# On a question of Luca and Schinzel over Segal-Piatetski-Shapiro sequences

Jean-Marc Deshouillers *,
Mehdi Hassani
and
Mohammad Nasiri-Zare J.-M. D.
Institut de Mathématiques de Bordeaux, UMR 5251
Université de Bordeaux , Bordeaux INP et CNRS
33405 TALENCE Cedex
FRANCE
<EMAIL_ADDRESS>M. H.
Department of Mathematics
University of Zanjan
University Blvd, 45137-38791, ZANJAN
IRAN
<EMAIL_ADDRESS>M. N.-Z.
Department of Mathematics
University of Zanjan
University Blvd, 45137-38791, ZANJAN
IRAN
<EMAIL_ADDRESS>
###### Abstract.

We extend to Segal-Piatetski-Shapiro sequences previous results on the Luca-
Schinzel question. Namely, we prove that for any real $c$ larger than $1$ the
sequence $(\sum_{m\leq n}\varphi(\lfloor m^{c}\rfloor)/\lfloor
m^{c}\rfloor)_{n}$ is dense modulo $1$, where $\varphi$ denotes Euler’s
totient function. The main part of the proof consists in showing that when $R$
is a large integer, the sequence of the residues of $\lfloor m^{c}\rfloor$
modulo $R$ contains blocks of consecutive values which are in an arithmetic
progression.

###### Key words and phrases:

Segal-Piatetski-Shapiro sequences, Euler’s totient function, distribution
modulo 1.

###### 2010 Mathematics Subject Classification:

11N64, 11K31, 11K38, 11B50

##  1\. Introduction

At the Czech-Slovak Number Theory Conference in Smolenice in August 2007, F.
Luca asked whether the sequences of arithmetic and geometric means of the
first values of the Euler totient function are uniformly distributed modulo 1;
A. Schinzel asked whether the weaker statement that those sequences are dense
modulo 1 was known; this question was positively answered in [6]. This opened
the way to extensions considering different multiplicative functions with
constant mean value, or mean values over different sequences of integers as
well as the original question of Luca in special cases (cf. the bibliography).
One of the latest results is that of the first and third named authors of this
paper

######  Theorem 1 (Theorem 1 of [7]).

Let $\varphi$ denote the Euler function and $G$ be a non constant polynomial
with integral coefficients and taking positive values at positive arguments.
The sequence

$\left(\sum_{m\leq n}\frac{\varphi(G(m))}{G(m)}\right)_{n\geq 1}$

is dense modulo 1.

Here, we extend this result to Segal-Piatetski-Shapiro sequences.

######  Theorem 2.

Let $\varphi$ denote the Euler function and $c$ be a real number number larger
than $1$. The sequence

$\left(\sum_{m\leq n}\frac{\varphi(\lfloor m^{c}\rfloor)}{\lfloor
m^{c}\rfloor}\right)_{n\geq 1}$

is dense modulo 1.

The case when $c$ is an integer is but a special case of Theorem 1. In this
paper, we only consider the case when $c>1$ is not an integer, i.e. the case
of Segal - Piatetski-Shapiro sequences.

A way to tackle the Luca-Schinzel question on a sequence $(a(n))_{n}$ is to
obtain the arithmetical independence of some consecutive values of the
sequence $(a(m))_{m}$ so that for each given $\varepsilon$, one can find $H$
consecutive values of the sequence $((\varphi(a(m+h))/a(m+h))_{m}$ which are
all less than $\varepsilon$ but the sum of which is larger than $1$. A natural
idea would be to expect a complete arithmetical independence of the sequence
$(\lfloor(m+1)^{c}\rfloor,\ldots,\lfloor(m+H)^{c}\rfloor))_{m}$. This is not
the case: it is proved in [1] that the sequence of the residues of $\lfloor
m^{c}\rfloor$ modulo $k$ is normal for no $k\geq 2$. It is even shown that in
the sequence of the residues of $\lfloor m^{c}\rfloor$ modulo $k$, there are
finite blocks of consecutive values which do not occur.

