Another generalization of Euler’s arithmetic function and Menon’s identity

Abstract

We define the k-dimensional generalized Euler function \(\varphi _k(n)\) as the number of ordered k-tuples \((a_1,\ldots ,a_k)\in {\mathbb {N}}^k\) such that \(1\le a_1,\ldots ,a_k\le n\) and both the product \(a_1\cdots a_k\) and the sum \(a_1+\cdots +a_k\) are prime to n. We investigate some of the properties of the function \(\varphi _k(n)\), and obtain a corresponding Menon-type identity.

Motivation

Jordan’s arithmetic function \(J_k(n)\) is defined as the number of ordered k-tuples \((a_1,\ldots ,a_k)\in {\mathbb {N}}^k\) such that \(1\le a_1,\ldots ,a_k\le n\) and the gcd \((a_1,\ldots ,a_k,n)=1\). It is well known that \(J_k(n)\) is multiplicative in n and \(J_k(n)=n^k \prod _{p\mid n} (1-1/p^k)\). If \(k=1\), then \(J_1(n)=\varphi (n)\) is Euler’s arithmetic function.

A Menon-type identity concerning the function \(J_k(n)\), obtained by Nageswara Rao [7], is given by

$$\begin{aligned} \sum _{\begin{array}{c} a_1,\ldots ,a_k=1\\ (a_1,\ldots , a_k,n)=1 \end{array}}^n (a_1-1,\ldots ,a_k-1,n)^k = J_k(n) \tau (n) \quad (n\in {\mathbb {N}}), \end{aligned}$$
(1.1)

where \(\tau (n)\) is the number of divisors of n. If \(k=1\), then (1.1) reduces to Menon’s original identity [6].

Euler’s arithmetic function and Menon’s identity have been generalized in various directions by several authors. See, e.g., the books [5, 8], the papers [4, 7, 9, 11, 12], and their references.

The function \(X(n)= \# \{(a,b)\in {\mathbb {N}}^2: 1\le a,b\le n, (ab,n)=(a+b,n)=1 \}\) is an analog of Euler’s \(\varphi \)-function, and was introduced by Arai and Gakuen [2]. It was shown by Carlitz [3] that the function X(n) is multiplicative and

$$\begin{aligned} X(n)= n^2 \prod _{p\mid n} \left( 1-\frac{1}{p} \right) \left( 1-\frac{2}{p} \right) \quad (n\in {\mathbb {N}}). \end{aligned}$$

Note that if n is even, then \(X(n)=0\). The function X(n) can also be given as

$$\begin{aligned} X(n)= \varphi (n)^2 \sum _{d\mid n} \frac{\mu (d)}{\varphi (d)} \quad (n\in {\mathbb {N}}), \end{aligned}$$

\(\mu \) denoting the Möbius function. The corresponding Menon-type identity

$$\begin{aligned} \sum _{\begin{array}{c} a,b=1 \\ (ab,n)=(a+b,n)=1 \end{array}}^n (a+b-1,n) = X(n) \tau (n) \quad (n\in {\mathbb {N}}) \end{aligned}$$
(1.2)

was deduced by Sita Ramaiah [9, Cor. 10.4]. In fact, (1.2) is a corollary of a more general identity involving Narkiwicz-type regular systems of divisors and k-reduced residue systems.

Recently, identity (1.2) was generalized by Ji and Wang [4, 12] to residually finite Dedekind domains, by using Narkiwicz-type regular systems of divisors, and to the ring of algebraic integers, concerning Dirichlet characters modulo n, respectively. Note that in paper [12] identity (1.2) is called the “Arai–Carlitz identity.” However, Arai and Carlitz only considered the function X(n) and did not deduce such an identity. We refer to (1.2) as the Sita Ramaiah identity.

