On the distribution of consecutive square-free numbers of the form $\mathbf{[\alpha n], [\alpha n]+1}$

TL;DR

This paper proves the existence of infinitely many consecutive square-free numbers of specific forms involving irrational numbers with certain properties, expanding understanding of their distribution.

Contribution

It establishes the infinite occurrence of consecutive square-free numbers of the form [αn], [αn]+1 for irrational α with bounded partial quotients or algebraic irrationals.

Findings

Infinitely many such pairs exist for specified irrational α.

The result applies to α with bounded partial quotients or algebraic irrationals.

The proof involves advanced number theory techniques.

Abstract

In the present paper we show that there exist infinitely many consecutive square-free numbers of the form $[\alpha n]$, $[\alpha n]+1$, where $\alpha>1$ is irrational number with bounded partial quotient or irrational algebraic number.

Equations

$$\sigma =\prod _p\left(1-\frac{2}{p^2}\right).$$

$${\omega }_{\alpha }(x)=\{\begin{array}{l}1,\text{mbox}if1-\frac{1}{\alpha }<x<1;\\ \frac{1}{2},\text{mbox}ifx=1-\frac{1}{\alpha }\text{mbox}orx=1;\\ 0,\text{mbox}otherwise;\end{array}$$

$$\sum\limits_{n\leq N}\mu^{2}(n)\mu^{2}(n+1)=\sigma N+\mathcal{O}\big{(}N^{2/3+\varepsilon}\big{)}\,,$$

$$\sum _{n\le N}{\mu }^2(n){\mu }^2(n+1)=\sigma N+\mathcal{O}\left(N^{2/3}(\log N{)}^{4/3}\right).$$

$$\sum\limits_{n\leq N}\mu^{2}(n)\mu^{2}(n+1)=\sigma N+\mathcal{O}\big{(}N^{7/11}(\log N)^{7}\big{)}\,.$$

$$\sum\limits_{n\leq N}\mu^{2}(n)\mu^{2}(n+1)=\sigma N+\mathcal{O}\big{(}N^{(26+\sqrt{433})/81+\varepsilon}\big{)}$$

$$\sum _{n\le N}{\mu }^2([\alpha n])=\frac{6}{{\pi }^2}N+\mathcal{O}\left(\frac{N\log \log N}{\log N}\right)$$

$$\sum _{n\le N}{\mu }^2([\alpha n])=\frac{6}{{\pi }^2}N+\mathcal{O}\left(N^{\frac{2}{3}+\epsilon }\right)$$

$$\sum _{n\le N}{\mu }^2([\alpha n])=\frac{6}{{\pi }^2}N+\mathcal{O}\left(A{N}^{\frac{5}{6}}{\log }^{5}N\right)$$

$$S(N,\alpha )=\sum _{n\le N}{\mu }^2([\alpha n]){\mu }^2([\alpha n]+1).$$

$$S(N,\alpha )=\sigma N+\mathcal{O}\left(N^{\frac{5}{6}+\epsilon }\right)$$

$${\omega }_{\alpha }(x)=\frac{1}{\alpha }+\psi (x)-\psi \left(x+\frac{1}{\alpha }\right)$$

$$\psi(t)=\sum\limits_{1\leq|k|\leq J}a(k)e(kt)+\mathcal{O}\Bigg{(}\sum\limits_{|k|\leq J}b(k)e(kt)\Bigg{)}\,,\quad a(k)\ll 1/|k|\,,\;\;b(k)\ll 1/J\,.$$

$$\Big{|}\sum_{n\leq X}e(\alpha n)\Big{|}\leq\min\left(X,\frac{1}{2||\alpha||}\right)\,.$$

$$\sum _{n\le X}\min \left(Y,\frac{1}{||\lambda n||}\right)\ll \frac{XY}{q}+(X+q)\log 2q.$$

$$S(N,\alpha )$$

$$=\sum _{m\le \alpha N}{\mu }^2(m){\mu }^2(m+1){\omega }_{\alpha }\left(\frac{m}{\alpha }\right).$$

$$S(N,\alpha )$$

$$=\frac{1}{\alpha }{S}_1(N,\alpha )+S_2(N,\alpha ),$$

$$S_1(N,\alpha )$$

$$S_2(N,\alpha )$$

$$S_{1}(N,\alpha)=\sigma\alpha N+\mathcal{O}\big{(}N^{(26+\sqrt{433})/81+\varepsilon}\big{)}\,,$$

$$J=\sqrt{\alpha N}.$$

$$S_2(N,\alpha )$$

$$\displaystyle+\mathcal{O}\Bigg{(}\sum\limits_{m\leq\alpha N}\mu^{2}(m)\mu^{2}(m+1)\sum\limits_{|k|\leq J}b(k)e\left(\frac{km}{\alpha}\right)\left[1+e\left(\frac{k}{\alpha}\right)\right]\Bigg{)}$$

