Truncated convolution of M\"obius function and multiplicative energy of an integer $n$

TL;DR

This paper derives upper bounds for moments of a truncated Dirichlet convolution involving the Möbius function and estimates the multiplicative energy of an integer's divisors, revealing their typical smallness and structural properties.

Contribution

It introduces new bounds for the moments of the truncated Möbius convolution and provides estimates for the multiplicative energy of divisor sets, advancing understanding of their behavior.

Findings

$M(n,j)$ is usually quite small for $j \

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Abstract

We establish an interesting upper bound for the moments of truncated Dirichlet convolution of M\"obius function, a function noted $M(n,z)$. Our result implies that $M(n,j)$ is usually quite small for $j \in \{1,\dots,n\}$. Also, we establish an estimate for the multiplicative energy of the set of divisors of an integer $n$.

Equations

$$M(n,z):=\sum _{\begin{array}{c}d\mid n\\ d\le z\end{array}}\mu (d).$$

$$\biggl{|}\sum_{\begin{subarray}{c}d\mid n\\ a\leq d\leq b\end{subarray}}\mu(d)\biggr{|}\leq\binom{\omega(n)}{\lfloor\frac{\omega(n)}{2}\rfloor}\quad(1\leq a\leq b\leq n)$$

$$L_t(n):={\int }_1^{n}M(n,z{)}^{t}dz$$

$$|L_t(n)|\le (1+{\delta }_{2,t})n\exp \left(Ct\frac{\omega (n{)}^{1-\frac{1}{t}}}{(1-1/t){\log }_{+}^{\frac{1}{t}}\omega (n)}\right),$$

$$|L_{t}(n)|\leq\left\{\begin{array}[]{ll}2n\exp\bigl{(}t\omega(n)^{1-\frac{1}{t}}\bigr{)}&\mbox{if}\ t=2\ \mbox{and}\ \omega(n)\leq 55,\\[5.69054pt] n\exp\bigl{(}t\omega(n)^{1-\frac{1}{t}}\bigr{)}&\mbox{otherwise}.\end{array}\right.$$

$${\mathcal{H}}_{\theta }(n):=|\{j\in [1,n]\cap \mathbb{N}:|M(n,j)|\ge 2^{\theta \omega (n)}\}|(\theta \in (0,1]).$$

$$\frac{\exp (W(x))}{W(x)}=x.$$

$$\alpha:=W\biggl{(}\frac{e}{\theta\log 2}\biggr{)}.$$

$$\mathcal{H}_{\theta}(n)\leq n\exp{\scriptstyle\Bigl{(}-\frac{\theta\log 2}{\alpha}\omega(n)\log\omega(n)+\bigl{(}\frac{\alpha-1}{1-\frac{\alpha-1}{\log\omega(n)}}\bigr{)}^{3}\frac{\omega(n)}{\exp\bigl{(}\frac{\alpha-1}{1+\frac{\alpha-1}{\log\omega(n)}}\bigr{)}\log\omega(n)}\Bigr{)}}.$$

$$\tau(n)\leq\biggl{(}\frac{\log(n\gamma(n))}{\omega(n)}\biggr{)}^{\omega(n)}\beta(n)\quad(n\geq 2)$$

$$\beta (n):=\prod _{p\mid n}\frac{1}{\log p}.$$

$$\frac{(\log z{)}^{\omega (n)}}{\omega (n)!}\beta (n)\le |\{1\le j\le z:\gamma (j)\mid \gamma (n)\}|\le \frac{(\log (z\gamma (n)){)}^{\omega (n)}}{\omega (n)!}\beta (n)$$

$$\tau (n,z):=\sum _{\begin{array}{c}d\mid n\\ d\le z\end{array}}1\le \frac{(\log (z\gamma (n)){)}^{\omega (n)}}{\omega (n)!}\beta (n).$$

$$\tau (n,z)\le \sum _{0\le j\le D(n,z)}\left(\genfrac{}{}{0ex}{}{\omega (n)}{j}\right)$$

$$D(n,z):=\underset{\begin{array}{c}d\mid n\\ d\le z\end{array}}{\max }\omega (d).$$

$$-\underset{\begin{array}{c}0\le j\le D(n,z)\\ 2\nmid j\end{array}}{\max }\left(\genfrac{}{}{0ex}{}{\omega (n)-1}{j}\right)\le M(n,z)\le \underset{\begin{array}{c}0\le j\le D(n,z)\\ 2\mid j\end{array}}{\max }\left(\genfrac{}{}{0ex}{}{\omega (n)-1}{j}\right)$$