It is natural to introduce and to study the following k-dimensional generalization of the function X(n), and to ask if the corresponding generalization of the Sita Ramaiah identity is true for it. These were not investigated in the literature, as far as we know. For \(k\in {\mathbb {N}}\) we define the function \(\varphi _k(n)\) as

$$\begin{aligned} \varphi _k(n):= \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1\\ (a_1+\cdots +a_k,n)=1 \end{array}}^n 1. \end{aligned}$$
(1.3)

Note that \(\varphi _1(n)=\varphi (n)\) is Euler’s function and \(\varphi _2(n)=X(n)\) of above. We investigate some of the properties of the function \(\varphi _k(n)\), and obtain a corresponding Menon-type identity. Our main results are included in Sect. 2, and their proofs are presented in Sects. 3 and 4.

We will use the following notations: \({\text {id}}_k(n)=n^k\), \(\mathbf{1}(n)=1\) (\(n\in {\mathbb {N}}\)), \(\omega (n)\) will denote the number of distinct prime factors of n, and “\(*\)” the Dirichlet convolution of arithmetic functions.

Main results

In this paper we prove the following results.

Theorem 2.1

For every \(k,n\in {\mathbb {N}}\),

$$\begin{aligned} \varphi _k(n) =&\ \varphi (n)^k \prod _{p\mid n} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \end{aligned}$$
(2.1)
$$\begin{aligned} =&\ n^k \prod _{p\mid n} \left( 1-\frac{1}{p} \right) \left( \left( 1-\frac{1}{p}\right) ^k - \frac{(-1)^{k}}{p^k} \right) . \end{aligned}$$
(2.2)

It is a consequence of Theorem 2.1 that the function \(\varphi _k(n)\) is multiplicative. Also, \(\varphi _k(n)=0\) if and only if k and n are both even. Further properties of \(\varphi _k(n)\) can be deduced. Its average order is given by the next result.

Theorem 2.2

Let \(k\ge 2\) be fixed. Then

$$\begin{aligned} \sum _{n\le x} \varphi _k(n)= \frac{C_k}{k+1}x^{k+1} +O\left( x^k (\log x)^{k+1}\right) , \end{aligned}$$

where

$$\begin{aligned} C_k= \prod _p \left( 1+\frac{1}{p^{k+1}}\left( \left( 1-\frac{1}{p}\right) \left( (p-1)^k-(-1)^k\right) -p^k \right) \right) . \end{aligned}$$

Corollary 2.3

(\(k=2\)) We have

$$\begin{aligned} \sum _{n\le x} X(n)= \frac{C_2}{3}x^3 +O\left( x^2 (\log x)^3\right) , \end{aligned}$$

where

$$\begin{aligned} C_2= \prod _p \left( 1-\frac{3}{p^2}+ \frac{2}{p^3} \right) \approx 0.286747. \end{aligned}$$

We have the following generalization of Menon’s identity.

Theorem 2.4

Let f be an arbitrary arithmetic function. Then for every \(k,n\in {\mathbb {N}}\),

$$\begin{aligned} \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ (a_1+\cdots +a_k,n)=1 \end{array}}^n f((a_1+\cdots +a_k-1,n)) = \varphi _k(n)\sum _{d\mid n} \frac{(\mu *f)(d)}{\varphi (d)}. \end{aligned}$$
(2.3)

Corollary 2.5

(\(f(n)=n\)) For every \(k,n\in {\mathbb {N}}\),

$$\begin{aligned} \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ (a_1+\cdots +a_k,n)=1 \end{array}}^n (a_1+\cdots +a_k-1,n) = \varphi _k(n)\tau (n). \end{aligned}$$
(2.4)

If \(k=1\), then (2.4) reduces to Menon’s identity and if \(k=2\), then it gives the Sita Ramaiah identity (1.2).

Proofs of Theorems 2.1 and 2.2

We need the following lemmas.

Lemma 3.1

Let \(n,d\in {\mathbb {N}}\), \(d\mid n\), and let \(r\in {\mathbb {Z}}\). Then

$$\begin{aligned} \sum _{\begin{array}{c} a=1\\ (a,n)=1\\ a\equiv r \, \text { (mod }d) \end{array}}^n 1 = {\left\{ \begin{array}{ll} \displaystyle \frac{\varphi (n)}{\varphi (d)} &{} \text { if }(r,d)=1, \\ 0 &{} \text { otherwise}. \end{array}\right. } \end{aligned}$$

Lemma 3.1 is known in the literature, usually proved by the inclusion–exclusion principle. See, e.g., [1, Th. 5.32]. The following generalization and a different approach of proof are given in our paper [11].

