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\begin{center}
\vskip 1cm{\Large\bf 
On The Distribution of Consecutive Generalized $r$-Free \\
\vskip .1in
Integers in Piatetski-Shapiro Sequences
}
\vskip 1cm
Angkana Sripayap and Teerapat Srichan\footnote{Corresponding author.}\\
Department of Mathematics \\
Faculty of Science\\ 
Kasetsart University\\
Bangkok 10900 \\
Thailand\\
\href{mailto:fscianr@ku.ac.th}{\tt fscianr@ku.ac.th}\\
\href{mailto:fscitrp@ku.ac.th}{\tt fscitrp@ku.ac.th}\\ 
\ \\
Pinthira Tangsupphathawat\\
Department of Mathematics \\
Faculty of Science and Technology\\
Phranakhon Rajabhat University\\
Bangkok 10220\\
Thailand\\
\href{mailto: t.pinthira@hotmail.com}{\tt  t.pinthira@hotmail.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
We use  Rieger's technique to generalize a previous result on the distribution of consecutive $r$-free integers in Piatetski-Shapiro sequences. 
\end{abstract}

\section{Introduction} 
Let $r$ be a fixed integer $\geq 2$. 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
$$\mathbb{N}^c=\{\lfloor n^c \rfloor\}_{n\in\mathbb{N}}\qquad (c>1, c\notin\mathbb{N}),$$
where $\lfloor z\rfloor$ is the integer part of $z\in\mathbb{R}$. The Piatetski-Shapiro sequence was  introduced by Piatetski-Shapiro \cite{piatetski} to study prime numbers in a sequence of the form $\lfloor f(n) \rfloor$, 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   \cite{cao,cao2,deshouillers,rieger,stux,zhang}. The distribution of consecutive square-free and cube-free integers  in Piatetski-Shapiro sequences   is also a topic of interest. In 2018 Dimitrov \cite{dimitrov} proved that for every fixed $1<c< 7/6$, there exist infinitely many consecutive square-free integers of the form $\lfloor n^c \rfloor, \ \lfloor  n^c \rfloor+1$ by showing that
\begin{align}																																						\label{Rdimitrov}
\sum_{\substack{x/2 <n\leq x\\ \lfloor n^c \rfloor, \, \lfloor n^c \rfloor+1 \, \text{are square-free}}}1=\frac{1}{2}\prod_ p\Big(1-\frac{2}{p^2}\Big)x+O\Big(x^{\frac{6c+1}{8}+\varepsilon}\Big),\qquad \text{for}\  1<c<\frac{7}{6}.
\end{align}

Very recently, Tangsupphathawat,  Srichan, and  Laohakosol \cite{tang}  used Rieger's technique in \cite{rieger} to improve the range of $c$ and the error term in Dimitrov's work in \eqref{Rdimitrov}. They showed that for  $1<c< 3/2,$ and sufficiently small $\varepsilon>0$,  
\begin{align}\label{pin}
\sum_{\substack{n\leq x\\ \lfloor n^c \rfloor, \, \lfloor n^c \rfloor+1 \, \textnormal{are square-free}}}1=\prod_ p\Big(1-\frac{2}{p^2}\Big)x+O\Big(x^{\frac{2c+1}{4}+\varepsilon}\Big)\qquad(x\to \infty).
\end{align}
In the case of cube-free numbers,  Zhang and Li    \cite{zhang} proved that, for any $\varepsilon$ less than $10^{-10}$, one has
\begin{align*}
\sum_{\substack{n\leq x\\ \lfloor n^c \rfloor \, \text{is cube-free}}}1
=\left(\frac{1}{\zeta(3)}+o(x^\varepsilon)\right)x \qquad \text{for}\  1<c< \frac{11}{6}.
\end{align*}
In 2018,  Dimitrov \cite{dimitrov1} used the method of Zhang and Li to prove that 
\begin{align}\label{rd}
\sum_{\substack{n\leq x\\ \lfloor n^c \rfloor, \lfloor n^c \rfloor+1\, \text{are cube-free}}}1=\prod_{p}\Big(1-{2\over p^3}\Big)x+O(x^{1-\delta^2/2}),\qquad \text{for $1<c<{31\over 17}$},
\end{align}
where $0<\delta<\min\{{31-17c\over 9c-9}, 10^{-10}\}$ 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 \cite{subbarao0} 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 $n=a^kb$, where $a,b\in\mathbb{N}$ and $b$ is $r$-free. They noticed that in the limiting case when $k\to \infty$, a $(k, r)$-integer becomes an $r$-free integer.
 
