A note on multiplicative automatic sequences

TL;DR

This paper proves that all completely multiplicative functions that are automatic are essentially Dirichlet characters, confirming a conjecture and answering a question in the intersection of automata theory and number theory.

Contribution

It establishes that q-automatic completely multiplicative functions must coincide with Dirichlet characters, providing a significant link between automata and multiplicative number theory.

Findings

Any q-automatic completely multiplicative function is a Dirichlet character.

Confirms a conjecture by Bell, Bruin, and Coons.

Answers a question posed by Allouche and Goldmakher.

Abstract

We prove that any $q$-automatic completely multiplicative function $f:\mathbb{N}\to\mathbb{C}$ essentially coincides with a Dirichlet character. This answers a question of J. P. Allouche and L. Goldmakher and confirms a conjecture of J. Bell, N. Bruin and M. Coons for completely multiplicative functions. Further, assuming two standard conjectures in number theory, the methods allows for removing the assumption of completeness.

Equations

$$K_q(f)=\left\{\{f(q^{i}n+r{\}}_{n\ge 0}|i\ge 1,0\le r\le q^i-1\right\}$$

$$\frac{f(q^{i_1}m{\prod }_{p\in {\mathcal{M}}_0}p+1)}{f(q^{i_2}m{\prod }_{p\in {\mathcal{M}}_0}p+1)}=1\ne 0,$$

$$|\{p\le x:p\equiv u(\mathrm{mod}v),(p-1)/K=P_2(\alpha ,\delta )|\gg x/(\log x{)}^2,$$

Full text

1. Introduction

Automatic sequences play important role in computer science and number theory. For a detailed account of the theory and applications we refer the reader to the classical monograph [AS03]. One of the applications of such sequences in number theory stems from a celebrated theorem of Cobham [Cob72], which asserts that in order to show the transcendence of the power series $\sum_{n\geq 1}f(n)z^{n}$ it is enough to establish that the function$f:\mathbb{N}\to\mathbb{C}$ is not automatic. In this note, rather than working within the general set up, we confine ourselves to functions with the range in $\mathbb{C}.$ There are several equivalent definitions of automatic (or more precisely, $q$-automatic) sequences. It will be convenient for us to use the following one.

Definition 1.1**.**

The sequence $f:\mathbb{N}\to\mathbb{C}$ is called $q$-automatic if the $q$-kernel of it defined as a set of subsequences

[TABLE]

is finite.

We remark that any $q-$automatic sequence takes only finitely many values, since it is a function on the states of finite automata. A function $f:\mathbb{N}\to\mathbb{C}$ is called completely multiplicative if $f(mn)=f(m)f(n)$ for all $m,n\in\mathbb{N}.$ The question of which multiplicative functions are $q$-automatic has been the subject of study by several authors including [Yaz01], [SP11], [BBC12], [SP03], and [AG18]. In particular, the following conjecture was made in [BBC12].

Conjecture 1.2** (Bell-Bruin-Coons).**

For any multiplicative $q$-automatic function $f:\mathbb{N}\to\mathbb{C}$ there exists eventually periodic function $g:\mathbb{N}\to\mathbb{C},$ such that $f(p)=g(p)$ for all primes $p.$

This conjecture is still open in general, although some progress has been made when $f$ is assumed to be completely multiplicative. In particular, Schlage-Puchta [SP11] showed that a completely multiplicative $q$-automatic sequence which does not vanish is almost periodic. Hu [Hu17] improved on that result by showing that the same conclusion holds under a slightly weaker hypothesis. Our first result confirms a strong form of Conjecture 1.2 when $f$ is additionally assumed to be completely multiplicative function.

Theorem 1.3**.**

Let $q\geq 2$ and let $f:\mathbb{N}\to\mathbb{C}$ be completely multiplicative $q$-automatic sequence. Then, there exists a Dirichlet character of conductor $Q$ such that either $f(n)=\chi(n),$ for all $(n,Q)=1$ or $f(p)=0$ for all sufficiently large $p.$

We remark that similar result has been very recently obtained independently by Li [Li] using combinatorial methods relying on the techniques developed in the theory of automatic sequences. Our proof is shorter and builds upon two deep number theoretic results. Further, assuming the generalized Riemann hypothesis (which in particular implies a strong form of the Artin primitive root conjecture for primes in progressions) together with the set of base-$q$ Wieferich primes having density zero, our method can be adapted to show the full conjecture (i.e. the assumption on complete multiplicativity can be removed.)