In Section 3, we are however going to prove that $\lfloor(m+h)^{c}\rfloor$ can
be locally approximated by a linear polynomial, more precisely

######  Theorem 3.

Let $\beta=2^{(c+2)}(c+2)^{2}/\|c\|$ and let $H$ be a positive integer. If $R$
is sufficiently large an integer, for any residue $r$ modulo $R$, there exists
$m\leq R^{\beta}$ such that

(1) $\forall h\in[1,H]\colon\lfloor(m+h)^{c}\rfloor\equiv
r+h\;(\operatorname{mod}R).$

In connection with [1], we notice that this result implies that for any $H$ a
positive integer and any non integral $c$ larger than $1$, if $R$ is large
enough, the sequence of the residues of $\lfloor n^{c}\rfloor$ modulo $R$
contains the block $(1,2,\ldots,H)$.

We show how to use this result for proving Theorem 2 in Section 4.

##  2\. General notation and results

###  2.1. Notation

For a real number $u$, we can write in a unique way $u=\lfloor
u\rfloor+\\{u\\}$, where $\lfloor u\rfloor\in\mathbb{Z}$ and
$\\{u\\}\in[0,1)$. We further let $\|u\|=\min(\\{u\\},1-\\{u\\})$.

For a real number $t$, we let $\operatorname{\mathrm{e}}(t)=\exp(2\pi it)$.

The letters $p$, $q$ denote prime numbers, the letter $h$ a non-negative
integer.

We use Vinogradov’s notation: if $f$ and $g$ are two real functions defined on
some interval $[a,+\infty)$, $g$ taking positive values, we write $f\ll g$ for
$f=O(g)$ and, when a parameter $u$ is involved, we write $f\ll_{u}g$ for
$f=O_{u}(g)$. Similarly, we define $f\gg g$ (_resp._ $f\gg_{u}g$) if there
exists a positive constant $C$, absolute, (_resp._ which may depend on $u$)
such that $f(x)\geq Cg(x)$ on $[a,+\infty)$.

For $c$ real and $\ell$ a positive integer, we let
$\gamma_{c}(\ell)=\frac{c(c-1)\cdots(c-\ell+1)}{\ell!}$; we further let
$\gamma_{c}(0)=1$. By Taylor’s expansion, we have, for $c>0,m>{~{}}0$ and
$h\geq 0$

(2) $(m+h)^{c}=\sum_{\ell=0}^{\lfloor
c\rfloor}\gamma_{c}(\ell)h^{\ell}m^{c-\ell}+r_{c}(m,h),$

with $0\leq r_{c}(m,h)\leq\gamma_{c}(\lfloor c\rfloor+1)h^{\lfloor
c\rfloor+1}m^{\\{c\\}-1}\leq h^{\lfloor c\rfloor+1}m^{\\{c\\}-1}$.

###  2.2. The Erdős-Turán-Koksma-Szüsz inequality

This inequality permits to give an upper bound for the discrepancy of a
sequence of elements in $\mathbb{R}^{s}$ in terms of finitely many
trigonometrical sums. The case when $s=1$ has been given by Erdős and Turán in
[10] and generalized to higher dimensions independently by Koksma [12] and
Szüsz [14]. We quote here the formulation given by Drmota and Tichy [8]. We
first recall the definition of the discrepancy.

An _interval_ $I$ in $[0,1)^{s}$ is a cartesian product $\prod_{1\leq i\leq
s}[a_{i},b_{i})$, with $0\leq a_{i}\leq b_{i}\leq 1$ for $1\leq i\leq s$; its
Lebesgue measure $\prod_{1\leq i\leq s}(b_{i}-a_{i})$ is denoted by
$\lambda(I)$. For an interval $I\subset[0,1)^{s}$, we denote by $\chi_{I}$ its
indicator (also called characteristic) function.