Recently, the second author \cite{srichan}  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 $(\kappa_1, \lambda_1)$ and  $(\kappa_2, \lambda_2)$ satisfying 
$$r\lambda_1-\kappa_1<1,\ r\lambda_2-\kappa_2>1,\  k\lambda_1-\kappa_1>1,\ \frac{r(\lambda_2- \lambda_1)-(\kappa_2-\kappa_1)}{\lambda_2(1+\kappa_1)-\lambda_1(1+\kappa_2)}>1,$$ 
we have
\begin{align*} 
\sum_{\substack{n\leq N\\ \lfloor n^c \rfloor \, \text{is a $(k,r)$-integer}}}1&=\frac{\zeta(k)}{\zeta(r)}N+O(N^{(c/r)+\delta(\kappa_1, \lambda_1, \kappa_2, \lambda_2)}\log N),
\end{align*}
for $1<c<\frac{r(\lambda_2- \lambda_1)-(\kappa_2-\kappa_1)}{\lambda_2(1+\kappa_1)-\lambda_1(1+\kappa_2)}$,
where $\delta(\kappa_1, \lambda_1, \kappa_2, \lambda_2)=\frac{(\lambda_2\kappa_1- \lambda_1\kappa_2)+r^{-1}(\kappa_2-\kappa_1)}{\lambda_2(1+\kappa_1)-\lambda_1(1+\kappa_2)}$.

In this work, we use use Rieger's technique \cite{rieger} to generalize the previous result on the distribution of consecutive $r$-free integers in Piatetski-Shapiro sequences, by proving the following result:
\begin{theorem}
\label{thm}
Let $k,r$ be integers with $1<r<k$. For a large $N\in\mathbb{N}$, let
$$T_c(N;k,r):=\sum_{\substack{n\leq N\\ \lfloor n^c \rfloor, \, \lfloor n^c \rfloor+1 \, \textnormal{are  $(k,r)$-integers}}}1$$
denote the number of positive integers $n\leq N$ such that $\lfloor n^c \rfloor$ and  $\lfloor n^c \rfloor+1$ are   $(k,r)$-integers.
For $1<c<{3\over 2}$, we have, as $N\to \infty$, 
\begin{align}\label{t1}
T_c(N;k,2)&=N\sum_{m=1}^\infty {\tau(m)\lambda_{k,2}(m)\over m}+
O(N^{c/2+1/4}\log N),
\end{align}
and for $1<c<2$, we have, as $N\to\infty$,
$$T_c(N;k,r)=N \sum_{m=1}^\infty {\tau(m)\lambda_{k,r}(m)\over m}+
O(N^{c/3+1/3}\log N),\, \qquad \text{if}\quad r\geq 3,
$$
where $\lambda_{k,r}(n)$ is a multiplicative function defined by
\begin{align}\label{lembda}
\lambda_{k,r}(p^a)=\begin{cases}
    1, & \text{if  } a \equiv \modd{0} {k};\\
   -1, & \text{if } a \equiv \modd{r} {k}; \\
    0, & \text{otherwise},
  \end{cases}
\end{align}
and $\tau(n)$ denotes the number of divisors of $n$.
\end{theorem}
As mentioned above,  in the limiting case when $k\to \infty$, a $(k, 2)$-integer becomes a  square-free integer, which leads to the following corollary.
\begin{corollary}\label{Cor1}
For $1<c<{3\over 2}$, we have, as $N\to \infty$, 
\begin{align}\label{cor1}
T_c(N;\infty,2)&=N\prod_p\Big(1-{2\over p^2}\Big)+
O(N^{c/2+1/4}\log N),
\end{align}
where the big $O$-term is independent of $k$.
\end{corollary}  
\begin{proof}
For a fixed  large $N$, every positive  integer  $a$, and a positive $r$-free integer  $b$, the inequality
