1 Introduction

An arithmetical function is a complex valued function defined on the set of positive integers. An arithmetical function f is said to be multiplicative if \(f(1)=1\) and

$$\begin{aligned} f(mn)=f(m)f(n) \end{aligned}$$
(1)

whenever \((m, n)=1\). Multiplicative functions constitute perhaps the most important class of arithmetical functions. There are in the literature various superclasses and subclasses of multiplicative functions, see e.g. [7, 10, 12, 14, 15]

A multiplicative function f is completely multiplicative if (1) holds for all m, n. The power function \(N_k(n)=n^k\) is an example of completely multiplicative functions. The function \(\lambda _k\) is another example of completely multiplicative functions, where \(\lambda _k(n)=k^{\Omega (n)}\) and \(\Omega (n)\) is the total number of prime factors of n with \(\Omega (1)=0\). See [16].

A multiplicative function f is strongly multiplicative if \(f(p^a)=f(p)\) for all primes p and integers a \((\ge 1)\), see [11, 12]. The function \(E_k\) is an example of strongly multiplicative functions, where \(E_k(n)=k^{\omega (n)}\) and \(\omega (n)\) is the number of distinct prime factors of n with \(\omega (1)=0\).

A multiplicative function is over-multiplicative if there exists an arithmetical function F with \(F(1)=1\) such that

$$\begin{aligned} f(mn)=f(m)f(n)F((m, n)) \end{aligned}$$
(2)

for all positive integers m, n, see [12]. Euler’s totient function \(\phi (n)\) is defined as the number of integers \(x\pmod {n}\) with \((x, n)=1\). Euler’s totient function \(\phi \) possesses the property

$$\begin{aligned} \phi (mn)\phi ((m, n))=\phi (m)\phi (n)(m, n) \end{aligned}$$
(3)

for all positive integers m, n, see [2]. Dedekind’s totient \(\psi (n)\) is defined as

$$\begin{aligned} \psi (n)=n\prod _{p\mid n}\left( 1+\frac{1}{p}\right) , \end{aligned}$$

where the product is over the distinct primes p dividing n. Dedekind’s totient \(\psi \) satisfies the arithmetical equation

$$\begin{aligned} \psi (mn)\psi ((m, n))=\psi (m)\psi (n)(m, n) \end{aligned}$$
(4)

for all positive integers m, n, see [8]. Therefore the functions \(\phi \) and \(\psi \) are over-multiplicative with \(F(n)=n/\phi (n)\) and \(F(n)=n/\psi (n)\).

Equation (2) is closely related to

$$\begin{aligned} f(mn)f((m, n))=f(m)f(n)g((m, n)), \end{aligned}$$
(5)

see [3, 8]. We consider this equation at the end of this paper.

The Dirichlet convolution of two arithmetical functions f and g is defined as

$$\begin{aligned} (f\star g)(n)=\sum _{d|n} f(d)g(n/d). \end{aligned}$$

The function \(\delta \), defined as \(\delta (1)=1\) and \(\delta (n)=0\) otherwise, serves as the identity under the Dirichlet convolution. An arithmetical function f possesses a Dirichlet inverse \(f^{-1}\) if and only if \(f(1)\ne 0\). The Dirichlet inverse of a completely multiplicative function f is of the form \(f^{-1}=\mu f\), where \(\mu \) is the Möbius function.

A multiplicative function f is said to be a totient if there exist completely multiplicative functions \(f_t\) and \(f_v\) such that \(f=f_t\star f_v^{-1}\). See [6, 10, 14, 16]. Totients can be characterized with various arithmetical equations, see [6]. For example, an arithmetical function f is a totient if and only if there is a completely multiplicative function h such that

$$\begin{aligned} f(mn) = f(m) \sum _{ \genfrac{}{}{0.0pt}2{d\mid n}{\gamma (d)\mid m} } f(n/d)h(d) \end{aligned}$$
(6)

for all positive integers m and n, where \(\gamma \) is the strongly multiplicative function with \(\gamma (p)=p\) for all primes p. In this case \(f_v=h\).