The _discrepancy_ of a finite set $X=\\{x_{1},x_{2},\ldots,x_{N}\\}$ of
elements of $\mathbb{R}^{k}$ is defined by

(3)
$D_{N}(X)=D_{N}(x_{1},x_{2},\ldots,x_{N})=\sup_{I\subset[0,1)^{s}}\left|\frac{1}{N}\sum_{n=1}^{N}\chi_{I}(\\{x_{n}\\})-\lambda(I)\right|.$

######  Lemma 1 (Multidimensional Erdős-Turán-Koksma-Szüsz inequality).

Let $s\geq 1$, $X=\\{x_{1},x_{2},\ldots,x_{N}\\}$ be a finite set of elements
of $\mathbb{R}^{s}$ and $K$ an arbitrary positive integer. We have

(4)
$D_{N}(X)\leq\left(\frac{3}{2}\right)^{s}\left(\frac{2}{K+1}+\sum_{0<\|k\|_{\infty}\leq
K}\frac{1}{r(k)}\left|\frac{1}{N}\sum_{n=1}^{N}\operatorname{\mathrm{e}}(k\cdot
x_{n})\right|\right),$

where, for $k=(k_{1},k_{2},\ldots,k_{s})\in\mathbb{Z}^{s}$, we let
$r(k)=\prod_{i=1}^{s}\max\\{1,|k_{i}|\\}$ and $u\cdot v$ denote the usual
scalar product of two elements $u$ and $v$ in $\mathbb{R}^{s}$.

###  2.3. Upper bound for some trigonometrical sums

The following lemma provides us with an upper bound for the trigonometric sums
useful in the application of the ETKSz inequality. The exponent in (5) is not
very strong, but the uniformity in the statement makes it very convenient for
our purpose. It is a rather straightforward consequence of van der Corput
inequalities.

######  Lemma 2.

Let $c$ be a real number in $(1,\infty)\backslash\mathbb{N}$. There exist two
constants $A$ and $B$ such that for any $N$ larger than $A$ and any non-zero
$(\lfloor c\rfloor+1)$-tuple $(a_{0},\ldots,a_{\lfloor c\rfloor})$ satisfying

(5) $\forall\ell\in[0,\lfloor c\rfloor]\colon\text{ either }a_{\ell}=0\text{
or }N^{-\|c\|/2}\leq|a_{\ell}|\leq N^{\|c\|/2},$

one has

$S:=\left|\sum_{n=N+1}^{2N}\operatorname{\operatorname{\mathrm{e}}}\left(\sum_{\ell=0}^{\lfloor
c\rfloor}a_{\ell}n^{c-\ell}\right)\right|\leq BN^{1-\theta},$

where $\theta=2^{-(c+2)}\|c\|$.

###### Proof.

We start with the case when all the coefficients are $0$ except $a_{\lfloor
c\rfloor}$. In this case, we use Kusmin-Landau inequality (cf. Theorem 2.1 of
[11]). If $N$ is large enough, we have, on $(N,2N]$,
$\|\frac{d}{dt}(a_{\lfloor c\rfloor}t^{\\{c\\}})\|\gg N^{-\|c\|/2+\\{c\\}-1}$
and thus $S\ll N^{1+\|c\|/2-\\{c\\}}\ll N^{1-\|c\|/2}\leq N^{1-\theta}$.

For the other cases, we use the following lemma, due to van der Corput, which
one can find as a combination of Theorem 2.2 (for $q=0$) and Theorem 2.8 (for
$q\geq 1$) of [11].

