A note on multiplicative automatic sequences, II

TL;DR

This paper proves that multiplicative automatic functions are either closely related to Dirichlet characters or become zero on large primes, confirming a conjecture in the field.

Contribution

It establishes a definitive characterization of q-automatic multiplicative functions, resolving a conjecture about their structure.

Findings

Such functions either match a Dirichlet character or vanish on large primes.

The result confirms a strong form of a conjecture by Bell, Bruin, and Coons.

Provides a complete classification of multiplicative automatic functions.

Abstract

We prove that any $q$-automatic multiplicative function $f:\mathbb{N}\to\mathbb{C}$ either essentially coincides with a Dirichlet character, or vanishes on all sufficiently large primes. This confirms a strong form of a conjecture of J. Bell, N. Bruin, and M. Coons.

Equations

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

$${\mathbb{P}}_{\_}>1=\{p\text{ prime}:|f(p^e)|>1\text{ for some e≥1}\}$$

$${\mathbb{P}}_{\_}<1=\{p\text{ prime}:|f(p^e)|<1\text{ for some e≥1}\},$$

$$\frac{f(q^{i_{\_}1}m{\prod }_{\_}p\in {\mathbb{P}}_{\_}>1\cup {\mathbb{P}}_{\_}<1\cup {\mathbb{P}}_{0}p+1)}{f(q^{i_{\_}2}m{\prod }_{\_}p\in {\mathbb{P}}_{\_}>1\cup {\mathbb{P}}_{\_}<1\cup {\mathbb{P}}_{0}p+1)}=1\ne 0,$$

$$\widehat{{\mathbb{P}}_1}:=\{p\text{ prime}:f(p^e)=1\text{ for some e≥1}\}.$$

$${\alpha }_{\_}p:=\min \{{\alpha }_{\_}p,\delta :f(p^{\delta })=1\},$$

$${\delta }_{\_}p:=\min \{\delta :{\alpha }_{\_}p,\delta ={\alpha }_{\_}p\}.$$

$$f(q^{i}m+r)=f(q^{j}m+r)\forall m\in \mathbb{N},$$

$$R:=\prod _{\_}rq<p_{\_}i\in S,i\le \varphi (q^{A+s_{\_}0})p_{\_}^{e_{\_}{p}_{\_}i}i\equiv r_{\_}{S}^{\varphi (q^{A+s_{\_}0})}\equiv 1\mathrm{mod}{q}^{A+s_{\_}0}$$

$$f(q^{A+Cn}{m}_{\_}0+r)=1\forall n\in \mathbb{N}.$$

$$f(q^{i}m+r)=f(q^{j}m+r)\forall m\in \mathbb{N}.$$

$${\nu }_{\_}p(q^{C{\mathrm{ord}}_{\_}p(q^C)n}-1)={\nu }_{\_}p(q^{C{\mathrm{ord}}_{\_}p(q^C)}-1)+{\nu }_{\_}p(n)=\gamma +{\nu }_{\_}p(n).$$

$$q^{A+C(n_{\_}\delta +n{\mathrm{ord}}_{\_}p(q^C))}{m}_{\_}0+r=(q^{A+C{n}_{\_}\delta }{m}_{\_}0+r)q^{nC{\mathrm{ord}}_{\_}p(q^C)}-r(q^{nC{\mathrm{ord}}_{\_}p(q^C)}-1).$$

$$p^{\alpha +\gamma }||q^{A+C(n_{\_}\delta +p^{\alpha }{\mathrm{ord}}_{\_}p(q^C))}{m}_{\_}0+r$$

$$p^{\delta }||q^{A+C(n_{\_}\delta +p^{\delta -\gamma +1}{\mathrm{ord}}_{\_}p(q^C))}{m}_{\_}0+r_{\_}0.$$

$$q^{t{p}^{\delta -\gamma }C{\mathrm{ord}}_{\_}p(q^C)}=(1+p^{\delta }B{)}^t=1+t{p}^{\delta }B+p^{2\delta }D.$$

$$q^{A+C(n_{\_}\delta +n_{\_}t{\mathrm{ord}}_{\_}p(q^C))}{m}_{\_}0+r\equiv (q^{A+C{n}_{\_}\delta }{m}_{\_}0+r)(1+tB{p}^{\delta })-rtB{p}^{\delta }\mathrm{mod}{q}^{\delta +1}.$$

