A note on multiplicative automatic sequences, II

We prove that any q ‐automatic multiplicative function f:N→C either essentially coincides with a Dirichlet character, or vanishes on all sufficiently large primes. This confirms a strong form of a conjecture of Bell, Bruin and Coons [Trans. Amer. Math. Soc. 364 (2012) 933–959].


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 [2]. One of the applications of such sequences in number theory stems from a celebrated theorem of Christol [5] (also cf. [6]), which asserts that in order to show the transcendence of the power series n 1 f (n)z n over F q [z] it is enough to establish that the function f : N → F q is not q-automatic. In this note, rather than working within the general set up, we confine ourselves to functions with range in 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.
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 : N → 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 [1,3,7,12,[14][15][16][17]. In particular, the following conjecture was made in [3]. Some progress towards this conjecture has been made when f is assumed to be completely multiplicative. In particular, Schlage-Puchta [16] showed that a completely multiplicative qautomatic sequence which does not vanish is almost periodic. Hu [10] improved on that result by showing that the same conclusion holds under a slightly weaker hypothesis. Allouche and Goldmakher [1] considered a related question classifying what they called 'mock' Dirichlet characters. Finally Li [14] and the authors [12] very recently proved Conjecture 1.2 when f is additionally assumed to be completely multiplicative. Moreover, the methods of [12] settled Conjecture 1.2 fully under the assumption of the Generalized Riemann Hypothesis. The key tool there was a variant of Heath-Brown's results [9] on Artin's primitive root conjecture. In this article, we develop an alternative approach, and unconditionally prove a strong form of Conjecture 1.2. Theorem 1.3. Let q 2 and let f : N → C be multiplicative q-automatic sequence. Then, there exists a Dirichlet character χ and an integer Q 1, such that either f (n) = χ(n), for all (n, Q) = 1 or f (p) = 0 for all sufficiently large p.
We remark that Konieczny [13], very recently and independently established a variant of Theorem 1.3 using different methods relying on the structure theory of automatic sequences.

Proof of the main result
Let 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 : N → C is finite and therefore if we define the sets then, from multiplicativity of f , we easily deduce |P >1 |, |P <1 | < ∞. We begin by assuming additionally that |P 0 | < ∞.
Proposition 2.1. Let f : N → C be a q-automatic multiplicative function and suppose |P 0 | < ∞. Then there exists a Dirichlet character χ, such that f (p k ) = χ(p k ) for all p / ∈ P >1 ∪ P <1 ∪ P 0 and any k 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 = i 2 such that f (q i1 n + 1) = f (q i2 n + 1) for all n 1.
for all m 1. The conclusion now immediately follows from the following result by Elliott and Kish (stated as Theorem 1 in the arXiv preprint.) We remark that 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 [11] to get the result.
In what follows we will assume that |P 0 | = ∞ and show that in this case f (p) = 0 for all sufficiently large primes p. We begin with a convenient reduction: defining g as we note that sequence (g(n)) n is also q-automatic, and has the same zero set as the sequence (f (n)) n . Thus, replacing f by g, it is therefore enough to prove the claim for binary valued functions f : N → {0, 1}.
To facilitate our discussion, we introduce the set Since the case of finite P 1 is trivial, we from now on assume that P 1 is infinite.
We note that if f is q-automatic, then there exists k 0 = k 0 (f ) with the following property: if for some i 1 and 0 r q i − 1 the equality f (q i n + r) = 0 holds for all integer n ∈ [1, k 0 ], then f (q i n + r) = 0 for all n 1. A useful consequence is the following: Corollary 2.3. Given any arithmetic progression {q l m + r} m 1 , with r < q l , so that f (q l m + r) = 1 for infinitely many m, there exists m 0 k 0 such that f (q l m 0 + r) = 1.
We next show that non-vanishing of f in certain progressions is either trivial or infinite.
Lemma 2.4. Assume | P 1 | = ∞. Given r, l ∈ N such that f (r) = 1 and q l > r, there exists infinitely many m such that f (q l m + r) = 1.
Proof. Let p i ∈ P 1 denote distinct primes such that p i > r and select exponents e i so that f (p ei i ) = 1, and denote P i = p ei i . We may then select a sequence of finite disjoint sets I j such that i∈Ij P i ≡ 1 mod q l . Letting n j = r · i∈Ij P i we find, using multiplicativity and that (r, P i ) = 1, that f (n j ) = 1, and that n j ≡ r mod q l . Consequently, we can write n j = q l m j + r, and the proof is concluded.
Let q 1 , . . . , q k0+1 ∈ P 0 denote distinct primes with q i > q for all 1 i k 0 + 1. Define Q := k0+1 i=1 q i . By the Chinese remainder theorem, there exists an interval I ⊂ Z/Q 2 Z with |I| = k 0 + 1 such that all integers n ∈ I have the property that at least one q i ||n. In particular, f (n) = 0 for all n ∈ I. Further, letting L = φ(Q 2 ), we note that q L ≡ 1 mod Q 2 .
We can now arrive at a contradiction by constructing a sequence of integer functions r i so that on one hand f (r i ) = 1 for all i, while at the same time, some r i mod Q 2 must fall in I and consequently f (r i ) = 0.
To construct such a sequence we argue as follows. Given r 1 such that f (r 1 ) = 1, select s 1 so that l 1 := s 1 L has the property that q l1 > r. By Lemma 2.4, there exist infinitely many m such that f (q l1 m + r 1 ) = 1, and thus, by Corollary 2.3, there exists m 1 k 0 such that r 2 := q l1 m 1 + r 1 has the property that f (r 2 ) = 1. We also note that r 2 ≡ m 1 + r 1 mod Q 2 . Proceeding inductively, given r i with properties as above, we can define r i+1 := q li m i + r i so that f (r i+1 ) = 1, m i k 0 , and r i+1 ≡ r i + m i mod Q 2 (in particular, take l i = Ls i with s i large enough so that q li > r i .) Now, as all m i k 0 , we find that {r i mod Q 2 } must intersect I, consequently f (r i ) = 0 for some i and we have arrived at a contradiction.
Remark 2.5. The proof can be modified to give a slightly stronger conclusion, namely if |P 0 | = ∞, then | P 1 | < ∞. From here one can easily get more refined information about the set of values of {f (p n )} n 1 . We leave the details to the interested reader.