aa r X i v : . [ m a t h . N T ] A ug ON MULTIPLICATIVE AUTOMATIC SEQUENCES
JAKUB KONIECZNY
Abstract.
We show that any automatic multiplicative sequence either coin-cides with a Dirichlet character or is identically zero when restricted to in-tegers not divisible by small primes. This answers a question of Bell, Bruinand Coons. A similar result was obtained independently by Klurman andKurlberg. Introduction
Automatic sequences — that is, sequences computable by finite automata —give rise to one of the most basic models of computation. As such, for any classof sequences it is natural to inquire into which sequences in it are automatic. Inparticular, the question of classifying automatic multiplicative sequences has beeninvestigated by a number of authors, including [SP11], [BBC12], [Hu17], [AG18],[Li19] and [KK19a]. The interplay between multiplicative automatic sequences isstudied also in [Yaz01], [SP03], [BCH14] and [LM19], among others.The two most recant papers [Li19], [KK19a] listed above give a classification ofcompletely multiplicative automatic sequences, but until now the question remainedopen for sequences which are multiplicative but not completely so. In particular,the authors of [BBC12] conjectured that a multiplicative automatic sequence agreeswith an eventually periodic sequence on the primes.We confirm this conjecture and give some additional structural results. A similarresult is also obtained in an upcoming preprint of Klurman and Kurlberg [KK19b].
Theorem 1.1. If a : N → C is an automatic multiplicative sequence then there ex-ists a threshold p ∗ and sequence χ which is either a Dirichlet character or identicallyzero such that a ( n ) = χ ( n ) for all n not divisible any prime p < p ∗ . The proof naturally splits into two cases, depending on how often a vanishes.These cases are addressed in Sections 2 and 3 respectively. Remark 1.2.
Not all multiplicative sequences satisfying the conclusion of theabove theorem are automatic. A full classification of automatic multiplicative se-quences appears to still be out of reach of the available techniques, even if barelyso. In principle, combining a slightly more precise version of this theorem discussedin subsequent sections and the classification of multiplicative periodic sequences in[LW76], we could completely classify multiplicative k -automatic sequences whichvanish on all integers not coprime to k . The behaviour of a on powers of primesdividing k remains problematic, as evidenced by the fact that when k is prime thenfor any Dirichlet character with modulus k r and any root of unity ξ , the sequence a χ ( n ) := ξ ν k ( n ) χ ( n/k ν k ( n ) ) is multiplicative and k -automatic. The last sequence isa mock Dirichlet character, investigated in [BBC12]. Mathematics Subject Classification.
Primary: 11B85; Secondary: 11N64, 68R15.
Basics and notion.
Throughout, N denotes the positive integers and N := N ∪ { } . A sequence a : N → C is multiplicative if a ( nm ) = a ( n ) a ( m ) for anycoprime m, n ∈ N , and it is completely multiplicative the assumption of coprimalitycan be dropped.For n ∈ N we let [ n ] = { , , . . . , n } (in particular, [0] = ∅ ). If p is a prime, α ∈ N and n ∈ N then ν p ( n ) denotes the largest power of p which divides n and p α k n means that α = ν p ( n ) (or, equivalently, that p α | n but p α +1 ∤ n ). If n, m ∈ N then n ⊥ m is shorthand for gcd( n, m ) = 1. For two quantities X and Y we write X = O ( Y ) or X ≪ Y if there exists an absolute constant c such that | X | < cY .We let Σ k = { , , . . . , k − } denote the set of digits in base k . For a set X welet X ∗ denote the set of words over X , including the empty word ǫ . If u ∈ Σ ∗ k then[ n ] k ∈ N denotes the integer obtained by interpreting u as a digital expansion inbase k . Conversely, if n ∈ N then ( n ) k ∈ Σ ∗ k denotes the expansion of n in base k without any leading zeros. More generally, for l ∈ N we let ( n ) lk denote the suffixof the word 0 ∞ ( n ) k of length l .A sequence a : N → C is k -automatic if there exists finite automaton A =( S, s , δ, τ ) which produces a . Here, S is a finite set of states, s ∈ S is the initialstate, δ is the transition function Σ k × S → S , ( u, s ) δ u ( s ), extended to a mapΣ ∗ k × S → S by δ uv = δ u ◦ δ v , τ is an output function τ : S → C , and finally thesequence produced by A is [ u ] k τ ( δ u ( s )), where u ∈ Σ k does not begin withany initial zeros. A sequence is automatic if it is k -automatic for some k ∈ N .We fix from now on the automatic multiplicative sequence a : N → C and anautomaton A = ( S, s , δ, τ ) which produces it. It is well-known that if k, l ∈ N aremultiplicatively dependent, i.e., log( k ) / log( l ) ∈ Q \{ } , then k -automatic sequencesare the same as l -automatic sequences. Hence, we may assume without loss ofgenerality that k is not a perfect power. Acknowledgements.