 $a^{\lfloor N\rfloor}b\leq N$ holds only when $a=1$. This indicates that in the interval $[1,N]$, every $(N, r)$-integer is an $r$-free integer.
 Thus, as $N\to \infty$, the function $T_c(N;\infty,2)$ counts the number of positive integers $n\leq N$ such that $\lfloor n^c \rfloor$ and  $\lfloor n^c \rfloor+1$ are   square-free integers. 
In view of (\ref{lembda}), we have 
\begin{align*}
\sum_{m=1}^\infty{ \lambda_{N,r}(m)\over m^s}={\zeta(Ns)\over \zeta(rs)},\quad s>{1\over r}. 
\end{align*}
When  $N\to \infty$, since $\zeta(Ns)= 1$, we have,
\begin{align*}
\sum_{m=1}^\infty{ \lambda_{N,2}(m)\over m^s}={1\over \zeta(2s)}.
\end{align*}
The main term of (\ref{cor1}) follows from  (\ref{t1}) of Theorem \ref{thm} and we get
\begin{align*}
\sum_{m=1}^\infty {\tau(m)\lambda_{N,2}(m)\over m}=\sum_{\substack{d,\, t\\ \gcd\gcd(d,t)=1}}{\mu(d)\mu(t)\over d^2t^2}=\prod_p\Big(1-{2\over p^2}\Big).
\end{align*}
Since the big  $O$-term of (\ref{t1}) is independent to $k$, Corollary \ref{Cor1} follows.
\end{proof}
Similarly,  since an $(\infty,3)$-integer becomes a cube-free integer, we obtain an improved result for (\ref{rd}) in the following corollary.
\begin{corollary}
For $1<c<2$, we have, as $N\to \infty$, 
\begin{align*}
T_c(N;\infty,3)&=N\prod_p\Big(1-{2\over p^3}\Big)+
O(N^{c/3+1/3}\log N).
\end{align*}
\end{corollary}
\section{Lemmas}
We collect now some lemmas needed later.
\begin{lemma}\label{lem2.1} \cite[Lemma 2.6]{subbarao}
Let
$$q_{k,r}(n)=\begin{cases} 1, &\textnormal{if}\  n\  \textnormal{is a}\  (k,r)\textnormal{-integer};\\
0, &\textnormal{if}\  n\  \textnormal{is not a} \ (k,r)\textnormal{-integer},
\end{cases}
$$ 
denote the characteristic function of the set of $(k,r)$-integers. Then
$$
q_{k,r}(n)=\sum_{a^kb^rc=n}\mu(b).
$$
\end{lemma}
\noindent Let
\begin{equation}\notag
d(r,k,n):= \sum_{n_1^r n_2^k =n}  1																								
\end{equation}
denote the number of ways of writing an integer $n$ in the form $n=n_1^rn_2^k$, and put 
$$D(r,k,x)=\sum_{n\leq x}d(r,k,n).$$ 
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 \cite[Section 14.3]{ivic}.
\begin{lemma}  																					\label{lemivic}
For a sufficiently large $x\in\mathbb{R}$, we have
$$D(r,k,x)\ll x^{1/r}.$$
\end{lemma}
\begin{lemma}\label{lem2.2}
For $x\geq 1$,  we have
$$
\sum_{m\leq x}\lambda_{k,r}(m)\ll x^{1/r}.
$$
\end{lemma}
\begin{proof} 
In view of  (\ref{lembda}),  Lemmas  \ref{lem2.1} and \ref{lemivic}, we have
\begin{align*}
&\sum_{m\leq x}\lambda_{k,r}(m)=\sum_{a^kb^r\leq x}\mu(b)\ll \sum_{a^kb^r\leq x}1\ll  x^{1/r}.
\end{align*}
\end{proof}