######  Lemma 3.

There exists an absolute constant $C$ satisfying the following. Let $q$ be a
non-negative integer, $f$ a real valued function defined on $[N,2N]$, with
$q+2$ continuous derivatives satisfying

$\lambda\leq|f^{(q+2)}(t)|\leq\alpha\lambda,$

for some $\lambda>0$ and some $\alpha\geq 1$. Then, one has

(6) $\sum_{n=N+1}^{2N}\operatorname{\mathrm{e}}(f(n))\leq
C\left(N(\alpha^{2}\lambda)^{\frac{1}{4Q-2}}+N^{1-\frac{1}{2Q}}\alpha^{\frac{1}{2Q}}+N^{(\frac{Q-1}{Q})^{2}}\lambda^{\frac{-1}{2Q}}\right),$

where $Q=2^{q}$.

Let $\ell_{0}$ be the smallest integer $\ell$ such that $a_{\ell}$ is non
zero. The function of interest is thus $f(n)=\sum_{\ell=\ell_{0}}^{\lfloor
c\rfloor}a_{\ell}n^{c-\ell}$. Since the non zero coefficients $a_{\ell}$,
including $a_{\ell_{0}}$ are small powers of $N$, the term
$a_{\ell_{0}}n^{c-\ell_{0}}$ dominates $f$ and its derivatives dominate the
derivatives of $f$. In the application of Lemma 3, we take $q=\lfloor
c\rfloor-\ell_{0}-1$; we first notice that $q$ is non-negative (we already
treated the case $\ell_{0}=\lfloor c\rfloor$); we also notice that
$q\leq\lfloor c\rfloor-1\leq c-1$ so that $1\leq Q\leq 2^{c-1}$. Furthermore,
$\alpha$ depends only of $c$, and (since our constants may depend on $c$), it
plays no role in our application of (6); moreover $\lambda$ is of the order
$|a_{\ell_{0}}|N^{\\{c\\}-1}$, so that

$\lambda\ll_{c}N^{\|c\|/2+\\{c\\}-1}\ll_{c}N^{-\|c\|/2}\text{ and
}\lambda^{-1}\ll_{c}N^{\|c\|/2+1-\\{c\\}}\ll_{c}N^{3\|c\|/2}.$

By (6), we have

$S\ll_{c}N^{1-\frac{\|c\|}{8Q-4}}+N^{1-\frac{2}{Q}}+N^{(\frac{Q-1}{Q})^{2}+\frac{3\|c\|}{4Q}}.$

We easily notice that each of the three terms in RHS of the last relation is
less than $N^{1-\theta}$. ∎

###  2.4. Contribution of large primes to $\varphi(n)/n$

######  Lemma 4.

Let $\alpha>0$. For any $C$ in $(0,\min(\alpha,1))$ there exists $N$ such that
for $n\geq N$, one has

(7) $\prod_{\begin{subarray}{c}p|n\\\ p\geq(\log
n)^{\alpha}\end{subarray}}\left(1-\frac{1}{p}\right)\geq C.$

###### Proof.

This is a simple consequence of classical results on the distribution of prime
numbers. Explicit value for $N$ in terms of $\alpha$ and $C$, which is not
relevant here, can be obtained from the evaluations given in [9].
Since the LHS of (7) is an increasing function of $\alpha$, it enough to
consider the case when $\alpha\in(0,1)$. Let $n$ be sufficiently large an
integer. We denote by $p_{0},p_{1},\ldots,p_{r},p_{r+1}$ the consecutive prime
numbers such that

(8) $p_{0}<\log^{\alpha}n\leq p_{1}<\cdots<p_{r+1}\text{ and }p_{1}\cdots
p_{r}\leq n<p_{1}\cdots p_{r+1}.$

By a direct lower bound and by Mertens Theorem, for $n$ large enough, we have

$\prod_{\begin{subarray}{c}p|n\\\ p\geq(\log
n)^{\alpha}\end{subarray}}\left(1-\frac{1}{p}\right)\geq\prod_{j=1}^{r}\left(1-\frac{1}{p_{j}}\right)\sim\frac{\log
p_{0}}{\log p_{r}}.$

From the first part of (8), we have $p_{0}\sim\log^{\alpha}n$. Taking
logarithms in the second part of (8), we have $\log
n\sim\theta(p_{r})-\theta(p_{0})$. The last relations we obtained imply
$p_{r}\sim\log n$ and so we have

$\log p_{0}/\log p_{r}\sim\alpha\log\log n/\log\log n=\alpha.$

Thus, for any $0<C<\min(\alpha,1)$, Relation (7) holds true when $n$ is large
enough. ∎

###  2.5. Elementary relations concerning integral and fractional parts

For convenience, we recall here some elementary relations.