The author is grateful to O. Klurman for sharing the afore-mentioned preprint and for helpful comments. This research is supported by ERCgrant ErgComNum 682150. 2.
Sparse case
Throughout this section, make the following assumption:(1) There exists infinitely many primes p such that a ( p α ) = 0 for some α ∈ N . We let Z ⊆ N denote the set of n ∈ N such that a ( n ) = 0. Proposition 2.1.
The set Z is a finite union of (possibly degenerate) geometricprogressions with ratio k l ( l ∈ N ). From here, it easily follows that a ( p ) = 0 for large primes. We also note thatthe fact that a is k -multiplicative imposes further restrictions on Z and on thebehaviour of a on Z . In fact, it is not hard to show that when k is composite, Z needs to be finite. For lack of other nontrivial observations we do not delve furtherinto this subject and devote the remainder of this section to proof of the aboveProposition 2.1 ULTIPLICATIVE AUTOMATIC SEQUENCES 3
Reduction to arid sets.
Our first step is to show that Z is, using the ter-minology borrowed from [BK19], an arid set. Definition 2.2.
A set A ⊆ N is a basic arid set of rank ≤ r if it takes the form(2) A = n [ u r v l r r u r − . . . u v l u ] k (cid:12)(cid:12)(cid:12) l , . . . , l r ∈ N o . for some u , . . . , u r ∈ Σ ∗ k and v , . . . , v r ∈ Σ ∗ k . A set A ⊆ N is arid of rank ≤ r ifit is a union of finitely many basic arid sets of rank ≤ r . If A ⊆ N then the rank of A is the smallest r such that A is contained in an arid set of rank ≤ r , or ∞ ifno such r exists. Proposition 2.3 ([BK19, Proposition 3.4.]) . One of the following is true: ( i ) The set Z is arid. ( ii ) There exists c = 0 and words w, v , v , u ∈ Σ ∗ k such that | v | = | v | , v = v and a ([ wvu ] k ) = c for all v ∈ { v , v } ∗ . In the following argument it will be convenient to use the notion of an IP + r set,that is, the set of the form(3) A = (cid:8) n + P i ∈ I n i (cid:12)(cid:12) I ⊆ [ r ] (cid:9) , where n ∈ N and n , . . . , n r ∈ N . We refer to n , . . . , n r as the sidelengths of A . Lemma 2.4.
Let m ∈ Z and let A ⊆ N be an IP + r set with sidelengths coprime to m . Then A mod m ) ≥ min( m, r + 1) .Proof. We proceed by induction on r , the case r = 0 being trivial. If r ≥ A as the sumset A + { , n r } of an IP + r − set A ′ and a two-element set.Either A ′ mod m ) = m , in which case we are done, or there exists n ∈ A ′ mod m such that n + n r A ′ mod m , in which case A mod m ) ≥ A ′ mod m )+1 ≥ r +1so we are also done. (cid:3) Remark 2.5.
The final step uses a trivial case of the Cauchy–Davenport theorem.
Proposition 2.6.