It is well known that Euler’s totient function \(\phi \) can be written as

$$\begin{aligned} \phi =N*\mu =N*\zeta ^{-1}, \end{aligned}$$

where \(N(n)=n\) and \(\zeta (n)=1\) for all positive integers n. Thus \(\phi \) is a totient in the sense of the above definition with \(\phi _t=N\) and \(\phi _v=\zeta \).

Dedekind’s totient \(\psi \) can be written as \(\psi =N*\vert \mu \vert \). It is another example of a totient, since \(\vert \mu \vert =\lambda ^{-1}\), where \(\vert \mu \vert (n)=\vert \mu (n)\vert \) and \(\lambda \) is Liouville’s function, which is a completely multiplicative function such that \(\lambda (p)=-1\) for all primes p. Note that \(\lambda =\lambda _{-1}\).

A totient f is said to be a level totient if \(f_t=\zeta \). See [6, 16]. The functions \(E_k\) are examples of level totients. In fact, it can be verified that \(E_k=E_1\star \lambda _{1-k}^{-1}=\zeta \star \lambda _{1-k}^{-1}\). See [16].

Totients belong to the class of rational arithmetical functions. In fact, totients are rational arithmetical functions of order (1, 1). See [9, 16].

We denote by \(\mathcal {C}, \mathcal {S}, \mathcal {O}, \mathcal {T}\), and \(\mathcal {L}\), respectively, the class of completely multiplicative functions, the class of strongly multiplicative functions, the class of over-multiplicative functions, the class of totients, and the class of level totients. The symbol \(\mathcal{C}\mathcal{L}\) refers to the class of usual products of completely multiplicative functions and level totients. For a class \(\mathcal {A}\) of arithmetical functions let \(\mathcal {A}^\bullet \) denote the class of those arithmetical functions \(f\in \mathcal {A}\) for which \(f(n)\ne 0\) for all n. In this paper we show that \(\mathcal {S}=\mathcal {L}\), \(\mathcal {O}=\mathcal {T}\), \(\mathcal {L}\subsetneq \mathcal{C}\mathcal{L}\subsetneq \mathcal {T}\) and \(\mathcal {L}^\bullet \subsetneq (\mathcal{C}\mathcal{L})^\bullet =\mathcal {T}^\bullet \).

2 Results

Theorem 2.1

\(\mathcal {S}=\mathcal {L}\).

Proof

Suppose that \(f\in \mathcal {S}\). Then \(f(p^a)=f(p)\) for all primes p and integers a \((\ge 1)\). Let \(f_v\) be a completely multiplicative function such that \(f_v(p)=1-f(p)\) for all primes p. Then \((\zeta \star f_v^{-1})(p^a)=(\zeta \star (\mu f_v))(p^a) =1-f_v(p)=f(p)=f(p^a)\) for all primes p and integers a \((\ge 1)\). Thus \(f=\zeta \star f_v^{-1}\), which means that \(f\in \mathcal {L}\).

Suppose that \(f\in \mathcal {L}\). Then for all primes p and all integers \(a\ge 1\), \(f(p^a)=(\zeta \star f_v^{-1})(p^a)=(\zeta \star (\mu f_v))(p^a)=1-f_v(p)\), which does not depend on a. Thus \(f(p^a)=f(p)\), that is, \(f\in \mathcal {S}\). \(\square \)

Proposition 2.1

(See [6]) A multiplicative function f is a totient if and only if for each prime p there exists a complex number z(p) such that

$$\begin{aligned} f(p^a)=f(p)\left[ z(p)\right] ^{a-1} \end{aligned}$$
(7)

for all \(a\ge 1\). In this case \(z(p)=f_t(p)\).

Theorem 2.2

\(\mathcal {O}=\mathcal {T}\).

Proof

Suppose that \(f\in \mathcal {O}\). Then there exists an arithmetical function F such that \( f(mn)=f(m)f(n)F((m, n)) \) for all m, n. Let \(m=p^{a-1}\) and \(n=p\), where p is a prime and a is an integer \((\ge 2)\). Thus \(f(p^a)=f(p^{a-1})f(p)F(p)\). Applying this recursion we obtain \(f(p^a)=f(p)\left[ f(p)F(p)\right] ^{a-1}\) for all primes p and integers a \((\ge 1)\). Thus, according to Proposition 2.1, \(f\in \mathcal {T}\).