Let $\ell$ denote a non-negative integer and $x,x_{1},\ldots,x_{\ell}$ real
numbers. We have

(9) $\lfloor\ell x\rfloor\geq\ell\lfloor x\rfloor\;\text{ and }\;\\{\ell
x\\}\leq\ell\\{x\\},$ (10) $\text{if
}\\{x_{1}\\}+\cdots+\\{x_{\ell}\\}<1,\;\text{ then }\;\lfloor
x_{1}+\cdots+x_{\ell}\rfloor=\lfloor x_{1}\rfloor\cdots+\lfloor
x_{\ell}\rfloor,$ (11) $\text{if }\;\ell\\{x\\}<1,\;\text{ then }\;\lfloor\ell
x\rfloor=\ell\lfloor x\rfloor.$

###### Proof.

We multiply $x=\lfloor x\rfloor+\\{x\\}$ by $\ell$, which leads to

$\displaystyle\ell x=\,$ $\displaystyle\ell\lfloor x\rfloor+\ell\\{x\\}$
$\displaystyle=\,$ $\displaystyle\ell\lfloor
x\rfloor+\lfloor\ell\\{x\\}\rfloor+\\{\ell\\{x\\}\\}$

which leads in turn to $\lfloor\ell x\rfloor=\ell\lfloor
x\rfloor+\lfloor\ell\\{x\\}\rfloor$ and $\\{\ell
x\\}=\\{\ell\\{x\\}\\}=\ell\\{x\\}-\lfloor\ell\\{x\\}\rfloor$, whence (9).
We have

$x_{1}+\cdots+x_{\ell}=(\lfloor x_{1}\rfloor+\cdots+\lfloor
x_{\ell}\rfloor)+(\\{x_{1}\\}+\cdots+\\{x_{\ell}\\});$

The first term in the RHS is an integer and the second term is in $[0,1)$,
which implies (10). Relation (11) is just the special case of (10) in which
all the summands are equal. ∎

######  Lemma 5.

Let $R$ and $r$ be integers with $0\leq r<R$, $u$ be a real number in $[0,1]$,
and $x$ be a real number such that

$\left\\{\frac{x}{R}\right\\}\in\left[\frac{r}{R},\frac{r+u}{R}\right).$

We have

$\lfloor x\rfloor\equiv r\,(\operatorname{mod}R)\;\text{ and }\;\\{x\\}<u.$

###### Proof.

The hypothesis implies that there exists a rational integer $t$ such that

$\frac{x}{R}-t\in\left[\frac{r}{R},\frac{r+u}{R}\right).$

This implies

$tR+r\leq x<tR+r+u,\;\text{ whence }\;\lfloor x\rfloor=tR+r\;\text{ and
}\;0\leq\\{x\\}<u.$

∎

##  3\. Local approximation by a linear polynomial

In this section, we keep the notation of Theorem 3, assuming, without loss of
generality that $0\leq r<R$.

###  3.1. A lemma on fractional parts

######  Lemma 6.

Let $m$ be an integer satisfying

  1. (i)

$m^{1-\\{c\\}}>4(cH)^{{c+1}}$,

  2. (ii)

$\left\\{\frac{m^{c}}{R}\right\\}\in\left[\frac{r}{R},\frac{r}{R}+\frac{1}{4R}\right),$

  3. (iii)

$\left\\{\frac{cm^{c-1}}{R}\right\\}\in\left[\frac{1}{R},\frac{1}{R}+\frac{1}{4RH}\right),$

  4. (iv)

$\forall\ell\in[2,\lfloor
c\rfloor]\colon\left\\{\frac{\gamma_{c}(\ell)m^{c-\ell}}{R}\right\\}\in\left[0,\frac{1}{4cRH^{\ell}}\right)$.