The set Z is arid.Proof. For the sake of contradiction, suppose that Z was not arid, and let c and w, v , v , u be as in Proposition 2.3.ii. Assumption (1) guarantees that we can finda prime power q = p α such that a ( q ) = 0 and p ∤ [ v ] k − [ v ] k and p ∤ k . Pick r := p α +1 and consider the set A := { [ wvu ] k | v ∈ { v , v } r } . It follows directly from the defining conditions in 2.3.ii that a ( n ) = c for all n ∈ A .On the other hand, A is an IP + r set with sidelengths of the form ([ v ] k − [ v ] k ) k l with l ∈ N , which are not divisible by p . By Lemma 2.4, A covers all residuesmodulo r . In particular, A contains an integer n exactly divisible by q , whence0 = c = a ( n ) = a ( n/q ) a ( q ) = 0 , which is the sought for contradiction. (cid:3) We are now left with the task of showing that arid sets cannot be the level sets ofmultiplicative sequences, except for arguably trivial cases of geometric progressionswhose ratio is a power of k . J. KONIECZNY
Base k geometric progressions. While arid sets are well-adjusted for study-ing combinatorial properties of base k expansions, in order to study arithmeticproperties it is convenient to work in a slightly more general setup. We define a generalised geometric progression of rank ≤ r as a set of the form(4) A = ( x + r X i =1 x i k α i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) α , . . . , α r ∈ N ) , where x , x , . . . , x r ∈ Q . For the sake of uniformity, define also α := 0 Likewise,we define a restricted generalised geometric progression of rank ≤ r as a set of theform(5) A = ( x + r X i =1 x i k α i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) α ∈ F , α ∈ F ( α ) , . . . , α r ∈ F r ( α , . . . , α r − ) ) , where x , . . . , x r ∈ Q and F i are maps N i − → P inf ( N ) for each i ∈ [ r ]. Here, P inf ( X ) denotes the set of all infinite subsets of a set X . In a fully analogousmanner we define restricted arid sets of rank ≤ r as sets of the form(6) A = n [ u r v l r r u r − . . . u v l u ] k (cid:12)(cid:12)(cid:12) l ∈ F , l ∈ F ( l ) , . . . , l r ∈ F r ( l , . . . , l r − ) o , where u , . . . , u r ∈ Σ ∗ k and v , . . . , v r ∈ Σ ∗ k and F i are like above.Given sequences F i as in (5), let us call a vector ( α , α , . . . , α s ) admissible if α = 0 and α i ∈ F i ( α , . . . , α i − ) for i ∈ [ s ], 0 ≤ s ≤ r . The elements ofthe restricted generalised geometric progression A given by (5) can naturally beindexed by the leaves of a regular rooted tree with vertex degree ∞ , whose verticesare admissisible sequences ( α , . . . , α s ), whose root is (0) and whose edges are givenby ( α , . . . , α s ) → ( α , . . . , α s , α s +1 ). It follows from basic Ramsey theory that ifthe leaves of such a tree are coloured by finitely many colours then there exists aninfinite regular subtree of depth r with monochromatic leaves. As a consequence,restricted generalised geometric progressions of a given rank are partition regular.The same observation, mutatis mutandis, applies to restricted arid sets.It is clear that any (restricted) arid set is a (restricted) generalised geometricprogression. The following lemma provides a partial converse to this statement. Lemma 2.7.
For any x , x , . . . , x r ∈ Q there exists B ∈ N and C > such thatthe following holds. Suppose that α , . . . , α r ∈ N is a sequence such that (7) x + r X i =1 x i k α i ∈ N and α i ≥ α i − + C for all i ∈ [ r ] , where α := 0 . Then there exist words u ∈ Σ Bk , u , . . . , u r ∈ Σ Bk and v , . . . , v r ∈ Σ Bk such that (8) x + r X i =1 x i k α i = [ u r v l r r u r − . . . u v l u ] k , where for i ∈ [ r ] the lengths l i ∈ N are uniquely determined by (9) 0 ≤ α i − B (cid:16)P ij =1 l j + 3 i − (cid:17) < B. ULTIPLICATIVE AUTOMATIC SEQUENCES 5
If additionally x i = 0 for all i ∈ [ r ] then additionally the expansion in (8) isnondegenerate in the sense that u r = 0 B and there is no i ∈ [ r ] such that v i = u i = v i − .Proof. This follows by inspection of the standard long addition procedure. Supposefirst that x , x , . . . , x r were all positive integers. Then the conclusion would holdwith v i = 0 B and u i = 0 ∗ ( x i ) k ∗ , where 0 ∗ denotes an unspecified string of zeros.If we drop the assumption of positivity, then the same conclusion holds except v i can also take the form ( k − B and u i need to be modified accordingly. Finally, if x i are rational then apply the above reasoning to the sequence M x , M x , . . . , M x r where M is multiplicatively rich enough that the latter sequence consists of onlyintegers, and use the fact that division by M takes periodic digital expansions toperiodic digital expansions. (cid:3) Remark 2.8. ( i ) The definition of l i in (9) is arranged so that if the right handside of (8) is construed as the B -block expansion of the sum on the left hand side(with each v i occupying one block and u i occupying three blocks) then the position α i falls in the middle block of u i for all i ∈ [ r ].( ii ) The constant B can be replaced by any multiple, and the constant C canalways be enlarged. We could have required that B = C , but we believe that woulddecrease the intuitive appeal of the result.Our definition rank guarantees that if A is a (restricted) arid set of rank ≤ r thenrank A ≤ r . It follows from Lemma 2.7 that if A ⊆ N is a (restricted) generalisedgeometric progression of rank ≤ r then also rank A ≤ r . In order to estimate rankof certain sets from below, we record some observations which essentially say thatthe above inequalities cannot be strict except for degenerate cases. In fact, thefollowing lemma would suffice. Lemma 2.9.