Suppose that \(f\in \mathcal {T}\). Then, according to Proposition 2.1,

$$\begin{aligned} f(p^a)=f(p)\left[ f_t(p)\right] ^{a-1} \end{aligned}$$
(8)

for all primes p and integers a \((\ge 1)\). Let F be a multiplicative function such that

$$\begin{aligned} F(p^a) =\left\{ \begin{array}{ll} \frac{f_t(p)}{f(p)} &{} \text{ if } f(p)\ne 0,\\ 0 &{} \text{ if } f(p)=0 \end{array} \right. \end{aligned}$$
(9)

for all primes p and integers a \((\ge 1)\). We show that (2) holds. Since f and F are multiplicative, it suffices to show that

$$\begin{aligned} f(p^{a+b})=f(p^{a})f(p^{b})F(p^{\min \{a, b\}}) \end{aligned}$$
(10)

for all primes p and integers a, b \((\ge 0)\). If \(a=0\) or \(b=0\), then (10) holds. Suppose that \(a\ne 0\) and \(b\ne 0\). We distinguish two cases: \(f(p)=0\), \(f(p)\ne 0\).

If \(f(p)=0\), then, according to (8), \(f(p^{a+b})=f(p^{a})=f(p^{b})=0\), and thus (10) holds. If \(f(p)\ne 0\), then, according to (8) and (9),

$$\begin{aligned} f(p^{a+b})= & {} f(p)\left[ f_t(p)\right] ^{a+b-1} =f(p)\left[ f_t(p)\right] ^{a-1}f(p)\left[ f_t(p)\right] ^{b-1}{\frac{f_t(p)}{f(p)}}\\= & {} f(p^{a})f(p^{b})F(p^{\min \{a, b\}}), \end{aligned}$$

and thus (10) holds.

Now, we have proved that (10) holds. Thus (2) holds, that is, \(f\in \mathcal {O}\). \(\square \)

Remark

It is easy to see that Equations (1)–(6) are recursive in character. For example, for a recursive character of Equation (2), see the proof of Theorem 2.2. The function values are totally determined by certain “initial values”. It is easy to see and well known that a multiplicative function is totally determined by its values at prime powers, and a completely multiplicative function is totally determined by its values at primes. A strongly multiplicative function is likewise totally determined by its values at primes. According to Proposition 2.1, a totient f is totally determined by the values of f and \(f_t\) at primes. It can be shown that a totient f is also totally determined by the values of f and \(f_v\) at primes or by the values of \(f_t\) and \(f_v\) at primes. A level totient f is totally determined by the values of f (or \(f_v\)) at primes. From the proof of Theorem 2.2 we see that an over-multiplicative function f is totally determined by the values of f and F at primes.

Theorem 2.3

\(\mathcal {L}\subsetneq \mathcal{C}\mathcal{L}\subsetneq \mathcal {T}\).

Proof

Since \(\zeta \in \mathcal {C}\), it follows that \(\mathcal {L}\subseteq \mathcal{C}\mathcal{L}\). It is clear that \(\phi \not \in \mathcal {S}=\mathcal {L}\), by Theorem 2.1. However, \(\phi =N\star \mu =N(\zeta \star \mu \frac{1}{N} )\), where \(N, \frac{1}{N}\in \mathcal {C}\). Thus \(\phi \in \mathcal{C}\mathcal{L}\) and therefore \(\mathcal {L}\) is a proper subclass of \(\mathcal{C}\mathcal{L}\).