$$p|rtB{t}_{\_}\delta +1-\frac{(q^{A+C{n}_{\_}\delta }{m}_{\_}0+r)}{p^{\delta }}$$

$$|\{p\in \widehat{{\mathbb{P}}_1}:{\alpha }_{\_}p=\alpha \}|=\infty ,$$

$$|\{p\in \widehat{{\mathbb{P}}_1}:{\alpha }_{\_}p\le \beta \}|<\infty .$$

$$M=M_{\_}\beta :=\{p\in \widehat{{\mathbb{P}}_1}:(p,q)=1,p>k_{\_}0,{\alpha }_{\_}p\le \beta \}$$

$$L=L_{\_}\beta :=\{p\in \widehat{{\mathbb{P}}_1}:(p,q)=1,p>k_{\_}0,{\alpha }_{\_}p>\beta \}.$$

$$r=\prod _{\_}p\in S{p}^{e_{\_}p}\cdot \prod _{\_}p\in M{p}^{e_{\_}p}$$

$$q^{C\Delta }\not\equiv 1\mathrm{mod}{p}_{\_}i$$

$$r_{\_}i:=\frac{r}{p_{\_}{i}^{e_{\_}{p}_{\_}i}}\cdot$$

$$k_{\_}i=\prod _{\_}p{p}^{f_{\_}p,i}=k_{\_}{i}^S{k}_{\_}{i}^M{k}_{\_}{i}^L$$

$$(r_{\_}j/k^S{)}^{\varphi (q)}:(r_{\_}i/k^S{)}^{\varphi (q)}=(p_{\_}{i}^{e_{\_}{p}_{\_}i}/p_{\_}{j}^{e_{\_}{p}_{\_}j}{)}^{\varphi (q)}$$

$$(r_{\_}i/k^S{)}^{\varphi (q)}\notin 1+q^{\beta +1}{\mathbb{Z}}_{\_}q.$$

$$Q:=\prod _{\_}i\le k_{\_}0p_{\_}i.$$

$$r_{\_}A\equiv -s{q}^A+p_{\_}s\mathrm{mod}{p}_{\_}^{2}s,1\le s\le k_{\_}0.$$

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 multiplicative if $f(mn)=f(m)f(n)$ for all pairs $(m,n)=1.$ The question of which multiplicative functions are $q$-automatic attracted considerable attention of several authors including [Yaz01], [SP11], [BBC12], [SP03], [Li], [KK] 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 an eventually periodic function $g:\mathbb{N}\to\mathbb{C},$ such that $f(p)=g(p)$ for all primes $p.$

Some progress towards this conjecture 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. Allouche and Goldmakher [AG18] considered related question classifying what they called “mock” Dirichlet characters. Finally Li [Li] and the authors [KK] very recently proved Conjecture 1.2 when $f$ is additionally assumed to be completely multiplicative. Moreover, the methods of [KK] settled the Conjecture 1.2 fully under the assumption of the Generalized Riemann Hypothesis. The key tool there was a variant of Heath-Brown’s results [HB86] on Artin’s primitive root conjecture. In this article, we develop alternative, more combinatorial approach, and prove a strong form of Conjecture 1.2.

Theorem 1.3**.**

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

In fact, our proof yields a more refined information about the set of values of $f$ (cf. Remark 2.8.) We remark that J. Koneczny, in forthcoming work (private comunication), established a variant of Theorem 1.3 using different methods relying on the structure theory of automatic sequences.

2. Proof of the main result

Let $\mathbb{P}_{0}$ the set of primes for which $f(p)=0.$ Since $f$ is $q$-automatic, it is well-known that the image of $f:\mathbb{N}\to\mathbb{C}$ is finite and therefore if we define the sets

[TABLE]

and

[TABLE]

then, from multiplicativity of $f$ we easily deduce $|\mathbb{P}_{\_}{>1}|,|\mathbb{P}_{\_}{<1}|<\infty.$ We begin by assuming additionally that $|\mathbb{P}_{0}|<\infty.$

Proposition 2.1**.**

Let $f:\mathbb{N}\to\mathbb{C}$ be a $q$-automatic multiplicative function and suppose that $|\mathbb{P}_{0}|<\infty.$ Then $f(p^{k})=\chi(p^{k})$ for all $p\notin\mathbb{P}_{\_}{>1}\cup\mathbb{P}_{\_}{<1}\cup\mathbb{P}_{0}$ and any $k\geq 1.$

Proof.