Then, for any $h$ in $[1,H]$, we have

$\lfloor(m+h)^{c}\rfloor\equiv r+h\;(\operatorname{mod}R).$

###### Proof.

Let $h$ be an integer in $[1,H]$.
By Relation ($i$), we have $0\leq c^{c}h^{\lfloor
c\rfloor+1}m^{\\{c\\}-1}<1/4$, which implies $\\{c^{c}h^{\lfloor
c\rfloor+1}m^{\\{c\\}-1}\\}<1/4$.
By Lemma 5, ($ii$) implies $\lfloor m^{c}\rfloor\equiv
r\;(\operatorname{mod}R)$ and $\\{m^{c}\\}<1/4$.
By Lemma 5, ($iii$) implies $\lfloor cm^{c-1}\rfloor\equiv
1\;(\operatorname{mod}R)$ and $\\{cm^{c-1}\\}<1/4H$ ; thanks to (9) and (11),
we obtain $\lfloor hcm^{c-1}\rfloor\equiv h\lfloor cm^{c-1}\rfloor\equiv
h\;(\operatorname{mod}R)$ and $h\\{cm^{c-1}\\}<1/4$.
In a similar way, we obtain $\lfloor h\gamma_{c}(\ell)m^{c-\ell}\rfloor\equiv
0\;(\operatorname{mod}R)$ as well as
$h^{\ell}\\{\gamma_{c}(\ell)m^{c-\ell}\\}<1/4c$.
Using Taylor’s expansion (2) and (10), we end the proof of Lemma 6. ∎

###  3.2. The use of Erdős-Turán-Koksma-Szüsz inequailty

In this subsection, we let $R$ be a (large) integer and

$\beta=2^{c+2}(c+2)^{2}/\|c\|,\;N=\lfloor R^{\beta}/2\rfloor,$ $s=\lfloor
c\rfloor+1\text{ and
}x_{n}=\left(\frac{(N+n)^{c}}{R},\frac{c(N+n)^{c-1}}{R},\ldots,\frac{\gamma_{c}(\lfloor
c\rfloor)(N+n)^{\\{c\\}}}{R}\right).$

Our aim is to show the following

######  Lemma 7.

As $R$ tends to infinity, we have

(12) $D_{N}(x_{1},x_{2},\ldots,x_{N})=o_{c,H}\left(R^{c+1}\right).$

We use ETKSz inequality, with

$K=\lceil N^{\theta/(c+2)}\rceil,\text{ where $\theta=2^{-(c+2)}\|c\|$ (cf.
Lemma \ref{upper})}.$

Up to a slight change of notation (the indices of the componants of $k$
running from $0$ to $\lfloor c\rfloor$ and the indices of the trigonometric
sums running from $N+1$ to $2N$), the trigonometric sums we have to consider
are

$S(N;k,R)=\sum_{n=N+1}^{2N}\operatorname{\mathrm{e}}\left(R^{-1}\sum_{\ell=0}^{\lfloor
c\rfloor}k_{\ell}\gamma_{c}(\ell)n^{c-\ell}\right),$

which are of the type studied in Lemma 2, with
$a_{\ell}=R^{-1}k_{\ell}\gamma_{c}(\ell)$. We have
$1\leq\gamma_{c}(\ell)\ll_{c}(1)$ and thus

