On The Distribution of Consecutive Generalized r-Free Integers in Piatetski-Shapiro Sequences

Angkana Sripayap and Teerapat Srichan^{1} Department of Mathematics Faculty of Science Kasetsart University Bangkok 10900 Thailand

                             fscianr@ku.ac.th
                             fscitrp@ku.ac.th

Pinthira Tangsupphathawat Department of Mathematics Faculty of Science and Technology Phranakhon Rajabhat University Bangkok 10220 Thailand

                          t.pinthira@hotmail.com

Abstract

We use Rieger’s technique to generalize a previous result on the distribution of consecutive r-free integers in Piatetski-Shapiro sequences.

1 Introduction

Let r be a fixed integer ≥ 2. A positive integer n is called r-free whenever it is not divisible by the r-th power of a prime. By convention, 2-free and 3-free integers are called square-free and cube-free, respectively. The Piatetski-Shapiro sequence of parameter c is defined by

ℕ^{c} = {⌊n^{c}⌋}_{n∈ℕ} (c > 1, c ∉ ℕ),

1Corresponding author.

where ⌊z⌋ is the integer part of z ∈ ℝ. The Piatetski-Shapiro sequence was introduced by Piatetski-Shapiro [9] to study prime numbers in a sequence of the form ⌊f(n)⌋, where f(n) is a polynomial. The study of the distribution of square-free and cube-free integers in Piatetski-Shapiro sequences has a long and rich history; see [2, 3, 4, 10, 12, 16]. The distribution of consecutive square-free and cube-free integers in Piatetski-Shapiro sequences is also a topic of interest. In 2018 Dimitrov [5] proved that for every fixed 1 < c < 7/6, there exist infinitely many consecutive square-free integers of the form ⌊n^{c}⌋, ⌊n^{c}⌋ + 1 by showing that

∑_{x/2<n≤x ⌊n^{c}⌋, ⌊n^{c}⌋+1 are square-free} 1 = \frac{1}{2} ∏_{p} (1 − \frac{2}{p^{2}} )x + O(x^{\frac{6c+1}{8} +ε}),  for 1 < c < \frac{7}{6}.  (1)

Very recently, Tangsupphathawat, Srichan, and Laohakosol [15] used Rieger’s technique in [10] to improve the range of c and the error term in Dimitrov’s work in (1). They showed that for 1 < c < 3/2, and sufficiently small ε > 0,

∑_{n≤x ⌊n^{c}⌋, ⌊n^{c}⌋+1 are square-free} 1 = ∏_{p} (1 − \frac{2}{p^{2}} )x + O(x^{\frac{2c+1}{4} +ε}) (x → ∞).  (2)

In the case of cube-free numbers, Zhang and Li [16] proved that, for any ε less than 10^{−10}, one has

∑_{n≤x ⌊n^{c}⌋ is cube-free} 1 = (\frac{1}{ζ(3)} + o(x^{ε})) x for 1 < c < \frac{11}{6}.

In 2018, Dimitrov [6] used the method of Zhang and Li to prove that

∑_{n≤x ⌊n^{c}⌋,⌊n^{c}⌋+1 are cube-free} 1 = ∏_{p} (1 − \frac{2}{p^{3}} )x + O(x^{1−δ^{2}/2}),  for 1 < c < \frac{31}{17},  (3)

where 0 < δ < min\{\frac{31−17c}{9c−9}, 10^{−10}\} is a sufficiently small constant.

From these articles, it is interesting to study similar problems of counting integers in Piatetski-Shapiro sequences which belong to larger classes such as the (k, r)-integers, defined below. In 1966, Subbarao and Harris [13] generalized the notion of r-free integers as follows: let k and r be fixed positive integers with 1 < r < k. A positive integer n is called a (k, r)- integer if n is of the form n = a^{k}b, where a, b ∈ ℕ and b is r-free. They noticed that in the limiting case when k → ∞, a (k, r)-integer becomes an r-free integer.

Recently, the second author [11] studied the distribution of (k, r)-integer, considered as generalized r-free integers, in Piatetski-Shapiro sequences. He proved that for all pairs of exponents (κ_{1}, λ_{1}) and (κ_{2}, λ_{2}) satisfying

rλ_{1} − κ_{1} < 1, rλ_{2} − κ_{2} > 1, kλ_{1} − κ_{1} > 1, \frac{r(λ_{2} − λ_{1}) − (κ_{2} − κ_{1})}{λ_{2}(1 + κ_{1}) − λ_{1}(1 + κ_{2})} > 1,

we have

∑_{n≤N ⌊n^{c}⌋ is a (k, r)-integer} 1 = \frac{ζ(k)}{ζ(r)}N + O(N^{(c/r)+δ(κ_{1},λ_{1},κ_{2},λ_{2})} log N),

for 1 < c < \frac{r(λ_{2}−λ_{1})−(κ_{2}−κ_{1})}{λ_{2}(1+κ_{1})−λ_{1}(1+κ_{2})}, where δ(κ_{1}, λ_{1}, κ_{2}, λ_{2}) = \frac{(λ_{2}κ_{1}−λ_{1}κ_{2})+r^{−1}(κ_{2}−κ_{1})}{λ_{2}(1+κ_{1})−λ_{1}(1+κ_{2})}.