Since $f$ is $q$-automatic, we have that the kernel $K_{\_}q(f)$ is finite. By the pigeonhole principle, 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 $n=m\prod_{\_}{p\in\mathbb{P}_{\_}{>1}\cup\mathbb{P}_{\_}{<1}\cup\mathbb{P}_{0}}p$ then

[TABLE]

for all $m\geq 1.$ The conclusion now immediately follows from Theorem $2$ of [EK17] (the result is stated only for completely multiplicative functions, but the proof is in fact also valid for multiplicative functions). Alternatively, one could also apply correlation formulas developed in [Klu17] to get the result. ∎

In what follows we will assume that $|\mathbb{P}_{0}|=\infty$ and show that in this case $f(p)=0$ for all sufficiently large primes $p.$ To this end, we perform the following two reductions:

• •

replace $f$ by $g=|f|$ which is also $q$-automatic

• •

change each value $|f(p^{k})|\neq\{0,1\}$ to $g(p^{k}):=1.$

These two operations preserve $q-$ automaticity of the modified multiplicative sequence $\{g(n)\}_{\_}{n\geq 1}$ and do not change the zero set $\mathbb{P}_{0}.$ It is therefore enough to prove the claim for the binary valued $f:\mathbb{N}\to\{0,1\}.$

To facilitate our discussion, we introduce the set

[TABLE]

Let $s_{\_}{0}<\infty$ denote the number of distinct sequences $\{f(q^{i}m+r)\}_{\_}{m\in\mathbb{N}}$ as $i,r\geq 0$ ranges over pairs of integers such that $r\in[0,q^{i}).$ Given prime $p$ and $\delta\geq 1,$ we define $\alpha_{\_}{p,\delta}$ by $q^{\alpha_{\_}{p,\delta}}||p^{\delta\phi(q)}-1$. Further, given $p\in\widehat{\mathbb{P}_{1}}$, let

[TABLE]

and let

[TABLE]

We also note that if $f$ is $q$-automatic, then there exists $k_{\_}0=k_{\_}0(f)$ with the following property: if for some $i\geq 1$ and $0\leq r\leq q^{i}-1$ the equality $f(q^{i}n+r)=0$ holds for all integer $n\in[1,k_{\_}0]$, then $f(q^{i}n+r)=0$ for all $n\geq 1.$

2.1. Preparatory lemmas.

We start with a few technical lemmas.

Lemma 2.2**.**

Assume that $|\widehat{\mathbb{P}_{1}}|=\infty$ and let $r>0$ be an integer such that $f(r)=1.$ Select $A$ such that $q^{A}>r$. Then there exists $i,j\in(A,A+s_{\_}{0}]$ such that $0<j-i\leq s_{\_}{0}$ with the property that

[TABLE]

and $f(q^{i}m_{\_}{0}+r)=1$ for some integer $m_{\_}{0}\in[1,k_{\_}{0}]$.

Proof.

By our assumption, there exists an infinite subset $S\subset\widehat{\mathbb{P}_{1}}$ and $r_{\_}{S}\in\mathbb{N}$ such that $p^{e_{\_}p}\equiv r_{\_}S\mod q^{A+s_{\_}{0}}$ and $f(p^{e_{\_}p})=1$ for all $p\in S.$ Select