Let B ∈ N and let u ∈ Σ Bk , u , . . . , u r ∈ Σ Bk and v , . . . , v r ∈ Σ Bk be nondegenerate in the sense of Lemma 2.7, and let A be the corresponding aridset given by (2) . Then rank A = r . This result is elementary and there are several ways to prove it. One possibil-ity is to run a combinatorial argument relaying on the intuition that given n =[ u r v l r r u r − . . . u v l u ] k it is essentially possible to recover r , u , . . . , u r , v , . . . , v r and l , . . . , l r , except for some basic operations which do not change the rank andexcept for some border cases corresponding to small l i . Another option is to useestimates on the rate of growth of arid sets. We choose yet another route, andderive Lemma 2.9 from the following stronger statement, whose proof we delegateto the appendix. Lemma 2.10.
Let A be a restricted generalised geometric progression of rank ≤ r given by (5) with x , . . . , x r = 0 . Then rank A = r . Multiplication and arid sets.
Recall that the set Z of nonzero places of a isclosed under products of coprime elements. More generally, if n, m ∈ Z and d ∈ N is such that d | n , n/d ⊥ n and n/d ⊥ m then also mn/d ∈ Z . This motivates theinterest in the following lemma. Lemma 2.11.
For any u, v, w ∈ Σ ∗ k with [ u ] k , [ w ] k = 0 there exists D ∈ N suchthat the following is true. For any prime p there exists Q ∈ N such that if l ∈ N and Q | l then ν p (cid:0) [ wv l u ] k (cid:1) ≤ ν p ( D ) . J. KONIECZNY
Proof.
For reasons which will become clear in the course of the argument, we willtake D := D D where D := [ wu ] k and D := (cid:12)(cid:12) k | u | [ v ] k − ( k | v | − u ] k (cid:12)(cid:12) . Notethat D = 0 since [ w ] k = 0 and D = 0 since D ≡ ( k | v | − u ] k k | u | . Theargument splits into two cases, depending on whether p divides k . Note that in fullgenerality we have[ wv l u ] k = [ w ] k k l | v | + | u | + [ v ] k k l | v | − k | v | − k | u | + [ u ] k . Suppose first that p ∤ k . For any ω ∈ N there exists Q ω such that[ wv l u ] k ≡ [ w ] k k | u | + [ u ] k = D (mod p ω )for all l ∈ N divisible by Q ω . In particular, letting ω > ν p ( D ) we conclude that ν p (cid:0) [ wv l u ] k (cid:1) ≤ ν p ( D ) ≤ ν p ( D )for all l divisible by Q ω .Secondly, suppose that p | k . Then for any ω ∈ N there exists Q ω such that( k | v | − wv l u ] k ≡ − [ v ] k k | u | + ( k | v | − u ] k ≡ ± D (mod p ω )for all l ∈ N with l ≥ Q ω (which in particular holds if Q ω | l ). Letting ω > ν p ( D )we conclude that ν p (cid:0) [ wv l u ] k (cid:1) ≤ ν p ( D ) ≤ ν p ( D )for all l divisible by Q ω . (cid:3) To simplify notation in the following result, for n, m ∈ N let gcd( m ∞ , n ) de-note the limit lim α →∞ gcd( m α , n ), or equivalently the product Q p | gcd( m,n ) p ν p ( n ) .Note we do not attribute any independent meaning to the symbol n ∞ outside ofgcd. It follows directly from Lemma 2.11 that, with notation therein, for each m ∈ N there exists an integer Q such that for any l ∈ N divisible by Q we havegcd (cid:0) m ∞ , [ wv l u ] k (cid:1) | D . Proposition 2.12.