In this work, we use use Rieger’s technique [10] to generalize the previous result on the distribution of consecutive r-free integers in Piatetski-Shapiro sequences, by proving the following result:

Theorem 1. Let k, r be integers with 1 < r < k. For a large N ∈ ℕ, let

T_{c}(N; k, r) := ∑_{n≤N ⌊n^{c}⌋, ⌊n^{c}⌋+1 are (k, r)-integers} 1

denote the number of positive integers n ≤ N such that ⌊n^{c}⌋ and ⌊n^{c}⌋ + 1 are (k, r)-integers. For 1 < c < \frac{3}{2}, we have, as N → ∞,

T_{c}(N; k, 2) = N ∑_{m=1}^{∞} \frac{τ(m)λ_{k,2}(m)}{m} + O(N^{c/2+1/4} log N),  (4)

and for 1 < c < 2, we have, as N → ∞,

T_{c}(N; k, r) = N ∑_{m=1}^{∞} \frac{τ(m)λ_{k,r}(m)}{m} + O(N^{c/3+1/3} log N), if r ≥ 3,

where λ_{k,r}(n) is a multiplicative function defined by

λ_{k,r}(p^{a}) = \begin{cases}1, & if a ≡ 0 (mod k); \\ −1, & if a ≡ r (mod k); \\ 0, & otherwise, \end{cases}  (5)

and τ(n) denotes the number of divisors of n.

As mentioned above, in the limiting case when k → ∞, a (k, 2)-integer becomes a squarefree integer, which leads to the following corollary.

Corollary 2. For 1 < c < \frac{3}{2}, we have, as N → ∞,

T_{c}(N; ∞, 2) = N ∏_{p} (1 − \frac{2}{p^{2}} ) + O(N^{c/2+1/4} log N),  (6)

where the big O-term is independent of k.

Proof. For a fixed large N, every positive integer a, and a positive r-free integer b, the inequality a^{⌊N⌋}b ≤ N holds only when a = 1. This indicates that in the interval [1, N], every (N, r)-integer is an r-free integer. Thus, as N → ∞, the function T_{c}(N; ∞, 2) counts the number of positive integers n ≤ N such that ⌊n^{c}⌋ and ⌊n^{c}⌋ + 1 are square-free integers. In view of (5), we have

∑_{m=1}^{∞} \frac{λ_{N,r}(m)}{m^{s}} = \frac{ζ(Ns)}{ζ(rs)}, s > \frac{1}{r}.

When N → ∞, since ζ(Ns) = 1, we have,

∑_{m=1}^{∞} \frac{λ_{N,2}(m)}{m^{s}} = \frac{1}{ζ(2s)}.

The main term of (6) follows from (4) of Theorem 1 and we get

∑_{m=1}^{∞} \frac{τ(m)λ_{N,2}(m)}{m} = ∑_{\begin{matrix}d, t \\ gcd gcd(d,t)=1 \end{matrix}} \frac{μ(d)μ(t)}{d^{2}t^{2}} = ∏_{p} (1 − \frac{2}{p^{2}} ).

Since the big O-term of (4) is independent to k, Corollary 2 follows.

Similarly, since an (∞, 3)-integer becomes a cube-free integer, we obtain an improved result for (3) in the following corollary.

Corollary 3. For 1 < c < 2, we have, as N → ∞,

T_{c}(N; ∞, 3) = N ∏_{p} (1 − \frac{2}{p^{3}} ) + O(N^{c/3+1/3} log N).

2 Lemmas

We collect now some lemmas needed later.

Lemma 4. [14, Lemma 2.6] Let

q_{k,r}(n) = \begin{cases}1, & if n is a (k, r)-integer; \\ 0, & if n is not a (k, r)-integer, \end{cases}

denote the characteristic function of the set of (k, r)-integers. Then

q_{k,r}(n) = ∑_{a^{k}b^{r}c=n} μ(b).

Let

d(r, k, n) := ∑_{n_{1}^{r}n_{2}^{k}=n} 1

denote the number of ways of writing an integer n in the form n = n_{1}^{r}n_{2}^{k}, and put

D(r, k, x) = ∑_{n≤x} d(r, k, n).

In the proof of our main result, we need the following estimate for the function D(r, k, x) whose proof can be found in [8, Section 14.3].

Lemma 5. For a sufficiently large x ∈ ℝ, we have

D(r, k, x) ≪ x^{1/r}.