$$E_s(n):=|\{(d_1,\dots ,d_s,d_{s+1},\dots ,d_{2s})\in d(n{)}^{2s}:d_1\cdots d_s=d_{s+1}\cdots d_{2s}\}|.$$

$$A(i,j)=\sum _{v=0}^j\left(\genfrac{}{}{0ex}{}{i+1}{v}\right)(-1{)}^v(j+1-v{)}^i(0\le j\le i-1,i,j\in \mathbb{Z}).$$

$$\tau (n{)}^{2s-1}{\left(\frac{A(2s-1,s-1)}{(2s-1)!}\right)}^{\omega (n)}<E_s(n)\le \tau (n{)}^{2s-1}{\left(\frac{1}{2^{2s-1}}\left(\genfrac{}{}{0ex}{}{2s}{s}\right)\right)}^{\omega (n)}$$

$$\frac{A(2s-1,s-1)}{(2s-1)!}\sim \sqrt{\frac{3}{\pi s}}\text{mbox}and\frac{1}{2^{2s-1}}\left(\genfrac{}{}{0ex}{}{2s}{s}\right)\sim \sqrt{\frac{4}{\pi s}}(s\to \infty ).$$

$$\int_{-\infty}^{\infty}\biggl{(}\frac{\sin(x)}{x}\biggr{)}^{2s}dx=\pi\frac{A(2s-1,s-1)}{(2s-1)!}\quad(s\in\mathbb{N}).$$

$$S_{i,j}=\sum _{v=0}^j(-1{)}^{j-v}(\genfrac{}{}{0ex}{}{j}{v})v^i.$$

$$\sum_{v=0}^{j}(-1)^{j-v}\binom{j}{v}v^{i}=\left\{\begin{array}[]{ll}0&\mbox{for}\ i=0,\dots,j-1,\\ i!&\mbox{for}\ i=j.\end{array}\right.$$

$$\left|\begin{array}{cccc}{x}_1^{u_1}& x_1^{u_2}& \cdots & x_1^{u_{\ell }}\\ x_2^{u_1}& x_2^{u_2}& \cdots & x_2^{u_{\ell }}\\ ⋮& ⋮& \cdots & ⋮\\ x_{\ell }^{u_1}& x_{\ell }^{u_2}& \cdots & x_{\ell }^{u_{\ell }}\end{array}\right|>0.$$

$$\operatorname{\rm{sgn}}(x):=\left\{\begin{array}[]{ll}-1&\mbox{if}\ x<0,\\ 0&\mbox{if}\ x=0,\\ 1&\mbox{otherwise}.\end{array}\right.$$

$$F_{\lambda ,m}(j):=\mathrm{sgn}(j)j^m(j\in \{-\lambda ,\dots ,\lambda \}).$$

$$\left\{\begin{array}[]{l}\mbox{$F_{\lambda,m}(x)$ is an odd function of degree $2\lambda-1$ with a leading term}\\ \mbox{of sign $(-1)^{\frac{2\lambda-m-2}{2}}$ if $m$ is even,}\\ \mbox{$F_{\lambda,m}(x)$ in an even function of degree $2\lambda$ with a leading term}\\ \mbox{of sign $(-1)^{\frac{2\lambda-m-1}{2}}$ if $m$ is odd}.\end{array}\right.$$

$$G_{\lambda ,m}(x):=F_{\lambda ,m}(x)-F_{\lambda ,m}(-x)$$

$$F_{\lambda ,m}(x):=\sum _{j=1}^{\lambda }{a}_{2j}{x}^{2j}.$$

$$\left(\begin{array}{cccc}{1}^2& 1^4& \cdots & 1^{2\lambda }\\ 2^2& 2^4& \cdots & 2^{2\lambda }\\ ⋮& ⋮& \cdots & ⋮\\ {\lambda }^2& {\lambda }^4& \cdots & {\lambda }^{2\lambda }\end{array}\right)\left(\begin{array}{c}{a}_2\\ a_4\\ ⋮\\ a_{2\lambda }\end{array}\right)=\left(\begin{array}{c}{1}^m\\ 2^m\\ ⋮\\ {\lambda }^m\end{array}\right).$$

Full text

AMS Subject Classification numbers: 11N37, 11N56, 11N64.

Key words: number of divisors function, Möbius function, multiplicative energy.