Assume that \(f\in \mathcal{C}\mathcal{L}\). Then \(f=g(\zeta \star \mu h)\), where \(g, h\in \mathcal {C}\). Thus \(f=g\star (\mu gh)\), where \(g, gh\in \mathcal {C}\), and therefore \(f\in \mathcal {T}\). This proves that \(\mathcal{C}\mathcal{L}\subseteq \mathcal {T}\). Next we show that \(\mu \in (\mathcal {T}\setminus \mathcal{C}\mathcal{L})\). Since \(\mu =\delta \star \mu \zeta \), where \(\delta , \zeta \in \mathcal {C}\), we have \(\mu \in \mathcal {T}\). Assume that \(\mu \in \mathcal{C}\mathcal{L}\), that is, \(\mu \in \mathcal{C}\mathcal{S}\), by Theorem 2.1. Then \(\mu =gh\), where \(g\in \mathcal {C}, h\in \mathcal {S}\), and thus for each prime p, \(g(p)h(p)=-1\) and \(g(p^2)h(p^2)=g(p)^2h(p)=0\), which is impossible. Therefore \(\mu \not \in \mathcal{C}\mathcal{L}\). So we have proved that \(\mu \in (\mathcal {T}\setminus \mathcal{C}\mathcal{L})\) and further that \(\mathcal{C}\mathcal{L}\) is a proper subclass of \(\mathcal {T}\). \(\square \)

Theorem 2.4

\(\mathcal {L}^\bullet \subsetneq (\mathcal{C}\mathcal{L})^\bullet =\mathcal {T}^\bullet \).

Proof

From Theorem 2.3 we can conclude that \(\mathcal {L}^\bullet \subseteq (\mathcal{C}\mathcal{L})^\bullet \). Since \(\phi \in (\mathcal{C}\mathcal{L})^\bullet \setminus \mathcal {L}^\bullet \), we see that \(\mathcal {L}^\bullet \) is a proper subclass of \((\mathcal{C}\mathcal{L})^\bullet \).

From Theorem 2.3 we also can conclude that \((\mathcal{C}\mathcal{L})^\bullet \subseteq \mathcal {T}^\bullet \). We prove that \(\mathcal {T}^\bullet \subseteq (\mathcal{C}\mathcal{L})^\bullet \). Assume that \(f\in \mathcal {T}^\bullet \), that is, \(f\in \mathcal {T}\) and \(f(n)\ne 0\) for all n. Since \(f(p^a)=f_t(p^a)-f_t(p^{a-1})f_v(p) =f_t(p)^{a-1}(f_t(p)-f_v(p))\), we see that \(f_t(p)\ne 0\) for all primes p. It is thus possible to define a completely multiplicative function g as \(g(p)=f_v(p)/f_t(p)\) for all primes p. Then \([f_t(\zeta \star \mu g)](p^a)=f_t(p)^a[1-f_v(p)/f_t(p)] =f_t(p^a)-f_t(p^{a-1})f_v(p)=f(p^a)\) for all primes p and integers a \((\ge 1)\). Thus \(f=f_t(\zeta \star \mu g)\in (\mathcal{C}\mathcal{L})^\bullet \). This proves that \(\mathcal {T}^\bullet \subseteq (\mathcal{C}\mathcal{L})^\bullet \) and further that \((\mathcal{C}\mathcal{L})^\bullet =\mathcal {T}^\bullet \). \(\square \)

Remark

A problem related to Equation (2) is to characterize the arithmetical functions f with \(f(1)=1\) satisfying Equation (5) for all positive integers m, n, where g is a completely multiplicative function. In fact, Apostol and Zuckerman [3] have shown that an arithmetical function f with \(f(1)=1\) satisfies (5) if and only if f is multiplicative and

$$\begin{aligned} f(p^{a+b}) f(p^{b})=f(p^{a})f(p^{b})g(p^{b}) \end{aligned}$$
(11)

for all primes p and integers \(a\ge b\ge 1\). Apostol and Zuckerman [3] assume that g is a completely multiplicative function. Their result holds even more generally, namely if g is a multiplicative function, see [13].

We obtain a more illustrative result if we assume that f possesses Property O which is defined as follows: an arithmetical function f satisfies Property O if for each prime p, \(f(p)=0\) implies \(f(p^a)=0\) for all \(a>1\). Under this condition, (5) is a characterization of totients if g is a completely multiplicative function. See [8]. If f is always nonzero, then (5) reduces to (2) with \(F=g/f\) and again, (5) is a characterization of totients.

Equation (5) has been studied in [1, 2, 5, 6]. For further material relating to this type of equations we refer to [4, 13].