Lemma 6. For x ≥ 1, we have

∑_{m≤x} λ_{k,r}(m) ≪ x^{1/r}.

Proof. In view of (5), Lemmas 4 and 5, we have

∑_{m≤x} λ_{k,r}(m) = ∑_{a^{k}b^{r}≤x} μ(b) ≪ ∑_{a^{k}b^{r}≤x} 1 ≪ x^{1/r}.

The proof of Theorem 1 makes use of the following estimate, due originally to Rieger [10], for the number of integers n up to x such that ⌊n^{c}⌋ belongs to an arithmetic progression.

Lemma 7. ([10]) For 1 < c < 2, let x be a positive real number, and let q and a be two integers such that 0 ≤ a < q ≤ x^{c}. Then

∑_{\begin{matrix}n≤x \\ ⌊n^{c}⌋ ≡ a (mod q) \end{matrix}} 1 = \frac{x}{q} + \begin{cases}O(\frac{x^{(c+4)/7}}{q^{1/7}} ), & for q < x^{c−5/4}; \\ O(\frac{x^{(c+1)/3}}{q^{1/3}} ), & for x^{c−5/4} ≤ q < x^{c−1/2}; \\ O(\frac{x^{c}}{q} ), & for x^{c−1/2} ≤ q < x^{c}. \end{cases}

The proofs of the next two lemmas make use of ideas similar to those in Lemma 2.1 of [1].

Lemma 8. Let A_{c}(x; k, r) and B_{c}(x; k, r) denote the number of 6-tuples (d_{1}, t_{1}, d_{2}, t_{2}, u, v) satisfying the conditions

d_{2}^{k}t_{2}^{r}v − d_{1}^{k}t_{1}^{r}u = 1,  d_{1}^{k}t_{1}^{r}u ≤ x^{c}.  (7)

I) If x^{c} < d_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r} ≤ x^{4c/3}, then

A_{c}(x; k, r) ≪ x^{4c/3r} log x.

II) If x^{4c/3} < d_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r} ≤ x^{2c}, then

B_{c}(x; k, r) ≪ x^{2c/3} log x.

Proof. I) For a fixed choice of d_{1}, t_{1}, d_{2} and t_{2} satisfying (7), we have d_{1}^{k}t_{1}^{r}u ≡ −1 ( mod d_{2}^{k}t_{2}^{r}), which fixes the value of u modulo d_{2}^{k}t_{2}^{r}. In view of (7), the total number of possibilities for u is O(1 + x^{c}/d_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r}). By (7), the value of v is fixed for a given choice of d_{1}, t_{1}, d_{2}, t_{2}, u . Then, by Lemma 5, we have

A_{c}(x; k, r) ≪ ∑_{x^{c}<d_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r}≤x^{4c/3}}(1 + \frac{x^{c}}{d_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r}})

≪ ∑_{x^{c}<m≤x^{4c/3}} τ(m)d(k, r; m)(1 + \frac{x^{c}}{m})

≪ ∑_{x^{c}<m≤x^{4c/3}} d(k, r; m)(log m + \frac{x^{c} log m}{m})

≪ x^{4c/3r} log x + x^{c/r} log x ≪ x^{4c/3r} log x.

II) From (7), we have uvd_{1}^{k}t_{1}^{r}d_{2}^{k}t_{2}^{r} ≤ x^{c}(x^{c} +1), whence uv ≤ x^{c}(x^{c} +1)x^{−4c/3} for every 6-tuple counted by B_{c}(x). From a divisor argument, the total number of choices for u, v is therefore bounded by O(x^{2c/3} log x). For ever such choice of u, v, the number of solutions in d_{1}, t_{1}, d_{2}, t_{2} of the equation d_{2}^{k}t_{2}^{r}v − d_{1}^{k}t_{1}^{r}u = 1 is O(log x); see [7].

3 Proof of Theorem 1

From q_{k,r}(n) = ∑_{d|n} λ_{k,r}(d) , we have

T_{c}(N; k, r) = ∑_{n≤N} q_{k,r}(⌊n^{c}⌋)q_{k,r}(⌊n^{c}⌋ + 1)

= ∑_{n≤N} ( ∑_{d|⌊n^{c}⌋} λ_{k,r}(d))( ∑_{t|⌊n^{c}⌋+1} λ_{k,r}(t))

= ( ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} + ∑_{d,t ; {gcd(d,t)=1 \atop dt>N^{c}}} )λ_{k,r}(d)λ_{k,r}(t) ∑_{n≤N ; {⌊n^{c}⌋ ≡ 0 (mod d) \atop ⌊n^{c}⌋+1 ≡ 0 (mod t)}} 1.