1 Introduction

Let $\mu(\cdot)$ be the Möbius function and consider

[TABLE]

The function $M(n,z)$ has been studied by various authors (see [5], [2], [7], [8], [6], [10], [4] for example). In [5] it is established that

[TABLE]

where $\omega(n)$ is the number of distinct prime divisors of $n$. A very interesting tool, known as the symmetrical chains, is used to establish a generalization of this property in [2]. In this paper, we are interested by the average size of $M(n,z)$ over $1\leq z\leq n$. More precisely, we consider the quantity

[TABLE]

for integer values of $t\geq 1$. Let’s remark that $L_{t}(n)=L_{t}(\gamma(n))$, where $\gamma(n):=\prod_{p\mid n}p$. From what we know, only the value of $L_{1}(n)$, which is $-\prod_{p\mid n}(1-p)$ for $n\geq 2$, is easy to evaluate. Let’s write $\log_{+}x:=\log\max(x,2)$ and $\delta_{i,j}$ for the Kronecker delta.

Theorem 1.1**.**

Let $t\geq 2$ be an integer and $n\geq 1$ be a squarefree integer. Then

[TABLE]

where $C=1.07073472\dots$ is defined in the statement of Lemma 2.4.

Theorem 1.2**.**

Let $t\geq 2$ be an integer and $n\geq 1$ be a squarefree integer. Then

[TABLE]

We then use Theorem 2 to get some control over the quantity

[TABLE]

To express our result, we need to define the function $W:[e,\infty)\rightarrow[1,\infty)$ implicitly by

[TABLE]

This function is linked to the Lambert $\mathcal{W}$ function by the relation W(x)=-\mathcal{W}\Bigl{(}\frac{-1}{x}\Bigr{)} in which we take the solution larger than 1.

Corollary 1.3**.**

Let $\theta\in(0,1]$ be fixed and write

[TABLE]

Let also $n\geq 2$ be a fixed squarefree integer. Then, assuming that $\alpha-1<\log\omega(n)$ and that $\omega(n)\geq 56$, we have

[TABLE]

We record some approximate values of $\alpha=\alpha(\theta)$ in Table 1.

[TABLE]

Table 1

In [12], it has been shown that the number of divisors function, noted $\tau(\cdot)$, satisfies the inequality

[TABLE]

where

[TABLE]

This inequality has been extensively worked out in the author’s Ph. D. Thesis [3]. It is worth mentioning that the function $\beta(n)$ is intimately linked to the value of $\tau(n)$ in more than one way. In particular, it follows from Theorem $5.3$ p.491 of [14] that

[TABLE]

so that

[TABLE]

In the special case where $n$ is squarefree, one prefers the estimate

[TABLE]

where

[TABLE]

For comparison, the argument in [5] allows one to establish that

[TABLE]

for every integer $n\geq 2$.

Let $s\geq 1$ be a fixed integer. For any integers $n\geq 1$ we write $d(n):=\{d:d\mid n\}$. We define the $s$-th multiplicative energy of $n$ to be

[TABLE]

In particular, we trivially have $E_{s}(n)\leq\tau(n)^{2s-1}$. In what follows, $A(i,j)$ are Eulerian numbers of the first kind that can be computed by using the formula

[TABLE]

Theorem 1.4**.**

Let $s,n\geq 2$ be positive integers. Then the inequality

[TABLE]

holds.

Remark 1.5**.**

It is possible to establish that

[TABLE]

The first relation is deduced from the identity

[TABLE]

The upper bound in (1.5) is in fact an equality in the case where $n$ is squarefree. In this direction, we will see that the proof gives a much more general result.

Throughout the paper, we denote the $k$-th prime number by $p_{k}$. Also, for each $k\geq 0$, we denote by $n_{k}$ the number $\prod_{j=1}^{k}p_{j}$ (so that $n_{0}=1$).

Acknowledgment. I thank Jean-Marie De Koninck for his interest in this article and Thomas J. Ransford for a discussion about the integrals (1.6) many years ago.

2 Preliminary lemmas

Lemma 2.1**.**

Let $i\geq 0$ and $j\geq 1$ be integers. Let also denote the number of surjections from a set of $i$ elements to a set of $j$ elements by $S_{i,j}$. It satisfies

[TABLE]

Proof.

It is a well known result. We remark that it implies that

[TABLE]

∎

Lemma 2.2**.**

Let $0\leq u_{1}<\cdots<u_{\ell}$ and $0<x_{1}<\cdots<x_{\ell}$ be two sequences of real numbers. Then the generalized Vandermonde determinant satisfies

[TABLE]

Proof.