Let u, v, w ∈ Σ ∗ k , v = ǫ , and [ u ] k , [ w ] k = 0 or [ v ] k = 0 . Thenthere exists l ∈ N such that [ wv l u ] k Z .Proof. For the sake of contradiction suppose that [ wv l u ] k ∈ Z for all l ∈ N .Replacing w and u with wv and vu , we may assume that [ u ] k , [ w ] k = 0. Let t ∈ N be a large parameter. Our strategy is to show that the elements of Z which can beconstructed taking products of t terms of the form [ wv l u ] k ( l ∈ N ) give rise to anarid set of rank ≥ t −
1, which leads to contradiction since rank
Z < ∞ .For l ∈ N let n ( l ) := [ wu l v ] k . It follows from Lemma 2.11 that there exists D ∈ N such that for any m ∈ N there exist Q ∈ N such that if l ∈ N and Q | l then gcd( m ∞ , n ( l )) | D . Using this observation iteratively we can find asequence of infinite sets F , F ( l ), F ( l , l ) , . . . , F t ( l , . . . , l t ) ⊆ N such that forany sequence l , . . . , l t ∈ N with l i ∈ F i ( l , . . . , l i − ) for all i ∈ [ t ] we have d i :=gcd (cid:16)Q i − j =1 n ( l j ) ∞ , n ( l i ) (cid:17) | D for all i ∈ [ t ]. Using partition regularity, we mayfurther assume that 1 = d , d , . . . , d t are independent of the choice of l , . . . , l t .Hence, for any admissible l , . . . , l t ∈ N we have d i | n ( l i ), n ( l i ) /d i ⊥ d i and n ( l i ) /d i ⊥ n ( l j ) /d j for all i, j ∈ [ t ] with i = j . Hence, Z contains the set A := ( t Y i =1 n ( l i ) /d i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) l ∈ F , . . . , l t ∈ F ( l , . . . , l t − ) ) . ULTIPLICATIVE AUTOMATIC SEQUENCES 7
It is clear from the definition of n ( l ) that there exists z, y ∈ Q \ { } and c ∈ N such that n ( l ) = zk cl + y . Hence, for any admissible l , . . . , l t we can expand theproduct in the definition of A as t Y i =1 n ( l i ) /d i = X I ⊆ [ t ] x I k α I , where x I = x | I | y t −| I | = 0 and α I = P i ∈ I l i (in particular, α ǫ = 0). Let s := 2 t − { , } t with { , , . . . , s } in a standard way. Replacing F i withsmaller sets if necessary we may assume that the sequence { α j } sj =0 is increasing,and indeed that α ′ j > α ′ j − + C for all j ∈ [ s ] where C > C and B be the constants from Lemma 2.7 we conclude thatthere exists words u ∈ Σ Bk , u , . . . , u s ∈ Σ Bk and v , . . . , v s ∈ Σ Bk obeying thenondegeneracy conditions [ u s ] = 0 and { v j , u j , v j − } > j ∈ [ s ], such that t Y i =1 n ( l i ) /d i = x + s X j =1 x j k α j = [ u s v m s s u s − . . . u v m u ] k , where m , . . . , m s ∈ N obey the asymptotic relation m j = ( α j − α j − ) /B + O (1) forall j ∈ [ s ]. Note that we could assume that u j , v j are independent of l , . . . , l t bypartition regularity. By the same token, we may also assume that m j := m j mod M ( j ∈ [ s ]) are independent of l , . . . , l t , where M is a multiplicatively rich constantsuch that δ Mv is idempotent for each v ∈ Σ ∗ k (we can take M = S !). If now followsthat Z contains the arid set n [ u s v m ′ s s u s − . . . u v m ′ u ] k (cid:12)(cid:12)(cid:12) m ′ j ∈ m j + M N for all j ∈ [ s ] o , whose rank is equal to s . In particular, rank Z ≥ s , as needed. (cid:3) Corollary 2.13.
