . , ℓ}.
A nested version of Theorem 1 is given below.
Theorem 2. Let k ⩾ 2 be an integer. Let {α 1 α 2 , α 2 } be a finite type subset of irrational numbers each greater than zero. Then the following asymptotic is satisfied: for some ε > 0.
Here, the interested reader is invited to investigate the following problem: Let Assuming that a 1 , a 2 , . . . , a n is of finite type (see Definition 1), show that #{p ⩽ x : ⌊a n ⌊a n−1 · · · ⌊a 1 p⌋⌋⌋ is k-free} = π(x) ζ(k) for some ε > 0. It might also be fruitful to investigate the possible power saving in the error term above.
1.1.1. Notation. We recall that for functions F and G where G is real non-negative, the notations F ≪ G and F = O(G) are equivalent to the statement that the inequality |F | ⩽ αG holds for some constant α > 0. Further we use F ∼ G to indicate (F/G)(x) tends to 1 as x → ∞. Given a real number x, we use the notation {x} for the fractional part of x, the notation ⌊x⌋ for the greatest integer not exceeding x and e(x) = e 2πix .
We use ∥x∥ to denote the distance from the real number x to the nearest integer. Λ(n) = log p if n = p r where p is a prime number (here and hereafter). Otherwise, Λ(n) = 0. µ(n) denotes the Mobius function. ϕ(n) denotes the Euler's totient function. τ (n) denotes the number of positive divisors of n. We also use π(x) to denote the number of primes not more than x.
If α is an irrational number of finite type τ , then by Dirichlet's approximation theorem (Lemma 2.1 of [25]) one has τ ⩾ 1. The celebrated theorems of Khinchin [17] and of Roth [21,22] state that τ = 1 for almost all (in the sense of the Lebesque measure) real numbers and for all irrational algebraic numbers respectively. Definition 2. A finite subset of real numbers {β 1 , β 2 , . . . , β ℓ } is said to be of finite type if there is τ > 0 such that the inequality has only finitely many solutions for h i ∈ Z.
If {β i } ℓ i=1 satisfies (1) for some τ > 0, then it follows from Dirichlet's theorem on rational approximations that τ ⩾ 1. Furthermore, such a set is linearly independent over Q.
Throughout this paper, we shall mostly use the weak form of the prime number theorem, that is Lemma 1. For every positive integer n ≥ 1, τ (n) < e C log 5n log log 5n for some constant C > 0.
Proof. Follows from [23,Theorem 2.11]. □ Lemma 2. If α − a q ⩽ 1 q 2 for some integers a and q such that (a, q) = 1, then Proof. This follows in a standard way using the main result of [12, §25]. □ is a finite sequence in R ℓ , then for any J ⊆ [0, 1) ℓ that is a Cartesian product of subintervals of [0, 1) and any H ⩾ 1, we have
Proof. For the proof see [18]. □ The following lemma is a classical result due to Vinogradov [26,Lemma 12].
Lemma 4. Let α, β and ∆ be real numbers such that Then there exists a periodic function Ψ(x), with period 1, satisfying has a Fourier expansion of the form for every |h| ⩾ 1 and some c fixed. Furthermore, a 0 = β − α.
Let I k denote the characteristic function of k-free integers. Since we have where z ⩽ x 1/k will be chosen later. It follows from Lemma 1 that for all i = 1, 2, . . . , ℓ there exists a positive constant Then, for all i = 1, 2, . . . , ℓ and p ≤ x where c = max{c 1 , . . . , c ℓ }. Set C = c(ℓ − 1). Then, by (4) and using partial summation in the last step, we get Therefore, Next, we will study the sum above appearing in (5)  and for all i = 1, . . . , ℓ. For a positive integer i, let p i denote the ith prime. Observing that k-FREE NUMBERS AND INTEGER PARTS OF αp where It follows from Erdős-Turán-Koksma Inequality that for all H ⩾ 1, Next, we shall prove the following lemma.