This is known as a result of Mitchell [11]. A modern proof uses the Lemma $A2$ of [9]. ∎

For $x\in\mathbb{R}$, we define the sign function by

[TABLE]

Lemma 2.3**.**

Let $\lambda\geq 1$ and $m\geq 0$ be integers. Let also $F_{\lambda,m}(x)$ be the polynomial of minimal degree that satisfies

[TABLE]

We assume that $0\leq m\leq 2\lambda-1$. We have

[TABLE]

Proof.

We first assume that $m\geq 1$ is odd. From Lagrange interpolation with $2\lambda+1$ points, we have $\deg F_{\lambda,m}(x)\leq 2\lambda$. Now, the polynomial

[TABLE]

has at least $2\lambda+1$ roots, so that $G_{\lambda,m}(x)$ is identically [math] and we deduce that $F_{\lambda,m}(x)$ is an even function. Therefore, we search for a polynomial of the type

[TABLE]

We get to the linear system

[TABLE]

By Cramer’s rule,

[TABLE]

so that we deduce from Lemma 2.2 that $\operatorname{\rm{sgn}}(a_{2\lambda})=(-1)^{\frac{2\lambda-m-1}{2}}$. The result follows from the fact that there is a unique such interpolating polynomial of degree at most $2\lambda$.

In the case where $m\geq 0$ is even, we simply observe that $F_{\lambda,m}(x):=\frac{F_{\lambda,m+1}(x)}{x}$. The proof is complete. ∎

Let’s define

[TABLE]

Lemma 2.4**.**

Let $t\geq 2$ and $k\geq 1$ be positive integers. Then

[TABLE]

Also,

[TABLE]

with $C=1.07073472\dots$. The constant $C$ is best possible and is attained only at $t=2$ and $k=2149$.

Proof.

We will begin with the proof of (2.1). We will prove this result by induction for every single values of $t\geq 2$. For $t=2,\dots,99$ we verify with a computer for every value of $k=1,\dots,56$. For $t\geq 100$ there is no need to verify since

[TABLE]

while $k^{1-1/t}\geq 0.74k$ for each $k=1,\dots,56$. We consider $t\geq 2$ as fixed. Now, for a fixed $k\geq 57$, we assume that the result holds for $k-1$. We will establish that

[TABLE]

which is clearly enough for the induction step with this value of $t$. We see that (2.4) holds if

[TABLE]

from the mean value theorem. Now, it is known that $p_{k}>k\log k$ for each $k\geq 1$, see [13]. Using this inequality, we have that (2.5) holds if

[TABLE]

We have used the fact that the function $\left(1-\frac{1}{x}\right)^{x}$ is strictly increasing for $x>1$. This concludes the proof of inequality (2.4) and thus the induction step for the fixed value of $t$. Inequality (2.1) is established.

We now turn to the proof of (2.2). The argument is very similar, that is we proceed by induction for every single value of $t\geq 2$. For $t=2,\dots,99$ we verify with a computer for each value of $k$ from 1 to what is written in Table 2.

[TABLE]

Table 2

For each $t\geq 100$, by using (2.3), it is enough to have

[TABLE]

which is easily seen to hold for $k=1,\dots,8$.

Let’s consider $t\geq 2$ as fixed. We assume that

[TABLE]

holds at $J=k-1$ and we want to show that it holds with $J=k(\geq 9)$ . It is enough to show that

[TABLE]

from the mean value theorem and the fact that $f_{t}(x):=\frac{x^{1-1/t}}{\log^{1/t}x}$ satisfies $f^{\prime\prime}_{t}(x)<0$ if x\geq 6>\exp\Bigl{(}\frac{2/t-1+\sqrt{5-4/t}}{2(1-1/t)}\Bigr{)} for each $t\geq 2$. Again, by using $p_{k}>k\log k$ for each $k\geq 1$, we deduce that (2.7) holds if

[TABLE]

which holds for $k$ greater that the corresponding value in Table 2 if $t=2,\dots,99$ or for $k\geq 9$ if $t\geq 100$. This completes the inductive step for the fixed value of $t\geq 2$ and the proof is complete. ∎

3 Proof of Theorem 1.1 and 1.2

We write

[TABLE]

where

[TABLE]

Now, for $n\geq 2$, we rearrange the terms according to the number $j=1,\dots,t$ of $d_{i}$ that are maximal at the same time and we use the fact that $M(n,n)=0$ to get to

[TABLE]

Thus, let $1=d_{1}<d_{2}<\cdots<d_{2^{\omega(n)}}=n$ be the sequence of divisors of $n$. We write

[TABLE]

Now, we deduce from (3.1) that

[TABLE]

Also, for any integer value of $\rho\geq 0$ and $\sigma\in\mathbb{R}_{\geq 0}$, we can write

[TABLE]

where we have used $\sigma=\frac{1}{\rho+1}$. We thus get to

[TABLE]

The results then follow from Lemma 2.4. The proof is complete.