[TABLE]

and let $r_{\_}{1}=rR$. Clearly $r_{\_}{1}\equiv r\mod q^{A+s_{\_}{0}}$ and for any integer $\l\in(A,A+s_{\_}{0}]$ we can write $r_{\_}{1}=q^{\l}m_{\_}{\l}+r.$ Since the sequence $\{f(n)\}_{\_}{n\geq 1}$ is $q-$automatic, by the pigeonhole principle there exist two indices $i,j\in(A,A+s_{\_}{0}]$ (say with $i<j$) such that $f(q^{i}m+r)=f(q^{j}m+r)$ for all $m\in\mathbb{N}$. Moreover, $f(q^{i}m_{\_}{i}+r)=f(r_{\_}{1})=f(r)f(R)=1.$ Since $r_{\_}1$ can be made arbitrary large, we see that any sequence $\{f(q^{i}m+r)\}_{\_}{m\in\mathbb{N}},$ $i\in(A,A+s_{\_}0]$ contains infinitely many ones. Hence there exists some $m_{\_}{0}\leq k_{\_}{0}$ such that $f(q^{i}m_{\_}{0}+r)=f(q^{j}m_{\_}0+r)=1.$ ∎

We next show that $f$ is non-vanishing along an exponentially growing subsequence.

Corollary 2.3**.**

Assume that $|\widehat{\mathbb{P}_{1}}|=\infty.$ Given an integer $r>qk_{\_}0^{2}$ such that $f(r)=1$ there exist integers $A,C\geq 0$ and $1\leq m_{\_}{0}\leq k_{\_}{0}$, with $C\leq s_{\_}{0}$, with the property that

[TABLE]

Proof.

We apply Lemma 2.2 to find $i,j\in(A,A+s_{\_}0],$ such that

[TABLE]

Without loss of generality we may assume that $A=i$; for convenience set $C=j-i$, we then have $0<C\leq s_{\_}{0}.$ Note that $f(q^{A+C}m+r)=f(q^{i}(q^{j-i}m)+r)=f(q^{j}m+r)=f(q^{i}m+r)=f(q^{A}m+r)$ for all $m\in\mathbb{N}.$ An easy induction argument now shows that for any $n\in\mathbb{N}$ we have $f(q^{A+Cn}m+r)=f(q^{A}m+r)$ for all $m\in\mathbb{N}.$ Lemma 2.2 also yields an integer $m_{\_}{0}\in(0,k_{\_}{0}]$ such that $f(q^{A}m_{\_}{0}+r)=1.$ This concludes the proof. ∎

In order to better illustrate the main idea of our proof, we first focus on the case $q$ being prime and then point out necessary modifications needed to treat the general case in Section 2.3.

Lemma 2.4**.**

Assume that $f(q^{A+Cn}m_{\_}{0}+r)=1$ for all $n\in\mathbb{N}$, and that $p^{\delta}||q^{A+Cn_{\_}{\delta}}m_{\_}{0}+r$ for some $n_{\_}{\delta}\geq 0$ with $\delta>\gamma$ where $p^{\gamma}||(q^{C})^{\operatorname{ord}_{\_}p(q^{C})}-1$, and $p\nmid m_{\_}{0}$. Then, given any $k\geq\gamma$, there exist $n_{\_}{k}$ such that $p^{k}||q^{A+Cn_{\_}{k}}m_{\_}{0}+r$. In particular, $f(p^{k})=1$ for all $k\geq\gamma$, and there exists $\alpha\in\mathbb{N}$ such that for any $\alpha_{\_}{1}>\alpha$, any element in $1+q^{\alpha}\mathbb{Z}_{\_}{q}/1+q^{\alpha_{\_}{1}}\mathbb{Z}_{\_}q$ is of the form $p^{k}$ for some (arbitrarily large) $k\in\mathbb{N}$. (With $\mathbb{Z}_{\_}q$ denoting the $q$-adic integers, we identify the unit subgroup $\{u\in\mathbb{Z}/q^{\alpha_{\_}{1}}\mathbb{Z}:u\equiv 1\mod q^{\alpha}\}$ with $1+q^{\alpha}\mathbb{Z}_{\_}{q}/1+q^{\alpha_{\_}{1}}\mathbb{Z}_{\_}q$.)

Proof.