$$=\sum _{1\le |k|\le J}a(k)\left[1-e\left(\frac{k}{\alpha }\right)\right]\sum _{m\le \alpha N}{\mu }^2(m){\mu }^2(m+1)e\left(\frac{km}{\alpha }\right)$$

$$\displaystyle+\mathcal{O}\Bigg{(}\sum\limits_{|k|\leq J}b(k)\left[1+e\left(\frac{k}{\alpha}\right)\right]\sum\limits_{m\leq\alpha N}\mu^{2}(m)\mu^{2}(m+1)e\left(\frac{km}{\alpha}\right)\Bigg{)}$$

$$\displaystyle=S_{3}(N,\alpha)+\mathcal{O}\Big{(}S_{4}(N,\alpha)\Big{)}\,,$$

$$S_3(N,\alpha )$$

$$S_4(N,\alpha )$$

$$S_3(N,\alpha )\ll \sum _{1\le |k|\le J}\frac{1}{k}\left|\sum _{m\le \alpha N}{\mu }^2(m){\mu }^2(m+1)e\left(\frac{km}{\alpha }\right)\right|.$$

$$S_4(N,\alpha )\ll \frac{1}{J}\sum _{1\le |k|\le J}\left|\sum _{m\le \alpha N}{\mu }^2(m){\mu }^2(m+1)e\left(\frac{km}{\alpha }\right)\right|+\sqrt{N}.$$

$$\Sigma =\sum _{1\le k\le J}\frac{1}{k}\left|\sum _{m\le \alpha N}{\mu }^2(m){\mu }^2(m+1)e(\lambda km)\right|.$$

$$\Sigma \ll N^{\frac{5}{6}+\epsilon },$$

Full text

On the distribution of consecutive square-free numbers of the form $\mathbf{[\alpha n],[\alpha n]+1}$

S. I. Dimitrov

Faculty of Applied Mathematics and Informatics, Technical University of Sofia

8, St.Kliment Ohridski Blvd. 1756 Sofia, BULGARIA

e-mail: sdimitrov@tu-sofia.bg

Abstract: In the present paper we show that there exist infinitely many consecutive square-free numbers of the form $[\alpha n]$, $[\alpha n]+1$, where $\alpha>1$ is irrational number with bounded partial quotient or irrational algebraic number.

Keywords: Consecutive square-free numbers, Asymptotic formula.

AMS Classification: 11L05 $\cdot$ 11N25 $\cdot$ 11N37.

1 Notations

Let $N$ be a sufficiently large positive integer. By $\varepsilon$ we denote an arbitrary small positive number, not necessarily the same in different occurrences. We denote by $\mu(n)$ the Möbius function and by $\tau(n)$ the number of positive divisors of $n$. As usual $[t]$ and $\{t\}$ denote the integer part, respectively, the fractional part of $t$. Let $||t||$ be the distance from $t$ to the nearest integer. Instead of $m\equiv n\,\pmod{k}$ we write for simplicity $m\equiv n\,(k)$. Moreover $e(t)$=exp($2\pi it$) and $\psi(t)=\{t\}-1/2$. Let $\alpha>1$ be irrational number with bounded partial quotient or irrational algebraic number.

Denote

[TABLE]

We define the characteristic function $\omega_{\alpha}(x)$ in the interval $(0,1]$ as follows

[TABLE]

and we extend it periodically to all real line.

2 Introduction and statement of the result

The problem for the consecutive square-free numbers arises in 1932 when Carlitz [3] proved that

[TABLE]

where $\sigma$ is denoted by (1).

Subsequently in 1949 Mirsky [9] improved the error term of (3) to

[TABLE]

Further in 1984 Heath-Brown [7] improved the error term of (4) to

[TABLE]

Finally in 2014 Reuss [10] improved the error term of (5) to

[TABLE]

and this is the best result up to now.

In 2008 Güloğlu and Nevans [6] showed that there exist infinitely many square-free numbers of the form $[\alpha n]$, where $\alpha>1$ is irrational number of finite type. More precisely they proved that the asymptotic formula

[TABLE]

holds.

On the other hand in 2009 Abercrombie and Banks[1] showed that for almost all $\alpha>1$ the asymptotic formula

[TABLE]

holds, however this result provides no specific value of $\alpha$.

Subsequently in 2013 Victorovich [13] proved that when $\alpha>1$ is irrational number with bounded partial quotient or irrational algebraic number, then the asymptotic formula

[TABLE]

holds. Here $A=A(N)=\max\limits_{1\leq m\leq N^{2}}\tau(m)$.

In 2018 the author [4] showed that for any fixed $1<c<22/13$ there exist infinitely many consecutive square-free numbers of the form $[n^{c}],[n^{c}]+1$.

Recently the author [5] proved that there exist infinitely many consecutive square-free numbers of the form $x^{2}+y^{2}+1$, $x^{2}+y^{2}+2$.