Remark 3.1**.**

The function $\mathcal{J}_{\rho}(\cdot)$ satisfies

[TABLE]

Indeed, let $n=q_{1}\cdots q_{\omega(n)}$ with $q_{1}<\cdots<q_{\omega(n)}$ the factorization of $n$. Thus, since $p_{r_{1}}\cdots p_{r_{l}}\leq q_{r_{1}}\cdots q_{r_{l}}$, it follows that the $i$-th term in the ordered sequence of divisors of $n_{\omega(n)}$ is at most equal to the $i$-th term in the corresponding sequence for $n$.

4 Proof of Corollary 1.3

Since the function $M(n,z)$ is constant for $z\in[j,j+1)$ ($j\in\mathbb{Z}_{\geq 0}$) and $M(n,n)=0$ for $n\geq 2$, we deduce that

[TABLE]

From Theorem 1.2 and the hypothesis $\omega(n)\geq 56$, we have

[TABLE]

Now, the idea is simply to optimize this last inequality over the even integers $t\geq 2$. Our strategy is to find the exact value $t_{0}\in(1,\infty)$ and to estimate the variation caused by $t=t_{0}+\xi$ with $|\xi|\leq 1$. We write

[TABLE]

so that

[TABLE]

Let’s write $t_{0}=c\log\omega(n)$. We have $f^{\prime}(t_{0})=0$ if and only if

[TABLE]

so that $t_{0}=\frac{\log\omega(n)}{\alpha-1}$ which is strictly larger than 1 by hypothesis. We verify that $f(t_{0})=\frac{\theta\log 2}{\alpha}\log\omega(n)$. Now, we have

[TABLE]

from the mean value theorem applied twice. The result follows from the estimate

[TABLE]

which holds since $t_{0}>1$. The proof is complete.

5 Proof of Theorem 1.4

We assume throughout the proof that $s\geq 2$ is a fixed integer. The function $E_{s}(n)$ is multiplicative, so it will be enough to show that

[TABLE]

for any prime $p$.

Now, for a fixed prime $p$, the function $E_{s}(p^{\alpha})$ counts the number of solutions to the system

[TABLE]

We clearly have $R_{s}(0)=1$ and also

[TABLE]

an identity that follows from $(x+1)^{2s}=(x+1)^{s}(x+1)^{s}$. In general, $R_{s}(\alpha)$ is the coefficient of $x^{0}$ in the expansion of

[TABLE]

from which we deduce that

[TABLE]

The last expression follows from

[TABLE]

that can be shown by using the identity $(x+1)^{2s}=((x^{2}+2x)+1)^{s}$.

Now, the idea of the proof is to show that $P_{s}(x)$ is an odd function with strictly positive coefficients (of $x^{j}$ with $j$ odd) so that it is clear that the function $\frac{P_{s}(x)}{x^{2s-1}}$ has a strictly negative derivative. With this in mind, we write

[TABLE]

so that we turn to

[TABLE]

where each $b_{2r}$ with $r=0,\dots,s-1$ is a strictly positive coefficient. By writing

[TABLE]

we deduce that it is enough to show that $\operatorname{\rm{sgn}}(c_{r})=(-1)^{r}$. We write

[TABLE]

where $Q_{s,2r+1}(x)$ is the Lagrange polynomial of degree at most $2s$ for which

[TABLE]

From Lemma 2.1 and the remark in the proof, we deduce that $\operatorname{\rm{sgn}}(c_{r})=(-1)^{s+1}\operatorname{\rm{sgn}}(e_{2s})$ where $e_{2s}$ is the leading term of $Q_{s,2r+1}(x)$. Now, since

[TABLE]

the function in Lemma 2.3. We deduce that $\operatorname{\rm{sgn}}(2e_{2s})=\operatorname{\rm{sgn}}(e_{2s})=(-1)^{s-r-1}$ so that $\operatorname{\rm{sgn}}(c_{r})=(-1)^{r}$ as wanted. The proof is complete.

6 Concluding remark

Let’s consider the quantity

[TABLE]

The methods used in the proof of Theorem 1.4 also apply to $T_{s}(\alpha)$. That is, the function $\frac{T_{s}(\alpha)}{(2\alpha+1)^{s-1}}$ is strictly decreasing for integer values of $\alpha\geq 0$ when $s\geq 3$, it is constant for $s=1$ or 2.