The set Z is a finite union of geometric progressions of the form (cid:8) xk cl (cid:12)(cid:12) l ∈ N (cid:9) where x ∈ N and c ∈ N .Proof. It is enough to notice that the only basic arid sets not containing patternsforbidden by Proposition 2.12 take the form (cid:8) [ w cl ] (cid:12)(cid:12) l ∈ N (cid:9) with c ∈ N . (cid:3) Dense case
We now assume that there exists a threshold p such that the following holds:(10) For all primes p ≥ p and all α ∈ N we have a ( p α ) = 0 . Our main aim is to show that a ( n ) coincides with a Dirichlet character for n devoidof small prime factors. We also record some observations concerning the behaviourof a on small primes.3.1. Large primes.
We first deal with large primes. From this point onwards, welet m , χ and p denote the objects in the following result; we may assume that p > m and that p > k . Proposition 3.1.
There exists a Dirichlet character χ with modulus m and athreshold p such that a ( n ) = χ ( n ) for all n ∈ N which are products of primes ≥ p . J. KONIECZNY
Relying on the following result, we can prove Proposition 3.1 by essentially thesame methods which were used by Klurman and Kurlberg [KK19a] for completelymultiplicative sequences. We will also need the fact that the k -kernel of any k -automatic sequence is finite. Here, the k - kernel of a is the set of all the sequences n a ( k α n + r ) with α ∈ N and 0 ≤ r < k α . Proposition 3.2.
There exists a threshold p such that if p ≥ p is a prime then a ( p α ) = a ( p ) α = 0 for all α ∈ N .Proof of Prop. 3.1 assuming Prop. 3.2. Suppose for the sake of clarity that p ≥ p . Because the k -kernel of a is finite, we can find integers β < γ such that a ( k β n + 1) = a ( k γ n + 1) for all n ∈ N . Denote by e a the multiplicative functiongiven by e a ( p α ) = a ( p α ) = a ( p ) α if p ≥ p and e a ( p α ) = 1 if p < p , α ∈ N . Clearly, e a is totally multiplicative. Because a takes on finitely many values, a ( p ) is a rootof unity for each p ≥ p . Letting Q be the product of all primes < p we obtain e a ( k β Qn + 1) = a ( k β Qn + 1) = a ( k γ Qn + 1) = e a ( k γ Qn + 1)for all n ∈ N , whence e a is a Dirichlet character by [EK17, Thm. 2]. (cid:3) In order to prove Proposition 3.2, it will be convenient to introduce an equiva-lence relation on Σ ∗ k where u ∼ v if | u | = | v | and words u , v give rise to the sametransition function in the automaton A , that is, δ u = δ v . Since transition func-tions are self-maps of S , the number of equivalence classes (cid:0) Σ lk / ∼ (cid:1) is boundeduniformly with respect to l ∈ N . Likewise, consider the equivalence relation on N given by n ∼ n if ( n ) lk ∼ ( n ) lk for all sufficiently large l ∈ N , or — equivalently— if ( n ) lk ∼ ( n ) lk for at least one l ∈ N with n , n < k l . Crucially, there are onlyfinitely many equivalence classes: N / ∼ ) < ∞ . Lemma 3.3.
There exists a threshold p such that for any p > p there exists apair n , n ∈ N with n n (mod p ) such that n ∼ n and pn ∼ pn for allsufficiently large l ∈ N .Proof. Since N / ∼ ) < ∞ , this follows from the pidgeonhole principle. (cid:3) For the sake of brevity, in the following argument and elsewhere we will say thata statement ϕ ( n ) is true for almost all n if the set of n for which it fails has Banachdensity 0: lim N →∞ X M ≥ N { n ∈ [ M, M + N ) | ¬ ϕ ( n ) } = 0 . Proof of Prop. 3.2.