The proof of Theorem \ref{thm} makes use of the following estimate, due originally to Rieger \cite{rieger}, for the number of integers $n$ up to $x$ such that $\lfloor n^c \rfloor$ belongs to an arithmetic progression. 
  \begin{lemma}  (\cite{rieger})																											\label{lemm}
     For $1<c<2$, let  $x$ be a positive real number, and let $q$ and $a$ be two integers such that $0\leq a<q\leq x^c$. Then
 \begin{align*}
\sum_{\substack{n\leq x\\ \lfloor n^c \rfloor \, \equiv\, \modds{a}  {q} }}1=\frac{x}{q }+\begin{cases}
O \Big(\frac{x^{(c+4)/7}}{q^{1/7}}\Big), & \text{ for }\, q<x^{c-5/4};\\[.1in]
O \Big(\frac{x^{(c+1)/3}}{q^{1/3}}\Big), &\text{ for }\, x^{c-5/4}\leq q<x^{c-1/2};\\[.1in]
O \Big(\frac{x^{c}}{q}\Big), &\text{ for }\, x^{c-1/2}\leq q<x^{c}.
\end{cases}
\end{align*}
\end{lemma}
The proofs of the  next two lemmas make use of ideas similar to those in   Lemma 2.1 of \cite{brudern}.
\begin{lemma}\label{lem2}
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
\begin{align}\label{n1}
d_2^kt_2^rv-d_1^kt_1^ru=1,\qquad d_1^kt_1^ru\leq x^c.
\end{align}
$\text{I}$) If
$x^{c}<d_1^kt_1^rd_2^kt_2^r\leq x^{4c/3},$
then
\begin{align*}
A_{c}(x; k, r)\ll x^{4c/3r}\log x. 
\end{align*}
$\text{II}$) If
$x^{4c/3}<d_1^kt_1^rd_2^kt_2^r\leq x^{2c}$,
then
\begin{align*}
B_{c}(x; k, r)\ll x^{2c/3}\log x. 
\end{align*}
\end{lemma}
\begin{proof}$\text{I}$)
For a fixed choice of $d_1, t_1, d_2$ and $t_2$ satisfying (\ref{n1}), we have $d_1^kt_1^ru \equiv -1 \, (\bmod \, d_2^kt_2^r)$, which fixes the value of $u$ modulo $d_2^kt_2^r$. In view of  (\ref{n1}), the total number of possibilities for $u$ is $O(1+x^c/d_1^kt_1^rd_2^kt_2^r)$. By (\ref{n1}), the value of $v$ is fixed for a given choice
of $d_1, t_1, d_2, t_2, u$ . Then, by Lemma \ref{lemivic}, we have
\begin{align*}
A_{c}(x; k, r)&\ll \sum_{x^{c}< d_1^kt_1^rd_2^kt_2^r \leq x^{4c/3}}(1+{x^c\over d_1^kt_1^rd_2^kt_2^r})\\
&\ll \sum_{x^{c}<m \leq x^{4c/3}}\tau(m)d(k, r; m)(1+{x^c\over m})\\
&\ll \sum_{x^{c}<m \leq x^{4c/3}}d(k, r; m)(\log m+{x^c\log m\over m})\\
&\ll x^{4c/3r}\log x+x^{c/r}\log x\ll x^{4c/3r}\log x. 
\end{align*}
$\text{II}$)
From (\ref{n1}), we have $uvd_1^kt_1^rd_2^kt_2^r\leq x^c(x^c+1)$, whence $uv\leq x^c(x^c+1)x^{-4c/3}$ 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 $d_2^kt_2^rv-d_1^kt_1^ru=1$ is $O(\log x)$; see  \cite{estermann}.
\end{proof}
\section{Proof of Theorem \ref{thm}}
From  $q_{k,r}(n)=\sum_{d\mid n}\lambda_{k,r}(d)$ , we have
\begin{align*}
T_c(N; k,r)&=\sum_{n\leq N}q_{k,r}(\lfloor n^c\rfloor)q_{k,r}(\lfloor n^c\rfloor+1)\\
&=\sum_{n\leq N}\Big(\sum_{d\mid \lfloor n^c\rfloor}\lambda_{k,r}(d)\Big)\Big(\sum_{t\mid \lfloor n^c\rfloor+1}\lambda_{k,r}(t)\Big)\\
&=\Big(\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}+\sum_{d,t\atop {\gcd(d,t)=1\atop dt> N^c}}\Big)\lambda_{k,r}(d)\lambda_{k,r}(t)\sum_{n\leq N\atop { \lfloor n^c\rfloor \, \equiv \, \modds{0}  {d}\atop \lfloor n^c\rfloor +1 \, \equiv \, \modds{0}  {t}}}1.
\end{align*}
In view of Lemma \ref{lem2}, we have 
\begin{align*}
T_c(N; k,r)&=\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}\lambda_{k,r}(d)\lambda_{k,r}(t)\sum_{n\leq N\atop { \lfloor n^c\rfloor \, \equiv \, \modds{0} {d}\atop \lfloor n^c\rfloor +1 \, \equiv\, \modds{0} {t}}}1+\sum_{d,t\atop {\gcd(d,t)=1\atop dt> N^c}}\lambda_{k,r}(d)\lambda_{k,r}(t)\sum_{n\leq N\atop { \lfloor n^c\rfloor \, \equiv \, \modds{0} {d}\atop \lfloor n^c\rfloor +1 \, \equiv \, \modds{0} {t}}}1\\
&=:\Sigma_1+O(N^{2c/3}\log N).
\end{align*}
By the Chinese remainder theorem, there is a positive integer $\alpha$, unique modulo $dt$, satisfying the congruence system $\alpha \equiv \modd{0}   {d}$ and $\alpha+1\equiv \modd{0}  {t}$. Thus,
\begin{align*}