$\text{either }a_{\ell}=0\text{ or }R^{-1}\leq|a_{\ell}|\ll_{c}R^{-1}K.$

When $a_{\ell}\neq 0$, we have

$|a_{\ell}|\ll_{c}R^{-1}K\ll_{c}K\ll_{c}N^{\theta/(c+2)}\;\text{ and
}\;|a_{\ell}^{-1}|\leq R\ll_{c}N^{\theta/(c+2)^{2}}.$

If $R$ is sufficiently large, we can use Lemma 2 and obtain

$\forall k\text{ with }0<\|k\|_{\infty}\leq K\,\colon\,S(N;k,R)\leq
N^{1-\theta}.$

We can now apply Lemma 1, with the most trivial bound $r(k)\geq{~{}}1$. We
obtain, when $R$, or equivalently $N$, is large enough

$D_{N}(X)\ll_{c}K^{-1}+K^{\lfloor
c\rfloor+1}N^{-\theta}\ll_{c}N^{-\theta/(c+2)}.$

###  3.3. End of the proof of Theorem 3

We first prove that, when $R$ is sufficiently large, there exists an integer
$m\in[N+1,2N]$ which satisfies Relations ($ii$), ($iii$) and ($iv$) of Lemma
6. For doing so, we wish to find an element $x_{m}$ in $X$ which belongs to an
interval $I$ of Lebesgue measure

$\lambda(I)=\frac{1}{4R}\times\frac{1}{4R}\times\prod_{\ell=2}^{\lfloor
c\rfloor}\frac{1}{4cRH^{\ell}}\gg_{c,H}R^{-\lfloor c\rfloor-1}.$

By the definition of the discrepancy recalled in (3), any interval in
$[0,1)^{s}$ with Lebesgue measure larger than $D_{N}(X)$ contains an element
from $X$. We have

$(\lambda(I))^{-1}\ll_{c,H}R^{{\lfloor c\rfloor+1}}\ll_{c,H}N^{\theta(\lfloor
c\rfloor+1)/(c+2)^{2}}=o_{c,H}(D_{N}(X)).$

Thus, if $N$ is large enough, there exists $m$ in $[N+1,2N]$ wich satisfies
Relations ($ii$), ($iii$) and ($iv$) of Lemma 6. To complete the proof of
Theorem 3, it is enough to notice that when $N$ is large enough, the integer
$m$ we have produced satisfies also Relation ($i$) and to apply Lemma 6. ∎

##  4\. Proof of Theorem 2

We follow a similar strategy to that of [6]: for an integer $H$ arbitrary
large, we construct an integer $m$ such

(13) $\forall
h\in[1,H]\;\colon\;\frac{1}{H}\leq\frac{\varphi(\lfloor(m+h)^{c}\rfloor)}{\lfloor(m+h)^{c}\rfloor)}\leq\frac{3}{H}.$

Theorem 2 easily follows from (13).

Without loss of generality, we may assume that $H\geq 21$ which implies that
for any $h$ in $[1,H]$, one has $\frac{\varphi(h)}{h}>\frac{3}{H}$ (notice
that $\varphi(h)$ is always at least $1$ and is larger than $3$ for $h\geq
H/3\geq 7$. Since $\varphi(p)/p$ tends to $1$ when the prime $p$ tends to
infinity and the infinite product of those terms diverges to $0$ (by Euler),
we can find finite families of prime numbers
$\left(\mathcal{P}_{h}\right)_{n}$ such that

  1. (i)

for $1\leq h<k\leq H$ the sets $\mathcal{P}_{h}$ and $\mathcal{P}_{k}$ are
disjoint,

  2. (ii)

for any $h$ all the elements of $\mathcal{P}_{h}$ are larger than $H$,

  3. (iii)

for any $h\;\colon\;2/H\leq(\varphi(h)/h)(\varphi(P_{h})/P_{h})\leq 3/H$,

where $P_{h}$ denotes the product of the primes in $\mathcal{P}_{h}$.