Lemma 3.2

[11, Lemma 2.1] Let \(n,d,e\in {\mathbb {N}}\), \(d\mid n\), \(e\mid n\) and let \(r,s\in {\mathbb {Z}}\). Then

$$\begin{aligned} \sum _{\begin{array}{c} a=1\\ (a,n)=1\\ a\equiv r \, \text { (mod }d) \\ a\equiv s \, \text { (mod }e) \end{array}}^n 1 = {\left\{ \begin{array}{ll} \displaystyle \frac{\varphi (n)}{\varphi (de)}(d,e) &{} \text { if }(r,d)=(s,e)=1\text { and }(d,e) \mid r-s, \\ 0 &{} \text { otherwise}. \end{array}\right. } \end{aligned}$$

In the case \(e=1\), Lemma 3.2 reduces to Lemma 3.1.

We need to define the following slightly more general function than \(\varphi _k(n)\):

$$\begin{aligned} \varphi _k(n,m):= \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1\\ (a_1+\cdots +a_k,m)=1 \end{array}}^n 1. \end{aligned}$$
(3.1)

If \(m=n\), then \(\varphi _k(n,n)=\varphi _k(n)\), given by (1.3).

Lemma 3.3

(recursion formula for \(\varphi _k(n,m)\)) Let \(k\ge 2\) and \(m\mid n\). Then

$$\begin{aligned} \varphi _k(n,m)= \varphi (n) \sum _{d\mid m} \frac{\mu (d)}{\varphi (d)} \varphi _{k-1}(n,d). \end{aligned}$$

Proof of Lemma 3.3

We have

$$\begin{aligned} \varphi _k(n,m)= & {} \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \end{array}}^n \sum _{d\mid (a_1+\cdots +a_k,m)} \mu (d) \\= & {} \sum _{d\mid m} \mu (d) \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1\\ a_1+\cdots +a_k\equiv 0 \text { (mod }d) \end{array}}^n 1 \\= & {} \sum _{d\mid m} \mu (d) \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1 \end{array}}^n \sum _{\begin{array}{c} a_k=1\\ (a_k,n)=1\\ a_k\equiv -a_1-\cdots -a_{k-1} \text { (mod }d) \end{array}}^n 1. \end{aligned}$$

By using Lemma 3.1 we deduce that

$$\begin{aligned} \varphi _k(n,m)= & {} \sum _{d\mid m} \mu (d) \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1\\ (a_1+\cdots +a_{k-1},d)=1 \end{array}}^n \frac{\varphi (n)}{\varphi (d)} \\= & {} \varphi (n) \sum _{d\mid m} \frac{\mu (d)}{\varphi (d)} \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1\\ (a_1+\cdots +a_{k-1},d)=1 \end{array}}^n 1 \\= & {} \varphi (n) \sum _{d\mid m} \frac{\mu (d)}{\varphi (d)} \varphi _{k-1}(n,d), \end{aligned}$$

where \(d\mid m\) and \(m\mid n\) imply that \(d\mid n\). \(\square \)

Proof of Theorem 2.1

Let \(k,n,m\in {\mathbb {N}}\) such that \(m\mid n\). We show that

$$\begin{aligned} \varphi _k(n,m) = \varphi (n)^k \prod _{p\mid m} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) . \end{aligned}$$
(3.2)

By induction on k. If \(k=1\), then \(\varphi _1(n,m)= \varphi (n)\), by its definition (3.1). Let \(k\ge 2\). Assume that (3.2) holds for \(k-1\) and prove it for k. We have, by using Lemma 3.3,

$$\begin{aligned} \varphi _k(n,m)= & {} \varphi (n) \sum _{d\mid m} \frac{\mu (d)}{\varphi (d)} \varphi _{k-1}(n,d) \\= & {} \varphi (n) \sum _{d\mid m} \frac{\mu (d)}{\varphi (d)} \varphi (n)^{k-1} \prod _{p\mid d} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \left. +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \\= & {} \varphi (n)^k \prod _{p\mid m} \left( 1+ \frac{\mu (p)}{\varphi (p)}\left( 1 - \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \left. \left. +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \right) \\= & {} \varphi (n)^k \prod _{p\mid m} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) , \end{aligned}$$

which proves formula (3.2). Now choosing \(m=n\), (3.2) gives identity (2.1), which can be rewritten as (2.2). \(\square \)