Define

[TABLE]

Motivated by these results and following the method of Victorovich [13] we shall prove the following theorem.

Theorem 1**.**

Let $\alpha>1$ be irrational number with bounded partial quotient or irrational algebraic number. Then for the sum $S(N,\alpha)$ defined by (7) the asymptotic formula

[TABLE]

holds. Here $\sigma$ is defined by (1).

3 Lemmas

Lemma 1**.**

For the function $\omega_{\alpha}(x)$ defined by (2) the formula

[TABLE]

holds.

Proof.

See ([2], p. 480 ). ∎

Lemma 2**.**

For every $J\geq 2$, we have

[TABLE]

Proof.

See [11]. ∎

Lemma 3**.**

If $X\geq 1$, then

[TABLE]

Proof.

See ([8], Ch. 6, §2). ∎

Lemma 4**.**

Suppose that $X,Y\geq 1$,  $\lambda=\frac{a}{q}+\frac{\theta}{q^{2}}$,  $q\geq 1$,  $(a,q)=1$,  $|\theta|\leq 1$. Then

[TABLE]

Proof.

See ([12], Lemma 1). ∎

4 Proof of the Theorem

The equality $m=[\alpha n]$ is tantamount to $\alpha n-1<m<\alpha n$,  $\frac{m}{\alpha}<n<\frac{m}{\alpha}+\frac{1}{\alpha}$, i.e. $\left\{\frac{m}{\alpha}\right\}>1-\frac{1}{\alpha}$. Then from (7) we get

[TABLE]

Now (4) and Lemma 1 give us

[TABLE]

where

[TABLE]

Estimation of $\mathbf{S_{1}(N,\alpha)}$

Bearing in mind (6) and (11) we obtain

[TABLE]

where $\sigma$ is denoted by (1).

Estimation of $\mathbf{S_{2}(N,\alpha)}$

Let

[TABLE]

From (12) and Lemma 2 it follows

[TABLE]

where

[TABLE]

Using (16) and Lemma 2 we find

[TABLE]

From (14), (17) and Lemma 2 we get

[TABLE]

In order to estimate the sums $S_{3}(N,\alpha)$ and $S_{4}(N,\alpha)$ we shall prove the following lemma.

Lemma 5**.**

Let $\alpha>1$ be irrational number with bounded partial quotient or irrational algebraic number. Then for the sum

[TABLE]

where $\lambda=\frac{1}{\alpha}$ , the estimation

[TABLE]

holds.

Proof.

Using (20) and the well-known identity $\mu^{2}(m)=\sum_{d^{2}|m}\mu(d)$ we write

[TABLE]

Splitting the range of $k$, $d$ and $t$ into dyadic subintervals we obtain

[TABLE]

where

[TABLE]

If $(d,\,t)>1$ then the sum $\Sigma_{0}(K,D,T)$ is empty. Suppose now that $(d,\,t)=1$. Then the congruence $rd^{2}+1\equiv 0\,(t^{2})$ is equivalent to $r\equiv r_{0}\,(t^{2})$, where $r_{0}$ is some integer with $1\leq r_{0}\leq t^{2}$. From the last consideration and (22) it follows

[TABLE]

Consider two cases.

Case 1.  $DT\leq{(\alpha N)}^{1/6}$.

The inequality (23) and Lemma 3 give us

[TABLE]

Replacing $m=kd^{2}t^{2}$ from (4) we get

[TABLE]

Since $\alpha$ (therefore $\lambda=\frac{1}{\alpha}$) is irrational number with bounded partial quotient or irrational algebraic number then $\lambda$ can be represented in the form $\lambda=\frac{a}{q}+\frac{\theta}{q^{2}}$, $(\alpha N)^{\frac{1}{2}-\varepsilon}\ll q\ll(\alpha N)^{\frac{1}{2}}$, $(a,q)=1$, $|\theta|\leq 1$. This follows for example from ([13], Ch.2, Lemma 1.5, Lemma 1.6). Bearing in mind these considerations, (14), (4), Lemma 4, the inequalities $DT\leq{(\alpha N)}^{1/6}$ and $K\leq J$ we find

[TABLE]

Case 2.  $DT>{(\alpha N)}^{1/6}$.

Using (14), (23), the trivial estimate, the inequalities $DT>{(\alpha N)}^{1/6}$ and $K\leq J$ we obtain

[TABLE]

From (21), (26) and (27) it follows

[TABLE]

The lemma is proved. ∎

On the one hand (18) and Lemma 5 give us

[TABLE]

On the other hand (19) and Lemma 5 imply

[TABLE]

By (4), (28) and (29) we find

[TABLE]

The end of the proof

Bearing in mind (4), (13) and (30) we obtain the asymptotic formula (8).

The theorem is proved.

Acknowledgments. The author thanks Professor Stephen Choi for his helpful comments and suggestions, that led to improvement of the reminder term in the asymptotic formula (8).