We now let $L$ be a prime number which is larger than all the elements of all
the $\mathcal{P}_{h}$ and we let

$R=H!\prod_{H<p\leq L}p.$

By the Chinese Remainder Theorem, we can find a residue $r$ modulo $R$
satisfying

$\displaystyle r\equiv-1\;(\operatorname{mod}P_{1})$ $\displaystyle\cdots$
$\displaystyle r\equiv-H\;(\operatorname{mod}P_{H})$ $\displaystyle r\equiv
0\;(\operatorname{mod}R/(P_{1}P_{2}\cdots P_{H})).$

By Theorem 3, if $R$ is large enough, i.e. if $L$ is large enough, there
exists $m\leq R^{\beta}$ such that

$\forall h\in[1,H]\;\colon\;\lfloor(m+h)^{c}\rfloor\equiv
r+h\;(\operatorname{mod}R)$

and thus $\gcd(\lfloor(m+h)^{c}\rfloor,R)=hP_{h}$. In other words,
$\lfloor(m+h)^{c}\rfloor$ is an integer which is less than $R^{\beta c+1}$,
divisible by $hP_{h}$ and such that any of its prime factor which does not
divide $hP_{h}$ is larger than $L$. We thus have, using Relation (iii) above

$\frac{2}{H}\prod_{\begin{subarray}{c}p|\lfloor(m+h)^{c}\rfloor\\\
p>L\end{subarray}}\frac{\varphi(p)}{p}\leq\varphi(\lfloor(m+h)^{c}\rfloor)/\lfloor(m+h)^{c}\rfloor)\leq\frac{3}{H}.$

In order to prove (13), and thus Theorem 2, it is enough to prove that for $L$
large enough, we have

(14) $\prod_{\begin{subarray}{c}p|\lfloor(m+h)^{c}\rfloor\\\
p>L\end{subarray}}\left(1-\frac{1}{p}\right)\geq 1/2.$

By the definition of $R$ and the prime number theorem, we have, when $L$ is
large enough

$\displaystyle\log\lfloor(m+h)^{c}\rfloor$ $\displaystyle\leq(\beta
c+1)\left(\sum_{p\leq L}\log p+\log H!\right)$ $\displaystyle\leq(\beta
c+2)L.$

When $L$ is large enough, and thus $m$ too, we have

$\displaystyle\prod_{\begin{subarray}{c}p|\lfloor(m+h)^{c}\rfloor\\\
p>L\end{subarray}}\left(1-\frac{1}{p}\right)$
$\displaystyle\geq\prod_{\begin{subarray}{c}p|\lfloor(m+h)^{c}\rfloor\\\
p>((\beta
c+2)^{-1}\log\lfloor(m+h)^{c}\rfloor\end{subarray}}\left(1-\frac{1}{p}\right)$
by the previous inequality
$\displaystyle\geq\prod_{\begin{subarray}{c}p|\lfloor(m+h)^{c}\rfloor\\\
p>(\log\lfloor(m+h)^{c}\rfloor)^{3/4}\end{subarray}}\left(1-\frac{1}{p}\right)$
since $m$ is large enough $\displaystyle\geq 1/2$ $\displaystyle\text{by Lemma
\ref{largep}}.$

Thus (14) is proved, as well as Theorem 2.

##  5\. Remarks and complements

We see this paper as an archetype of the more general study of the
distribution modulo 1 of mean vaues of sequences $(f(a(n)))_{n}$, where $f$ is
a regular multiplicative function with constant mean value and $a$ is a
function in a Hardy field with polynomial growth.

In particular, it is challenging to determine the values of $c$ for which the
sequence $\left(\sum_{m\leq n}\varphi(\lfloor m^{c}\rfloor)/\lfloor
m^{c}\rfloor\right)_{n\geq 1}$ is uniformly distributed modulo $1$.

##  6\. Acknowledgement

The first-named author has benefitted from the support of the joint FWF-ANR
project Arithrand: FWF: I 4945-N and ANR-20-CE91-0006.

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