By the lifting exponent lemma (cf. [wik]) we have that for all $n\geq 1,$

[TABLE]

Therefore, we have

[TABLE]

Note that for any $0\leq\alpha<\delta-\gamma$ we therefore have

[TABLE]

and

[TABLE]

Select $n_{\_}t=tp^{\delta-\gamma}$ for some $1\leq t<p$ and write $q^{p^{\delta-\gamma}C\operatorname{ord}_{\_}p(q^{C})}=1+p^{\delta}B$ with $(B,p)=1.$ By the binomial theorem, we have

[TABLE]

Consequently,

[TABLE]

Select $t=t_{\_}{\delta+1},$ such that

[TABLE]

and note that for such defined $t_{\_}{\delta+1}$ we have $p^{\delta+1}|q^{A+C(n_{\_}\delta+n_{\_}{t_{\_}{\delta+1}}\operatorname{ord}_{\_}p(q^{C}))}m_{\_}0+r.$ Replacing now $\delta\to\delta+1$ in the statement of the lemma and running the same proof again the first claim follows. Moreover, by choosing appropriate $n_{\_}k$ such that for $k\geq\gamma$ we have $p^{k}||q^{A+Cn_{\_}k}m_{\_}0+r$, we ensure that $f(p^{k})=1.$ Finally, using the lifting exponent lemma together with the binomial theorem as before for the sequence $\{p^{n\phi(q)}\}_{\_}{n\geq 1}$ yields the last claim of the lemma. ∎

2.2. Concluding the proof.

We are ready to prove the first key proposition.

Proposition 2.5**.**

Let $f:\mathbb{N}\to\{0,1\}$ be multiplicative $q-$automatic sequence and suppose that $|\widehat{\mathbb{P}_{1}}|=\infty.$ Then

[TABLE]

for some $\alpha\in\mathbb{N}.$

Proof.

Suppose that $|\{p\in\widehat{\mathbb{P}_{1}}:\alpha_{\_}{p}=\alpha\}|<\infty$ for all $\alpha\in\mathbb{N}$, and consequently, for any $\beta\in\mathbb{N}$ we have

[TABLE]

Let $S=(\widehat{\mathbb{P}_{1}}\cap[1,k_{\_}{0}])\setminus\{q\},$ and given cut off parameter $\beta\in\mathbb{N},$ we define the sets of “medium” and “large” primes as follows:

[TABLE]

and

[TABLE]

By (2.3), we find that both sets $M$ and $S$ are finite.

Let

[TABLE]

with the exponents $e_{\_}{p}$ chosen as follows.

• •

For $p\in S$, take $e_{\_}{p}=0$ if the set of exponents $\{\delta:f(p^{\delta})=1\}$ is finite. Otherwise chose $e_{\_}{p}$ so that $p^{e_{\_}{p}}>k_{\_}{0}$ and $f(p^{e_{\_}{p}})=1.$ In both cases, there exists $D=O(1)$ with the property that for any $p\in S$ such that $p^{\delta}|q^{n}m_{\_}{0}+r$ for some $1\leq m_{\_}{0}\leq k_{\_}{0}$ and $n$ such that $f(q^{n}m_{\_}{0}+r)=1$, we have $\delta\leq D$.

• •

For $p\in M$, let $e_{\_}{p}=\delta_{\_}{p}$ and observe that $f(p^{e_{\_}{p}})=1$.

We first choose $\Delta$ such that $q^{\Delta}\equiv 1\mod\prod_{\_}{p\in S}p^{D}.$ We then select primes $p_{\_}{1},\ldots,p_{\_}{\l}\in M$ with $\l$ to be specified later, such that $\alpha_{\_}{p_{\_}{i}}<\alpha_{\_}{p_{\_}{i+1}}$, and so that for any $C\in[1,k_{\_}{0}]$,

[TABLE]

for all $i\in[1,l]$. Note that such primes exist if we take $\beta$ sufficiently large. Further, we can assume that $\beta$ is large enough so that $\alpha_{\_}{p_{\_}{l}}<\beta$. Finally, for $i\in\{1,\ldots,l\}$ we define

[TABLE]

We note that $f(r_{\_}{i})=1$ for $1\leq i\leq l.$ Corollary 2.3 now implies that there exists $A_{\_}{i},C_{\_}{i},m_{\_}{i}$ with $C_{\_}{i}\leq s_{\_}{0}$, $m_{\_}{i}\leq k_{\_}{0}$, and $f(q^{A_{\_}{i}+C_{\_}{i}n}m_{\_}{i}+r_{\_}{i})=1$ for all $n\in\mathbb{N}$. Now, if $k_{\_}{i}=k_{\_}{i}(n)=q^{A_{\_}{i}+C_{\_}{i}n}m_{\_}{i}+r_{\_}{i}$, for $n>\beta+1$, write