Take any prime p with p > p , where p is the threshold inLemma 3.3, and let n , n be the pair whose existence is guaranteed by said lemma.Let l be a large integer, to be determined in the course of the argument, and put u i := ( n i ) lk , u ′ i := ( pn i ) lk . It is a general fact that if w ∈ Σ ∗ k the for almost all n theexpansion ( n ) k contains w . Hence, for almost all n there exists a decomposition( n ) k = x n u y n , for some x n , y n ∈ Σ ∗ k where x n is nonempty and does not start with any zerosand y n starts with at least l zeros. Letting x ′ n and y ′ n denote the expansions of p [ x n ] k and p [ y n ] k with x ′ n not starting with any zeros and | y ′ n | = | y n | , we get thedecomposition ( pn ) k = x ′ n u ′ y ′ n . ULTIPLICATIVE AUTOMATIC SEQUENCES 9 If p ∤ n then clearly a ( pn ) = a ( p ) a ( n ). On the other hand, if p | n then a ( pn ) = a ([ x ′ n u ′ y ′ n ] k ) = a ([ x ′ n u ′ y ′ n ] k )= a ( p ) a ([ x n u y n ] k ) = a ( p ) a ([ x n u y n ] k ) = a ( p ) a ( n ) . Hence, we have shown that a ( pn ) = a ( p ) a ( n ) for almost all n .Let α ∈ N . Integers n such that p α k n and n ⊥ q for all q < p constitute apositive proportion of all integers, whence there exists many n such that a ( p α +1 ) = a ( pn ) a ( n/p α ) = a ( p ) a ( n ) a ( n/p α ) = a ( p ) a ( p α ) . It now follows by induction that a ( p α ) = a ( p ) α . (cid:3) Small primes.
In this section we address the behaviour of a on small primes.Unfortunately, we can only obtain a weaker analogue of Proposition 3.1. Proposition 3.4.
For any prime p ∤ k , the sequence a ( p α ) is eventually periodic.Proof. Recall that there exists (many) pairs of distinct integers n , n ∈ N suchthat n ∼ n . Note also that if n ∼ n and n ′ ∼ n ′ then also k α n + n ′ ∼ k α n + n ′ for sufficiently large α . Hence, we can assume that d := n − n is divisible by anyprime p < p and also by a large power of k . Let v = ( n ) lk and v = ( n ) lk where l is a large integer.For any α ∈ N and any β sufficiently large in terms of α , there exists a prime q such that ( qp α ) k ∈ v Σ βk , that is, the expansion of qp α starts with 1 v andcontains β other digits. (This follows from the classical fact for any ε > N , there exists a prime between N and N + εN ; in fact, by thePrime Number Theorem there are ∼ εN/ log N such primes.) Let δ = ν p ( d ) andsuppose that α > δ . Then a ( p α ) = a ( qp α ) /a ( q ) = a ( qp α + dk β ) /a ( q )= a ( qp α − δ + dk β /p δ ) a ( p δ ) /a ( q ) = χ ( qp α − δ + dk β /p δ ) a ( p δ ) /χ ( q ) , where in the last transition we use the fact that any prime < p divides exactly oneof qp α − δ and ( d/p δ ) k β . We may also assume (using the Prime Number Theoremin arithmetic progressions) that q ≡ m and dk β /p δ ≡ d/p δ mod m , whence a ( p α ) = χ ( p α − δ + d/p δ ) a ( p δ ) . It remains to notice that the expression on the right hand side is periodic in p α . (cid:3) Corollary 3.5.
There exists a periodic sequence b : N → C and threshold n suchthat a ( n ) = b ( n ) for all n ≥ n coprime to k .Proof. Partitioning N into arithmetic progressions, we may assume that for eachprime p < p , either n is divisible by a large power of p or n is not divisible by p .Repeating the same reasoning as in Proposition 3.4 we conclude that a ( n ) = χ (cid:16) ( n + d ) / Q p p δ p (cid:17) a (cid:0) p δ p (cid:1) , where δ p = ν p ( d ) and the product runs over all primes p < p with p ∤ k . (cid:3) Appendix A. Rank of generalised geometric progressions
Lemma A.1.