2. Proof of the main result

We begin with a simple albeit important remark. Since $f$ is $q$-automatic the image of $f:\mathbb{N}\to\mathbb{C}$ is finite and therefore for any prime $p,$ $f(p)=0$ or $f(p)$ is a root of unity.

Proposition 2.1**.**

Let $f:\mathbb{N}\to\mathbb{C}$ be a $q$-automatic completely multiplicative function and let $\mathscr{M}_{0}=\left\{p|\ f(p)=0\right\}.$ If $|\mathscr{M}_{0}|<\infty,$ then $f(p)=\chi(p)$ for all $p\notin\mathscr{M}_{0}.$

Proof.

Since $f$ is $q$-automatic there exist positive integers $i_{1}\neq i_{2},$ such that $f(q^{i_{1}}n+1)=f(q^{i_{2}}n+1)$ for all $n\geq 1.$ If $n=m\prod_{p\in\mathscr{M}_{0}}p,$ then

[TABLE]

for all $m\geq 1.$ The conclusion now immediately follows from Theorem $2$ of [EK17].∎

Let $\overline{1,n}=[1,n]\cap\mathbb{Z}.$ Since $f$ is $q$-automatic, there exists $k_{0}=k_{0}(f),$ such that for all $i\geq 1$ and $0\leq r\leq q^{i}-1,$ the equalities $f(q^{i}n+r)=0$ for $n\in\overline{1,k_{0}}$ imply $f(q^{i}n+r)=0$ for all $n\geq 1.$

Lemma 2.2**.**

Suppose that $|\mathscr{M}_{0}|=\infty.$ For any $q,k_{0}<p_{1},p_{2},\dots,p_{k_{0}}\in\mathscr{M}_{0},$ there exists $r=r(q,p_{1},\ldots,p_{k_{0}})$ such that $(r,qp_{i})=1$ for all $i\in\overline{1,k_{0}}$ and $f(n\prod_{i\leq k_{0}}p_{i}+r)=0$ for all $n\geq 1.$ We may further assume that $r\equiv 3\pmod{16}$, and $(r-1,\prod_{i\leq k_{0}}p_{i})=1.$

Proof.

For an integer parameter $A\geq\log_{q}p_{k_{0}},$ which we shall choose later, by the Chinese remainder theorem there exists $r_{A}$ such that $(r_{A},q)=1$ and $r_{A}\equiv-sq^{2A}\pmod{p_{s}}$ for all $s\in\overline{1,k_{0}}.$ Since $p_{s}|q^{2A}s+r_{A}$ we have $f(q^{2A}n+r_{A})=0$ for all $n\in\overline{1,k_{0}}.$ The latter implies that $f(q^{2A}n+r_{A})=0$ for all $n\geq 1.$ We claim that $f(r_{A})=0.$ Indeed, if this is not the case we choose a prime $p$, such that $f(p)=1$ and consider $m=p^{\phi(q^{2A})}r_{A}.$ Clearly $m\equiv r_{A}\pmod{q^{2A}}$ and consequently $0=f(m)=(f(p))^{\phi(q^{2A})}f(r_{A})=1$, a contradiction. Note, that the same argument works for $n\prod_{i\leq k_{0}}p_{i}+r_{A}$ in place of $r_{A}$ and therefore we conclude that $f(n\prod_{i\leq k_{0}}p_{i}+r_{A})=0$ for all $n\geq 1.$ Setting $r=r_{A}$ finishes the proof. ∎

Next, without loss of generality we may assume that there exist three sufficiently large primes $t,t^{\prime},t^{\prime\prime}>\max(q,k_{0})$ such that $f(t)=f(t^{\prime})=f(t^{\prime\prime})=1.$ We will require the following consequence of a result due to Heath-Brown [HB86].

Lemma 2.3**.**

Given distinct primes $t,t^{\prime},t^{\prime\prime}>\max(q,k_{0})$ and $r=r(q,p_{1},\ldots,p_{k_{0}})$ as in Lemma 2.2, there exists infinitely many primes $q_{i}\equiv r\pmod{16\prod_{i\leq k_{0}}p_{i}}$ such that at least one of $t,t^{\prime},t^{\prime\prime}$ (say $t$) is a primitive root modulo $q_{i}$. Moreover, by passing to a subsequence we may assume that for such primes $(q_{i}-1,q_{j}-1)=2$ for $i\neq j,$ and for each $\l\in\overline{1,k_{0}}$ we have $(\l/q_{i})=1$ for all $i\in\overline{1,k_{0}}$.