\Sigma_1&=\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}\lambda_{k,r}(d)\lambda_{k,r}(t)\sum_{n\leq N\atop \lfloor n^c\rfloor \equiv \modds{\alpha} {dt}}1.
\end{align*}
In view of Lemma \ref{lemm}, we have
\begin{align*}
\Sigma_1&=N\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}+O\Big(N^{(c+4)/7}\Big|\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^{c-5/4}}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over (dt)^{1/7}}\Big|\Big)
\\
&+O\Big(N^{(c+1)/3}\Big|\sum_{d,t\atop {\gcd(d,t)=1\atop N^{c-5/4}< dt\leq N^{c-1/2}}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over (dt)^{1/3}}\Big|\Big)+O\Big(N^{c}\Big|\sum_{d,t\atop {\gcd(d,t)=1\atop N^{c-1/2}<dt\leq N^c}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}\Big|\Big).
\end{align*}
In view of Lemma \ref{lem2.2}, we note that,
\begin{align*}
\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^{c-5/4}}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over (dt)^{1/7}}&= \sum_{m\leq N^{c-5/4}}{\tau(m)\lambda_{k,r}(m)\over m^{1/7}}\\
&\ll \sum_{m\leq N^{c-5/4}}{\lambda_{k,r}(m)\over m^{1/7-\varepsilon}}
&\ll \begin{cases}
    (N^{c-5/4})^{\varepsilon+1/r-1/7}, & \text{if  } r<7;\\
   \log N, & \text{if }r\geq 7, 
  \end{cases}
\end{align*}
\begin{align*}
\sum_{d,t\atop {\gcd(d,t)=1\atop N^{c-5/4}< dt\leq N^{c-1/2}}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over (dt)^{1/3}}&= \sum_{N^{c-5/4}< m\leq N^{c-1/2}}{\tau(m)\lambda_{k,r}(m)\over m^{1/3}}\\
&\ll \sum_{N^{c-5/4}< m\leq N^{c-1/2}}{\lambda_{k,r}(m)\over m^{1/3-\varepsilon}}
&\ll \begin{cases}
    (N^{c-1/2})^{\varepsilon+1/r-1/3}, & \text{if  } r=2;\\
   \log N, & \text{if }r\geq 3,
  \end{cases}
\end{align*}
and 
\begin{align*}
\sum_{d,t\atop {\gcd(d,t)=1\atop N^{c-1/2}<dt\leq N^c}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}&= \sum_{N^{c-1/2}<m\leq N^c}{\tau(m)\lambda_{k,r}(m)\over m}\\
&\ll \sum_{N^{c-1/2}<m\leq N^c}{\lambda_{k,r}(m)\over m^{1-\varepsilon}}\ll (N^{c-1/2})^{\varepsilon+1/r-1}.
\end{align*}
Thus, 
\begin{align*}
\Sigma_1&=N\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}+\begin{cases}
    O(N^{c/2+1/4})\log N, & \text{if  } r=2;\\
   O(N^{c/3+1/3})\log N, & \text{if } r\geq 3. 
  \end{cases}
\end{align*}
Note that 
\begin{align}\label{P1}\notag
\sum_{d,t\atop {\gcd(d,t)=1\atop dt\leq N^c}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}&=\Big(\sum_{d,t\atop {\gcd(d,t)=1}}+\sum_{d,t\atop {\gcd(d,t)=1\atop dt> N^c}}\Big)\Big({\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}\Big)\\\notag
&=\sum_{d,t\atop {\gcd(d,t)=1}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}+O\Big(\sum_{m> N^c}{\tau(m)\lambda_{k,r}(m)\over m}\Big)\\
&=\sum_{d,t\atop {\gcd(d,t)=1}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}+O\Big( N^{c/r-c}\log N\Big).
\end{align}
Since ${c\over r}-c<0$, the sum on the right hand side of (\ref{P1}) converges when $r>1$.
Thus,
\begin{align*}
\Sigma_1&=N\sum_{d,t\atop {\gcd(d,t)=1}}{\lambda_{k,r}(d)\lambda_{k,r}(t)\over dt}+\begin{cases}
    O(N^{c/2+1/4})\log N, & \text{if  } r=2;\\
   O(N^{c/3+1/3})\log N, & \text{if } r\geq 3, 
  \end{cases}
\end{align*}
which completes the proof of Theorem \ref{thm}.


\section{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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\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11L07;
Secondary 11N37.

\noindent \emph{Keywords: }
generalized $r$-free integer, Piatetski-Shapiro sequence.

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\vspace*{+.1in}
\noindent
Received October 27 2021;
revised version received January 12 2022; February 2 2022. 
Published in {\it Journal of Integer Sequences},
February 2 2022.

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