For any x , . . . , x r ∈ Q there exists a constant C ≥ such that thefollowing is true. Let α , . . . , α r ∈ N and suppose that (11) P ri =1 x i k α i = 0 . Then there exists a partition [ r ] = I ∪ · · · ∪ I s and γ , . . . , γ s ∈ N such that (12) P ri ∈ I j x i k α i for all j ∈ [ s ] , and | γ j − α i | < C for all i ∈ I j , j ∈ [ s ] .Proof. We proceed by induction on r , the case r = 1 being trivial. Suppose that theclaim was false and let C ( α , . . . , α r ) denote the smallest value of C such that thereexists a partition described above. Then there exists a sequence α n , . . . α nr ∈ N ( n ∈ N ) such that C ( α n , . . . , α nr ) → ∞ as n → ∞ .Put γ n ∗ := max i ∈ [ r ] α ni . Passing to a subsequence if necessary, we may assumethat for each i ∈ [ r ], the limit δ i := lim n →∞ ( γ n ∗ − α ni ) exists in N ∪ {∞} . Let I ∗ be the set of those n for which δ i < ∞ . Note that if i ∈ I ∗ then α ni = γ n ∗ − δ i for allsufficiently large n ; passing to a subsequence again we may assume that the aboveholds for all n . It follows that X i ∈ I x i k α ni = k γ n ∗ lim m →∞ r X i =1 x i k α mi − γ m ∗ = 0for all n . Rearranging the indices if necessary, we may assume without loss ofgenerality that I ∗ is the terminal segment { r ′ + 1 , . . . , r } . The above constructionguarantees that P r ′ i =1 x i k α ni = 0for all n . Hence, by the inductive assumption there exists a constant C ′ such thatfor each n there exists a decomposition [ r ′ ] = I ∪ · · · ∪ I s ′ and γ n , . . . , γ ns ′ ∈ N such that P ri ∈ I j x i k α ni for all j ∈ [ s ′ ] , and (cid:12)(cid:12) γ nj − α ni (cid:12)(cid:12) < C ′ for all i ∈ I j , j ∈ [ s ′ ]. Passing to a subsequence again, we mayassume that the decomposition I , . . . , I s ′ is independent of n . Letting s = s ′ + 1, I s = I ∗ and γ s = γ ∗ we conclude that C ( α , . . . , α n ) ≤ max ( { δ i | i ∈ I s } ∪ { C ′ } ) . This contradicts the choice of α n , . . . , α nr and finishes the argument. (cid:3) Proof of Lemma 2.9.
By partition regularity, it will suffice to show that A is notcontained in any generalised geometric progression B of rank ≤ s < r . For the sakeof contradiction suppose otherwise and write B as(13) B = n y + P sj =1 y j k β j (cid:12)(cid:12)(cid:12) β , . . . , β s ∈ N o . Assume also that s is as small as possible. For each admissible sequence ~α =( α , α , . . . , α r ) pick ~β ( ~α ) = ( β ( ~α ) , . . . , β s ( ~α )) such that r X i =0 x i k α i = s X j =0 y i k β i ( ~α ) . ULTIPLICATIVE AUTOMATIC SEQUENCES 11
By Lemma A.1, there exists a constant
C > ~α thereexists a partition [ − s, r ] = I ∪ I ∪ . . . I t such that for each l ∈ [ t ] we have X i ∈ I l ∩ N x i k α i = X j ∈− I l ∩ N y j k β j ( ~α ) , and the diameter of the set { α i | i ∈ I l ∩ N } ∪ { β j | j ∈ − I l ∩ N } is ≤ C . Usingpartition regularity again, we may assume that the partition I , . . . , I t is indepen-dent of ~α .Since the sets F i ( α , . . . , α i − ) are infinite, we can always assume that | α i − α i ′ | >C for all i = i ′ . In particular, I l ∩ N ) ≤ l ∈ [ t ]. On the other hand, if I l ∩ N = ∅ for some l ∈ [ t ] then A would be contained in the generalised geometricprogression of rank < s obtained by setting β j = 0 for j ∈ − I l ∩ N , which contra-dicts the choice of s . Hence, I l ∩ N ) = 1 for each l ∈ [ t ]. Since x i = 0 for i ∈ [ r ]it follows that − I l ∩ N ) ≥ l ∈ [ t ]. This contradicts the assumptionthat s < r and finishes the argument. (cid:3) Remark A.2.
A careful inspection of the above proof shows that the restrictedgeneralised geometric progression A given by (5) is contained in the generalisedgeometric progression B given by (13) if and only if there exists a partition [0 , r ] = J ∪ J ∪ · · · ∪ J s and integers 0 = δ , δ , . . . , δ s such that x i = P j ∈ J i y j k δ j for each i ∈ [0 , r ]. References [AG18] J.-P. Allouche and L. Goldmakher. Mock characters and the Kronecker symbol.
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Einstein Institute of Mathematics Edmond J. Safra Campus, The He-brew University of Jerusalem Givat Ram. Jerusalem, 9190401, IsraelFaculty of Mathematics and Computer Science, Jagiellonian University in Krak´ow, Lojasiewicza 6, 30-348 Krak´ow, Poland
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