Are there multiplicative functions which are not rational?

Question

Vaidyanathaswamy calls an arithmetic function rational if it is the convolution of some finite collection of functions which are either completely multiplicative or inverse to a completely multiplicative function. Since multiplicative functions are closed under convolution and (Dirichlet) inversion, rational functions are all multiplicative. But are there multiplicative functions which are not rational? I expect so, but I haven't developed good intuition for convolutions.

Answer 1

Not all multiplicative functions are rational. For simplicity take arithmetic functions with complex values. It is easy to show that $f$ is rational of order $(m,n)$ (meaning the Dirichlet product of $m$ completely multiplicative and and $n$ inverse completely multiplicative functions) if and only if for every prime $q$ there exist $c_1(q), \ldots, c_m(q) \in \mathbb{C}$ such that $$f(q^k) = \sum_{i = 1}^m c_i(q) f(q^{k-i}) \mbox{ for all } k > n,$$ where we put $f(q^j) = 0$ if $j < 0$. (See Proposition 10.3 of my paper "Ring structures on groups of arithmetic functions" in Journal of Number Theory, which generalizes it to any ring.) This means that the $f(q^k)$ for fixed $q$ have to satisfy a linear recurrence relation. All you need to do is for some prime $q$ choose any sequence $a_{1}, a_{2}, \ldots$ of complex numbers that does not satisfy any such recurrence relation and define $f(q^n) = a_{n}$ for all $n$, then define $f(p^n)$ however you like for $p \neq q$, and extend $f$ by multiplicativity. Moreover, any multiplicative arithmetic function that is not rational can be constructed in this way.

Alternatively, if you want to avail yourself of exponentiation of multiplicative arithmetic functions, note that if $f$ is multiplicative, then $f^\alpha$ (exponentiation in the sense of Dirchlet convolution) is defined and multiplicative for any complex number $\alpha$, but if $f$ is rational and not the identity $\epsilon$ under convolution, then $f^\alpha$ is rational implies that $\alpha \in \mathbb{Q}$, so $f^\alpha$ is multiplicative but not rational for all $\alpha \in \mathbb{C}-\mathbb{Q}$. See Proposition 9.1 of the same paper (which generalizes the result to integral domains $R$, using exponents in the universal binomial ring over $R$.)

Answer 2

Yes, assuming by "inverse" you mean Dirichlet inverse.

An arithmetic function $f(n)$ is multiplicative iff its Dirichlet series at least formally admits an Euler product

$$\sum_{n=1}^{\infty} \frac{f(n)}{n^s} = \prod_p \left( \sum_{k=0}^\infty \frac{f(p^k)}{p^{ks}} \right).$$

It is completely multiplicative if in addition each factor in the Euler product takes the special form

$$\sum_{k=0}^\infty \frac{f(p^k)}{p^{ks}} = \frac{1}{1 - \frac{f(p)}{p^s}}.$$

If $f$ is rational, then each factor in the Euler product is a rational function of $p^s$ (presumably this is the motivation behind the terminology), but it's easy to pick some arbitrary Euler factors which don't have this property.