Proof.

Let $v=16(\prod_{i\leq k_{0}}p_{i})\prod_{2<p\leq k_{0}}p$ and chose $u$ such that $u\equiv 3\pmod{16}$ and $u\equiv r\pmod{\prod_{i\leq k_{0}}p_{i}}$, with $r$ as in Lemma 2.2. Moreover, by quadratic reciprocity we may further select $u\pmod{\prod_{2<p\leq k_{0}}p}$ such that $(u/p)=1$ for all primes $p\leq k_{0},$ and $(u-1,\prod_{2<p\leq k_{0}}p)=1.$ In particular, we have $(-3/p)=-1$ for any prime $p\equiv u\mod v$. Applying Lemma 3 of [HB86], with $u,v$ as above and $k=1$ (and $K=2^{k}=2$) there exists $\alpha\in(1/4,1/2]$ and $\delta>0$ such that

[TABLE]

with the implied constant possibly depending on $\alpha$, with $P_{2}(\alpha,\delta)$ denoting the union of the set of primes, together with the set of almost primes $n=t_{1}t_{2}$ with $t_{1}<t_{2}$ both primes, and $t_{1}\in[n^{\alpha},n^{1/2-\delta}]$. Heath-Brown’s argument then shows that at least one of $t,t^{\prime},t^{\prime\prime}$ is a primitive root for infinitely many primes $p\equiv u\pmod{v}$. Whether the primes $q_{i}$ produced have the properties that $(q_{i}-1)/2$ is prime, or that $(q_{i}-1)/2=t_{1}t_{2}$, we may pass to an infinite subsequence of primes $q_{1}<q_{2}<\ldots$ (satisfying $q_{1}>q$) so that $(q_{i}-1,q_{j}-1)=2$ for $i\neq j$ (for the latter case of almost primes, note that both $t_{1}$ and $t_{2}$ are growing.) ∎

Proposition 2.4**.**

Suppose that $|\mathscr{M}_{0}|=\infty.$ Then $f(p)=0$ for all sufficiently large primes $p.$

Proof.

Replacing $f$ by $|f|,$ which is also $q$-automatic, it is enough to prove the claim for the binary valued $f:\mathbb{N}\to\{0,1\}.$ By Lemma 2.3, we may select prime $t$ with $f(t)=1$, which is a primitive root modulo infinitely primes $q_{1}<q_{2}\dots<q_{k_{0}}$ (satisfying $q_{1}>\max(k_{0},q)$) such that $q_{i}\equiv r\pmod{16\prod_{j\leq k_{0}}p_{k_{0}}}$ and consequently $f(q_{i})=0.$ From the proof of Lemma 2.2 it follows that there exists $r_{A},$ such that $f(n\prod_{i\leq k_{0}}q_{i}+r_{A})=$ for all $n\geq 1.$ Since $t$ is a primitive root modulo $q_{j}$ for $j\in\overline{1,k_{0}}$, there exists $\gamma_{j}$ such that $t^{\gamma_{j}}\equiv r_{A}\pmod{q_{j}}$ for $j\in\overline{1,k_{0}}.$ By the construction and Lemma 2.3 we have $(r_{A}/q_{i})=(-iq^{2A}/q_{i})=-1$ and thus all $\gamma_{i}$ have the same parity. Consequently, by the Chinese remainder theorem we can choose $\gamma\in\mathbb{N},$ such that $\gamma\equiv\gamma_{j}\pmod{q_{j}-1}$ for all $j\in\overline{1,k_{0}}.$ For $\gamma$ defined this way we have $t^{\gamma}\equiv r_{A}\pmod{\prod_{j\leq k_{0}}q_{j}}.$ Hence, $f(t^{\gamma})$ must be zero. On the other hand $f(t^{\gamma})=f(t)^{\gamma}=1$, and this contradiction finishes the proof.∎

Combining Proposition 2.1 and Proposition 2.4 yields the conclusion of Theorem 1.3