In view of Lemma 8, we have

T_{c}(N; k, r) = ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} λ_{k,r}(d)λ_{k,r}(t) ∑_{n≤N ; {⌊n^{c}⌋ ≡ 0 (mod d) \atop ⌊n^{c}⌋+1 ≡ 0 (mod t)}} 1 + ∑_{d,t ; {gcd(d,t)=1 \atop dt>N^{c}}} λ_{k,r}(d)λ_{k,r}(t) ∑_{n≤N ; {⌊n^{c}⌋ ≡ 0 (mod d) \atop ⌊n^{c}⌋+1 ≡ 0 (mod t)}} 1

=: Σ_{1} + O(N^{2c/3} log N).

By the Chinese remainder theorem, there is a positive integer α, unique modulo dt, satisfying the congruence system α ≡ 0 (mod d) and α + 1 ≡ 0 (mod t). Thus,

Σ_{1} = ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} λ_{k,r}(d)λ_{k,r}(t) ∑_{{n≤N \atop ⌊n^{c}⌋≡α (mod dt)}} 1.

In view of Lemma 7, we have

Σ_{1} = N ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} + O(N^{(c+4)/7}| ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c−5/4}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{(dt)^{1/7}} |)

+ O(N^{(c+1)/3}| ∑_{d,t ; {gcd(d,t)=1 \atop N^{c−5/4}<dt≤N^{c−1/2}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{(dt)^{1/3}} |) + O(N^{c}| ∑_{d,t ; {gcd(d,t)=1 \atop N^{c−1/2}<dt≤N^{c}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt}|).

In view of Lemma 6, we note that,

∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c−5/4}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{(dt)^{1/7}} = ∑_{m≤N^{c−5/4}} \frac{τ(m)λ_{k,r}(m)}{m^{1/7}}

≪ ∑_{m≤N^{c−5/4}} \frac{λ_{k,r}(m)}{m^{1/7−ε}}  ≪ \begin{cases}(N^{c−5/4})^{ε+1/r−1/7}, & if r < 7; \\ log N, & if r ≥ 7, \end{cases}

∑_{d,t ; {gcd(d,t)=1 \atop N^{c−5/4}<dt≤N^{c−1/2}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{(dt)^{1/3}} = ∑_{N^{c−5/4}<m≤N^{c−1/2}} \frac{τ(m)λ_{k,r}(m)}{m^{1/3}}

≪ ∑_{N^{c−5/4}<m≤N^{c−1/2}} \frac{λ_{k,r}(m)}{m^{1/3−ε}}  ≪ \begin{cases}(N^{c−1/2})^{ε+1/r−1/3}, & if r = 2; \\ log N, & if r ≥ 3, \end{cases}

and

∑_{d,t ; {gcd(d,t)=1 \atop N^{c−1/2}<dt≤N^{c}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} = ∑_{N^{c−1/2}<m≤N^{c}} \frac{τ(m)λ_{k,r}(m)}{m}

≪ ∑_{N^{c−1/2}<m≤N^{c}} \frac{λ_{k,r}(m)}{m^{1−ε}} ≪ (N^{c−1/2})^{ε+1/r−1}.

Thus,

Σ_{1} = N ∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} + \begin{cases}O(N^{c/2+1/4}) log N, & if r = 2; \\ O(N^{c/3+1/3}) log N, & if r ≥ 3. \end{cases}

Note that

∑_{d,t ; {gcd(d,t)=1 \atop dt≤N^{c}}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} = ( ∑_{{d,t \atop gcd(d,t)=1}} + ∑_{d,t ; {gcd(d,t)=1 \atop dt>N^{c}}} )(\frac{λ_{k,r}(d)λ_{k,r}(t)}{dt})

= ∑_{{d,t \atop gcd(d,t)=1}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} + O( ∑_{m>N^{c}} \frac{τ(m)λ_{k,r}(m)}{m})

= ∑_{{d,t \atop gcd(d,t)=1}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} + O(N^{c/r−c} log N).  (8)

Since \frac{c}{r} − c < 0, the sum on the right hand side of (8) converges when r > 1. Thus,

Σ_{1} = N ∑_{{d,t \atop gcd(d,t)=1}} \frac{λ_{k,r}(d)λ_{k,r}(t)}{dt} + \begin{cases}O(N^{c/2+1/4}) log N, & if r = 2; \\ O(N^{c/3+1/3}) log N, & if r ≥ 3, \end{cases}

which completes the proof of Theorem 1.

4 Acknowledgments

This work was financially supported by Office of the Permanent Secretary, Ministry of Higher Education, Science, Research and Innovation, Grant No. RGNS 63-40.

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2020 Mathematics Subject Classification: Primary 11L07; Secondary 11N37. Keywords: generalized r-free integer, Piatetski-Shapiro sequence.

Received October 27 2021; revised version received January 12 2022; February 2 2022. Published in Journal of Integer Sequences, February 2 2022.

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