Proof of Theorem 2.2

Let \(\varphi _k= {\text {id}}_k * g_k\), that is, \(g_k=\varphi _k * \mu {\text {id}}_k\). Here the function \(g_k(n)\) is multiplicative and for any prime power \(p^\nu \) (\(\nu \ge 1\)),

$$\begin{aligned} g_k(p^{\nu }) =\varphi _k(p^{\nu })- p^k\varphi _k(p^{\nu -1}). \end{aligned}$$

We obtain from (2.2) that for \(\nu \ge 2\),

$$\begin{aligned} g_k(p^{\nu })= & {} p^{(\nu -1) k} \left( 1-\frac{1}{p} \right) \left( (p-1)^k-(-1)^k \right) \nonumber \\&- p^k p^{(\nu -2)k} \left( 1-\frac{1}{p} \right) \left( (p-1)^k-(-1)^k \right) =0, \end{aligned}$$
(3.3)

and for \(\nu =1\),

$$\begin{aligned} g_k(p)= & {} \varphi _k(p) -p^k = \left( 1-\frac{1}{p} \right) \left( (p-1)^k-(-1)^k \right) - p^k \nonumber \\= & {} -(k+1)p^{k-1} +\cdots +(-1)^k k, \end{aligned}$$
(3.4)

a polynomial in p of degree \(k-1\), with leading coefficient \(-(k+1)\). Actually, we have

$$\begin{aligned} -(k+1)p^{k-1}< g_k(p) < 0 \end{aligned}$$
(3.5)

for every integer \(k\ge 2\) and every prime \(p\ge 2\). To see this, note that by Lagrange’s mean value theorem,

$$\begin{aligned} k(p-1)^{k-1}< p^k-(p-1)^k < kp^{k-1}, \end{aligned}$$

and from (3.4) we deduce that

$$\begin{aligned} g_k(p)> & {} \left( 1-\frac{1}{p} \right) \left( p^k -kp^{k-1}- (-1)^k \right) - p^k \\= & {} - (k+1) p^{k-1} + kp^{k-2}- (-1)^k \left( 1-\frac{1}{p} \right) > -(k+1)p^{k-1}. \end{aligned}$$

On the other hand, \(p^k-(p-1)^k>1\), \((p-1)^k-(-1)^k<p^k\) imply that

$$\begin{aligned} g_k(p)< \left( 1-\frac{1}{p}\right) p^k -p^k <0. \end{aligned}$$

According to (3.5), \(|g_k(p)|< (k+1)p^{k-1}\) holds true for every \(k\ge 2\) and every \(p\ge 2\), and by (3.3) we deduce that

$$\begin{aligned} |g_k(n)| \le (k+1)^{\omega (n)} n^{k-1} \quad (n\in {\mathbb {N}}). \end{aligned}$$

To obtain the desired asymptotic formula we apply elementary arguments. We have

$$\begin{aligned} \sum _{n\le x} \varphi _k(n)= & {} \sum _{d\le x} g_k(d) \sum _{\delta \le x/d} \delta ^k \\= & {} \sum _{d\le x} g_k(d) \left( \frac{1}{k+1} \left( \frac{x}{d}\right) ^{k+1} +O \left( \left( \frac{x}{d}\right) ^k\right) \right) \\= & {} \frac{x^{k+1}}{k+1} \sum _{d=1}^{\infty } \frac{g_k(d)}{d^{k+1}} + O \left( x^{k+1} \sum _{d>x} \frac{|g_k(d)|}{d^{k+1}} \right) + O\left( x^k \sum _{d\le x} \frac{|g_k(d)|}{d^k}\right) . \end{aligned}$$