Proof. Since {α i } ℓ i=1 satisfies (1) for some τ > 0, there exists a positive constant A ≥ 1 such that (max{|h 1 |, . . . , |h ℓ |}) −τ ⩽ A ||h 1 α 1 + h 2 α 2 + · · · + h ℓ α ℓ || for all (h 1 , . . . , h ℓ ) ∈ Z ℓ such that max 1⩽i⩽ℓ {|h i |} > 0. Let h = (h 1 , . . . , h ℓ ) ∈ Z ℓ be such a tuple and set Let 1 ⩽ Q < x/2 to be determined later. By Dirichlet's rational approximation theorem, there exists r q ∈ Q such that 1 ⩽ q ⩽ x Q and On the other hand, it follows from (10) that Combining (11) and (12), we get Then it follows from Lemma 2 that where for the sake of brevity we set M = max{|h 1 D 1 |, . . . , |h ℓ D ℓ |}. By [13, Lemma 2.4], there exists 1 ⩽ Q < x/2 such that the left hand side of (14) is At this point, we can assume that x − 1 2τ M 1 2 D 1 2τ < 1, because otherwise the required upper bound holds trivially. Therefore, the second term is beaten by the first term giving the proof of Lemma 5. □ We next proceed by plugging this upper bound into (9). We also use the upper bound |h i | ⩽ H together with the upper bounds D ⩽ z kℓ and D i ⩽ z k(ℓ−1) . Then the difference in the first line of (9) is where in the last step we use integral test. Here we note that the implied constant depends on ℓ. Coupling (8), (9), (15) and (16), we arrive at for every H ⩾ 1 and every (d 1 , . . . , d ℓ ) such that d i ⩽ z ⩽ x 1/k for each i. Noting π(x) ≪ x and choosing 1 ⩽ H ⩽ x by [13,Lemma 2.4], the left hand side of (17) is On summing this over all tuples (d 1 , . . . , d ℓ ) of positive integers where d i ⩽ z for all i = 1, . . . , ℓ, we observe from (5) that for all 1 ⩽ z ⩽ x 1/k , and using the following inequality it follows by the mean value theorem that Therefore, the contribution of the sums running over d i ⩽ z for all i = 1, . . . , ℓ is

S. Ç AM Ç ELIK
where C = C(ℓ, α) is positive. On the right hand side of (18), the first term beats the third term as τ ≥ 1 and the second term whenever z ⩽ x 1 k(ℓ−1)τ +kℓ which one can assume since otherwise (18) holds trivially. Using now [13,Lemma 2.4] to choose optimal z ⩽ x 1/k , the left hand side of (18) is for some constant C ′ depending on ℓ and α, therefore the claim follows.
Let 1 ⩽ z ⩽ x 1/k be a number to be determined. Using (3), it follows that As we did before, we have where the implied constant depends only on α 1 and α 2 . This yields We now proceed to derive the main term. Writing and using partial summation to get for any 1 ⩽ z ⩽ x 1/k . Let us now concentrate on the error term and proceed to show that it is ≪ x 1−ε for some ε > 0. Using observation (7), together with Lemma 3 one ends up with where H 1 is a positive number to be determined. So, it boils down to estimate the exponential sum above. To do this, we let K be a sufficiently large number and we write where uniformly for all t ∈ R, we have . Therefore, the left hand side of (21) is Given 0 ⩽ i ⩽ K − 1, let β i = i/K, γ i = (i + 1)/K and 0 < ∆ < 1/K be a number to be chosen. By Lemma 4, there exists a periodic function Ψ i (x), with period 1, has a Fourier expansion of the form where a 0 = 1/K and for every |h| ⩾ 1 and some c fixed. Let ψ i (x) be 1 if β i ⩽ {x} ⩽ γ i and ψ i (x) = 0 otherwise. It follows that Ψ i (x) and ψ i (x) agree on [0, 1] except possibly for two subintervals of [0, 1] of length ⩽ ∆. Therefore, where I is a union of two intervals and is of length ∆. Since α 2 is of finite type, following the proof of Theorem 5.1 in [8] together with a partial summation argument, it follows that for some 0 < ε ′′ < 1/5, one has uniformly for all 0 < ∆ < 1/K. Therefore, we see that the left hand side of (23) is Letting H 2 be a positive integer to be determined, we split the sum running over h 2 at H 2 . For |h 2 | > H 2 , estimating the innermost exponential sum by π(x), and using the upper bounds a h ≪ 1/(∆h 2 ) and a h ≪ 1/|h|, we obtain that the left hand side of (23) is We are therefore left with the estimation of whenever max{|h 1 |, |h 2 |} > 0. To estimate the exponential sum, by Dirichlet's theorem we pick up a rational number a/q satisfying where 0 < κ < 1 is to be determined. Since {α 1 α 2 , α 2 } is of finite type, similar to how we obtain (13) x 1−κ τ d k τ max{|h 1 |, |h 2 d k |} ≪ q ⩽ x 1−κ for some τ ⩾ 1. Then by Lemma 2, the exponential sum (26) is At this point, we assume that 0 < max{|h 1 |, |h 2 |} ⩽ x ε ′ where ε ′ is a sufficiently small number to be determined in terms of κ. Then, uniformly for 0 < max{|h 1 |, |h 2 |} ⩽ x ε ′ .