[TABLE]

where for a subset $X$ of primes, we set $k_{\_}{i}^{X}=\prod_{\_}{p\in X}p^{f_{\_}{p,i}}.$

For each $k_{\_}{i}^{S}$ we have $k_{\_}{i}^{S}=\prod_{\_}{p\in S}p^{f_{\_}{p,i}}$ with $f_{\_}{p,i}\leq D$. Thus the set of possible $k_{\_}{i}^{S}$’s is finite, and there exists $i\neq j$ such that $k_{\_}{i}^{S}=k_{\_}{j}^{S}=:k^{S}$ (provided $l$ is chosen large enough; note that $l$ only depends on the number of possible $k_{\_}{i}^{S}$, which in turn depends on $S$, and hence $l=O(1)$).

For such defined $i\neq j,$ by the construction

[TABLE]

and since $\alpha_{\_}{i}<\alpha_{\_}{j}<\beta,$ we have that at least one of $(r_{\_}{i}/k^{S})^{\phi(q)}$, $(r_{\_}{j}/k^{S})^{\phi(q)}$ does not belong to $1+q^{\beta+1}\mathbb{Z}_{\_}q.$ Without loss of generality we may assume

[TABLE]

Clearly $k_{\_}{i}$ is coprime to all $p\in M\setminus\{p_{\_}{i}\}$ and thus the only possibility is $k_{\_}{i}^{M}=p_{\_}{i}^{e}$ for some $e\geq 0$. We note that replacing $n$ by $n+\Delta$ does not change $k_{\_}i^{S}(n)=k_{\_}i^{S}(n+\Delta)=k_{\_}{i}^{S}$, whereas by the choice (2.4) $k_{\_}{i}$ cannot remain divisible by $p_{\_}{i},$ In particular we can chose $n$ so that $k_{\_}{i}^{S}=k^{S}$ and $k_{\_}{i}^{M}=1$. Hence, provided $n$ is large enough, we find that $r_{\_}{i}/k^{S}=k_{\_}{i}^{L}\mod q^{\beta+1}$, which contradicts (2.5) since $(k_{\_}{i}^{L})^{\phi(q)}\in 1+q^{\beta+1}\mathbb{Z}_{\_}q$. ∎

We also need the following simple lemma.

Lemma 2.6**.**

If $|\{p\in\widehat{\mathbb{P}_{1}}:\alpha_{\_}{p}=\alpha\}|=\infty$ for some $\alpha\in\mathbb{Z}^{+}$, then for any $\alpha_{\_}{1}>\alpha$ all elements in $1+q^{\alpha}\mathbb{Z}_{\_}q/1+q^{\alpha_{\_}{1}}\mathbb{Z}_{\_}q$ can be written as finite products of coprime elements $p^{\delta_{\_}{p}}$ for $p\in\widehat{\mathbb{P}_{1}}$.

Proof.

The proof is essentially the same as of Lemma 2.4. ∎

We can now conclude the proof.

Proposition 2.7**.**

Let $f:\mathbb{N}\to\{0,1\}$ be multiplicative $q-$automatic sequence and $|\mathbb{P}_{0}|=\infty.$ Then $f(p)=0$ for sufficiently large $p.$

Proof.

If $\widehat{\mathbb{P}_{1}}$ is finite, then to conclusion is immediate. Otherwise, Proposition 2.5 implies that $|\{p\in\widehat{\mathbb{P}_{1}}:\alpha_{\_}{p}=\alpha\}|=\infty$ for some $\alpha\in\mathbb{Z}^{+}.$ Given $q<p_{\_}{1},\ldots,p_{\_}{k_{\_}{0}}\in\mathbb{P}_{0}$, i.e., primes such that $f(p_{\_}{i})=0$, let

[TABLE]

Given any $A$ such that $q^{A}>100Q^{2},$ select $r_{\_}{A}$ modulo $Q^{2}$ via the Chinese remainder theorem to satisfy

[TABLE]

For such $r_{\_}A,$ we have $0<r_{\_}{A}<Q^{2}<q^{A}$ and $p_{\_}{s}||r_{\_}{A}+q^{A}s.$ Hence $f(q^{A}s+r_{\_}A)=0$ for $1\leq s\leq k_{\_}{0}.$ Moreover, for $m<q^{A}/Q^{2}-1$, we have