Here the main term is \(\frac{C_k}{k+1} x^{k+1}\) by using the Euler product formula. To evaluate the error terms consider the Piltz divisor function \(\tau _{k+1}(n)\), representing the number of ordered \((k+1)\)-tuples \((a_1,\ldots ,a_{k+1})\in {\mathbb {N}}^{k+1}\) such that \(a_1\cdots a_{k+1}=n\). We have \(\tau _{k+1}(p^\nu )\ge \tau _{k+1}(p)=k+1\) for every prime power \(p^{\nu }\) (\(\nu \ge 1\)), and \(\tau _{k+1}(n)\ge (k+1)^{\omega (n)}\) for every \(n\in {\mathbb {N}}\).

We obtain

$$\begin{aligned} \sum _{d>x} \frac{|g_k(d)|}{d^{k+1}}\le \sum _{d>x} \frac{(k+1)^{\omega (d)}}{d^2} \le \sum _{d>x} \frac{\tau _{k+1}(d)}{d^2} \ll \frac{(\log x)^k}{x}, \end{aligned}$$

and

$$\begin{aligned} \sum _{d\le x} \frac{|g_k(d)|}{d^k} \le \sum _{d\le x} \frac{(k+1)^{\omega (d)}}{d} \le \sum _{d\le x} \frac{\tau _{k+1}(d)}{d} \ll (\log x)^{k+1}, \end{aligned}$$

by using known elementary estimates on the Piltz divisor function. See, e.g., [10, Lemma 3]. This completes the proof. \(\square \)

Proof of Theorem 2.4

Let \(M_k(n)\) denote the sum on the left-hand side of (2.3). We have by the convolutional identity \(f=(\mu *f)*\mathbf{1}\),

$$\begin{aligned} M_k(n)= & {} \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ (a_1+\cdots +a_k,n)=1 \end{array}}^n \sum _{d\mid (a_1+\cdots +a_k-1,n)} (\mu *f)(d)= \sum _{d\mid n} (\mu *f)(d) \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ (a_1+\cdots +a_k,n)=1\\ a_1+\cdots +a_k\equiv 1 \text { (mod }d) \end{array}}^n 1 \\= & {} \sum _{d\mid n} (\mu *f)(d) \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ a_1+\cdots +a_k\equiv 1 \text { (mod { d})} \end{array}}^n \sum _{\delta \mid (a_1+\cdots +a_k,n)} \mu (\delta ), \end{aligned}$$

that is

$$\begin{aligned} M_k(n)= \sum _{d\mid n} (\mu *f)(d) \sum _{\delta \mid n} \mu (\delta ) N_k(n,d,\delta ), \end{aligned}$$
(4.1)

where

$$\begin{aligned} N_k(n,d,\delta ):= \sum _{\begin{array}{c} a_1,\ldots , a_k=1 \\ (a_1\cdots a_k,n)=1 \\ a_1+\cdots +a_k\equiv 1 \text { (mod }d) \\ a_1+\cdots +a_k\equiv 0 \text { (mod }\delta ) \end{array}}^n 1. \end{aligned}$$

Next we evaluate the sum \(N_k(n,d,\delta )\), where \(d\mid n\), \(\delta \mid n\) are fixed. If \((d,\delta )>1\), then \(N_k(n,d,\delta )=0\), the empty sum. So, assume that \((d,\delta )=1\). If \(k=1\), then by using Lemma 3.2 we deduce

$$\begin{aligned} N_1(n,d,\delta ):= \sum _{\begin{array}{c} a_1=1 \\ (a_1,n)=1 \\ a_1\equiv 1 \text { (mod }d) \\ a_1 \equiv 0 \text { (mod }\delta ) \end{array}}^n 1= {\left\{ \begin{array}{ll} \frac{\varphi (n)}{\varphi (d)} &{} \text { if }\delta =1,\\ 0 &{} \text { otherwise}, \end{array}\right. } \end{aligned}$$
(4.2)

since for each term of the sum \(\delta \mid a_1\) and \(\delta \mid n\), which gives \(\delta \mid (a_1,n)=1\), so \(\delta =1\).

Lemma 4.1

(Recursion formula for \(N_k(n,d,\delta )\)) Let \(k\ge 2\), \(d\mid n\), \(\delta \mid n\), \((d,\delta )=1\). Then

$$\begin{aligned} N_k(n,d,\delta ) = \frac{\varphi (n)}{\varphi (d)\varphi (\delta )}\sum _{j\mid d} \mu (j) \sum _{t\mid \delta } \mu (t) N_{k-1}(n,j,t). \end{aligned}$$
(4.3)

Proof of Lemma 4.1

We have

$$\begin{aligned} N_k(n,d,\delta )= \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1 \end{array}}^n \sum _{\begin{array}{c} a_k=1\\ (a_k,n)=1 \\ a_k\equiv 1-a_1-\cdots -a_{k-1}\text { (mod }d) \\ a_k\equiv -a_1-\cdots -a_{k-1} \text { (mod }\delta ) \end{array}}^n 1. \end{aligned}$$

Using that \((d,\delta )=1\) and applying Lemma 3.2 we deduce that

$$\begin{aligned} N_k(n,d,\delta )= & {} \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1\\ (a_1+\cdots +a_{k-1}-1,d)=1\\ (a_1+\cdots +a_{k-1},\delta )=1 \end{array}}^n \frac{\varphi (n)}{\varphi (d)\varphi (\delta )} \\= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )} \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1 \end{array}}^n \sum _{j\mid (a_1+\cdots +a_{k-1}-1,d)} \mu (j) \sum _{t\mid (a_1+\cdots +a_{k-1},\delta )} \mu (t) \\= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )} \sum _{j\mid d} \mu (j) \sum _{t\mid \delta } \mu (t) \sum _{\begin{array}{c} a_1,\ldots , a_{k-1}=1 \\ (a_1\cdots a_{k-1},n)=1 \\ a_1+\cdots +a_{k-1}\equiv 1 \text { (mod } j) \\ a_1+\cdots +a_{k-1}\equiv 0 \text { (mod }t) \end{array}}^n 1 \\= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )} \sum _{j\mid d} \mu (j) \sum _{t\mid \delta } \mu (t) N_{k-1}(n,j,t). \end{aligned}$$

\(\square \)

Lemma 4.2

Let \(k\ge 2\), \(d\mid n\), \(\delta \mid n\), \((d,\delta )=1\). Then

$$\begin{aligned} N_k(n,d,\delta )= & {} \frac{\varphi (n)^k}{\varphi (d)\varphi (\delta )} \prod _{p\mid d} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \nonumber \\&\times \prod _{p\mid \delta } \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) . \end{aligned}$$
(4.4)

Proof of Lemma 4.2

By induction on k. If \(k=2\), then by the recursion (4.3) and (4.2),

$$\begin{aligned} N_2(n,d,\delta )= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )}\sum _{j\mid d} \mu (j) \sum _{t\mid \delta } \mu (t) N_1(n,j,t) \nonumber \\= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )}\sum _{j\mid d} \mu (j) \sum _{\begin{array}{c} t\mid \delta \\ t=1 \end{array}} \mu (t) \frac{\varphi (n)}{\varphi (j)} \end{aligned}$$
(4.5)
$$\begin{aligned}= & {} \frac{\varphi (n)^2}{\varphi (d)\varphi (\delta )} \sum _{j\mid d} \frac{\mu (j)}{\varphi (j)}= \frac{\varphi (n)^2}{\varphi (d)\varphi (\delta )} \prod _{p\mid d} \left( 1-\frac{1}{p-1} \right) . \end{aligned}$$
(4.6)