[TABLE]

Thus, as $f(q^{A}n+r_{\_}{A,M})=0$ for $1\leq n\leq k_{\_}{0},$ we find that in fact $f(q^{A}n+r_{\_}{A,M})=0$ for all $n\geq 1.$ Define the set

[TABLE]

and note that $m\mod q^{A}\in N_{\_}A$ implies that $f(m)=0.$ We next show that $N_{\_}{A}$ has a certain multiplicative invariance property: let $t$ be any prime such that $f(t)=1$. Then, if $n\in\mathbb{Z}$ has the property that $t\nmid n$ and $tn\mod q^{A}\in N_{\_}{A}$, we find that

[TABLE]

which forces $f(n)=0$. In other words, if $n\mod q^{A}\in t^{-1}N_{\_}{A}$, then $f(n)=0,$ provided that $t\nmid n.$

Now choose $e\in\mathbb{N}$ such that $q^{e}>Q^{2}$ and $A$ so that $A-e>\alpha$, with $\alpha$ as in Lemma 2.6. Applying Lemma 2.6 yields a finite set $Y\subset\widehat{\mathbb{P}_{1}}$ such that any element $m\in 1+q^{A-e}\mathbb{Z}_{\_}q/1+q^{A}\mathbb{Z}_{\_}q$ can be written as

[TABLE]

with $\epsilon_{\_}{p}\in\{0,1\}$ and $f(p^{\epsilon_{\_}{p}\delta_{\_}{p}})=1.$

Let $t$ be a sufficiently large prime so that $(t,q)=1$ and $t\not\in Y$ holds. Then $t\equiv n\mod q^{A-e}$ for some $n\in N_{\_}{A}$, and consequently there exists $m\in 1+q^{A-e}\mathbb{Z}_{\_}q/1+q^{A}\mathbb{Z}_{\_}q$ such that $tm\equiv n\mod q^{A}$.

In particular, $tm\in N_{\_}{A}$ and thus, chosing $\epsilon_{\_}{p}\in\{0,1\}$ such that $m\equiv\prod_{\_}{p\in Y}p^{\epsilon_{\_}{p}\delta_{\_}{p}}\mod q^{A}$, we arrive at

[TABLE]

This concludes the proof. ∎

2.3. The case when $q$ is composite.

We now give a brief outline how to modify the argument for the non-prime case. Let $q=\prod_{\_}{i=1}^{k}q_{\_}{i}^{e_{\_}{i}}$ with $q_{\_}{i}$ prime. Take $e$ large enough that $q_{\_}{1}^{e}>Q^{2}$. Then the image of set (cf. the proof of Proposition 2.7)

[TABLE]

under the reduction modulo $q_{\_}{1}^{Ae_{\_}{i}-e}\prod_{\_}{i=2}^{k}q_{\_}{i}^{Ae_{\_}{i}}$ map is onto. The same argument as before (i.e., define $\alpha_{\_}{p,\delta}$ by $q_{\_}{1}^{\alpha_{\_}{p,\delta}}||p^{\delta\phi(q_{\_}{1}^{e_{\_}{1}})}-1$ and arguing as before in the proof of Proposition 2.5) yields a finite set $Y\subset\widehat{\mathbb{P}_{1}}$ such that any element $m\in 1+q_{\_}{1}^{A-e}\mathbb{Z}_{\_}{q_{\_}{1}}/1+q_{\_}{1}^{A}\mathbb{Z}_{\_}{q_{\_}{1}}$ can be written as

[TABLE]

with $\epsilon_{\_}{p}\in\{0,1\}$ and $f(p^{\epsilon_{\_}{p}\delta_{\_}{p}})=1$. The same argument used to prove Proposition 2.7 now applies.

Remark 2.8**.**

Our proof gives somewhat stronger conclusion, namely if $|\mathbb{P}_{0}|=\infty$ then $|\widehat{\mathbb{P}_{1}}|<\infty.$ From here one can easily get a more refined information about the set of values of $\{f(p^{n})\}_{\_}{n\geq 1}.$ We leave the details to the interested reader.