Hence, the formula is true for \(k=2\). Assume it holds for \(k-1\), where \(k\ge 3\). Then we have, by the recursion (4.3),

$$\begin{aligned} N_k(n,d,\delta )= & {} \frac{\varphi (n)}{\varphi (d)\varphi (\delta )} \sum _{j\mid d} \mu (j) \sum _{\begin{array}{c} t\mid \delta \\ (t,j)=1 \end{array}} \mu (t) \frac{\varphi (n)^{k-1}}{\varphi (j)\varphi (t)} \\&\times \prod _{p\mid j} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \\&\times \prod _{p\mid t} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-3} \frac{1}{(p-1)^{k-3}}\right) , \end{aligned}$$

where the condition \((t,j)=1\) can be omitted, since \(j\mid d\), \(t\mid \delta \) and \((d,\delta )=1\). We deduce that

$$\begin{aligned} N_k(n,d,\delta )= & {} \frac{\varphi (n)^k}{\varphi (d)\varphi (\delta )} \sum _{j\mid d} \frac{\mu (j)}{\varphi (j)} \prod _{p\mid j} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \quad \left. +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \\&\times \sum _{t\mid \delta } \frac{\mu (t)}{\varphi (t)} \prod _{p\mid t} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-3} \frac{1}{(p-1)^{k-3}}\right) \\= & {} \frac{\varphi (n)^k}{\varphi (d)\varphi (\delta )} \prod _{p\mid d} \left( 1- \frac{1}{p-1}\left( 1-\frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. \left. +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \right) \\&\times \prod _{p\mid \delta } \left( 1- \frac{1}{p-1}\left( \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \right. \\&\qquad \qquad \qquad \qquad \qquad \quad \left. \left. +(-1)^{k-3} \frac{1}{(p-1)^{k-3}}\right) \right) , \end{aligned}$$

giving (4.4), which completes the proof of Lemma 4.2. \(\square \)

Now we continue the evaluation of \(M_k(n)\). According to (4.1) and Lemma 4.2, we have

$$\begin{aligned} M_k(n)= & {} \sum _{d\mid n}(\mu *f)(d) \sum _{\begin{array}{c} \delta \mid n\\ (\delta ,d)=1 \end{array}}\mu (\delta ) \frac{\varphi (n)^k}{\varphi (d)\varphi (\delta )} \prod _{p\mid d} \left( 1- \frac{1}{p-1}\right. \nonumber \\&\left. \quad +\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \\&\quad \times \prod _{p\mid \delta } \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \\= & {} \varphi (n)^k \sum _{d\mid n} \frac{(\mu *f)(d)}{\varphi (d)} \prod _{p\mid d} \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \\&\left. \qquad \qquad \qquad \qquad \quad +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \\&\quad \times \sum _{\begin{array}{c} \delta \mid n\\ (\delta ,d)=1 \end{array}} \frac{\mu (\delta )}{\varphi (\delta )} \prod _{p\mid \delta } \left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots \right. \\&\qquad \qquad \qquad \qquad \left. +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) , \end{aligned}$$

where the inner sum is

$$\begin{aligned}&\prod _{\begin{array}{c} p\mid n\\ p\not \mid d \end{array}} \left( 1+\frac{\mu (p)}{\varphi (p)}\left( 1- \frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-2} \frac{1}{(p-1)^{k-2}}\right) \right) \\&\quad = \prod _{p\mid n} \left( 1-\frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \\&\qquad \times \prod _{p\mid d} \left( 1-\frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) ^{-1}. \end{aligned}$$

This leads to

$$\begin{aligned} M_k(n)= & {} \varphi (n)^k \prod _{p\mid n} \left( 1-\frac{1}{p-1}+\frac{1}{(p-1)^2}-\cdots +(-1)^{k-1} \frac{1}{(p-1)^{k-1}}\right) \\&\times \sum _{d\mid n} \frac{(\mu *f)(d)}{\varphi (d)} \\= & {} \varphi _k(n) \sum _{d\mid n} \frac{(\mu *f)(d)}{\varphi (d)}, \end{aligned}$$

by using (2.1), finishing the proof of Theorem 2.4.

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Tóth, L. Another generalization of Euler’s arithmetic function and Menon’s identity. Ramanujan J (2021). https://doi.org/10.1007/s11139-020-00353-z

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Keywords

  • Euler’s arithmetic function
  • Menon’s identity
  • Asymptotic formula

Mathematics Subject Classification

  • 11A07
  • 11A25
  • 11N37