Automatic sequences defined by Theta functions and some infinite products

TL;DR

This paper investigates conditions under which infinite products involving rational functions and theta functions produce $q$-automatic sequences, establishing finiteness results for such polynomials over the rationals.

Contribution

It proves that for fixed q and degree d, only finitely many polynomials over Q generate $q$-automatic sequences from these infinite products.

Findings

Finiteness of polynomials producing $q$-automatic sequences

Characterization of sequences from infinite products involving rational functions

Conditions for automaticity in power series derived from theta functions

Abstract

Let $p(x) \in C(x)$ be a rational function satisfying the condition $p(0)=1$ and $q$ an integer larger than $1$, in this article we will consider the power expansion of the infinite product $$f(x)=\prod_{s=0}^{\infty}f(x^{q^{s}})=\sum_{i=0}^{\infty}c_ix^i,$$ and study when the sequence $(c_i)_{i \in \mathbf{N}}$ is $q$-automatic. The main result is that for given integers $q \geq 2$ and $d \geq 0$, there exist finitely many polynomials of degree $d$ defined over the field of rational numbers $\mathbf{Q}$, such that $\prod_{s=0}^{\infty}f(x^{q^{s}})=\sum_{i=1}^{\infty}c_ix^i$ is a $q$-automatic power series.

Equations

$$f(x)=\prod _{s=0}^{\infty }p(x^{q^s})=\sum _{i=0}^{\infty }{c}_i{x}^i,$$

$$Ker((a_i{)}_{i\in \mathbf{N}})=\left\{(a_{q^{l}n+b}{)}_{n\in \mathbf{N}}|l\in \mathbf{N},0\le b<q^l\right\}$$

$$\forall n\in \mathbf{Z},b_n=a_{q^{l}n+b}.$$

$$O_{l,b}(\sum _{n=0}^{\infty }{a}_n{x}^n)=\sum _{n=0}^{\infty }{a}_{q^{l}n+b}{x}^n.$$

$$\sum _{i=0}^m{a}_i(x)f(x^{k^i})=0.$$

$$(a_n{)}_{n\in \mathbf{N}}\to \sum _{n=0}^{\infty }{a}_n{x}^n.$$

$$\sum _{i=0}^{\infty }{a}_i{x}^i=\sum _{i=0}^{k-1}{x}^i(\sum _{j=0}^{\infty }{a}_{kj+i}{x}^{kj}).$$

$$\sum _{i=0}^{\infty }{a}_i{x}^i=\sum _{(b_n{)}_{n\in \mathbf{N}}\in B^{\prime }}{c}_b\sum _{i=0}^{\infty }{b}_i{x}^{ki},$$

$$c_b=\{\begin{array}{l}{x}^i,\text{if}{R}_{1,i}((a_n{)}_{n\in \mathbf{N}},(b_n{)}_{n\in \mathbf{N}})\\ 0,\text{otherwise.}\end{array}$$

$$\{\begin{array}{l}f(x)={\sum }_{(b_n{)}_{n\in \mathbf{N}}\in B}{c}_b^1{\sum }_{i=0}^{\infty }{b}_i{x}^{k^{N+1}i},\\ f(x^k)={\sum }_{(b_n{)}_{n\in \mathbf{N}}\in B}{c}_b^2{\sum }_{i=0}^{\infty }{b}_i{x}^{k^{N+1}i},\\ ...\\ f(x^{k^N})={\sum }_{(b_n{)}_{n\in \mathbf{N}}\in B}{c}_b^N{\sum }_{i=0}^{\infty }{b}_i{x}^{k^{N+1}i};\end{array}$$

$$a,b,a,b,a,b,a,b...$$

$$F(x)=A(x^2)+xB(x^2)$$

$$A(x)=(1+x)A(x^2)$$

$$B(x)=(1+x)B(x^2)$$

$$F(x)=(1+x^2)(1+x^4)A(x^8)+x(1+x^2)(1+x^4)B(x^8)$$

$$F(x^2)=(1+x^4)A(x^8)+x^2(1+x^4)B(x^8)$$

$$F(x^4)=A(x^8)+x^{4}B(x^8)$$

$$(x^8-x^6+x^4-x^2)((1+x^2)F(x^2)-F(x))=(x^4-x^3+x^2-x)(1+x^4)((1+x^4)F(x^4)-F(x^2))$$

$$a,b,b,a,b,a,a,b,b,a,a,b,a,b,b,a...$$

$$F(x)=F(x^2)+xG(x^2)$$

$$G(x)=G(x^2)+xF(x^2)$$

$$G(x^2)=G(x^4)+x^{2}F(x^4)$$

$$x^{2}G(x^4)=F(x^2)-F(x^4)$$

$$x^{2}G(x^2)=x^{2}G(x^4)+x^{4}F(x^4)$$

$$(x^4-1)F(x^4)+(1+x)F(x^2)-xF(x)=0$$

$$\sum _{s=0}^m{a}_s(x)G(x^{q^s})=0.$$

$$G(x)=p(x)G(x^q).$$

$$\sum _{s=0}^m{a}_s(x)\prod _{r=s}^{m}p(x^{q^{m-r}})=0.$$

$$p(x^{q^{m-1}})|a_m(x)$$

$$f(x)=\prod _{s=0}^{\infty }p(x^{q^s})$$

Full text

Automatic sequences defined by Theta functions and some infinite products

Shuo LI

1 Introduction

Let $p(x)\in C(x)$ be a rational function satisfying the condition $p(0)=1$ and $q$ an integer larger than $1$, in this article we will consider the expansion in power series of the infinite product

[TABLE]

and study when the sequence $(c_{i})_{i\in\mathbf{N}}$ is $q$-automatic. This topic has been studied by many authors, such as [Dum93], [DN15] and [CR18] , using analytical approach, here we want to review this topic by a basic algebraic approach.

The main result is that for given integers $q\geq 2$ and $d\geq 0$, there exist finitely many polynomials of degree $d$ defined over the field of rational numbers $\mathbf{Q}$, such that $f(x)=\prod_{s=0}^{\infty}p(x^{q^{s}})=\sum_{i=0}^{\infty}c_{i}x^{i}$ is a $q$-automatic power series.

2 Definitions and generality

• Definition

Let $(a_{n})_{n\in\mathbf{N}}$ be a sequence, we say it is $q$-automatic if the set

[TABLE]

is finite. This set will be called the $q$-kernel of $(a_{n})_{n\in\mathbf{N}}$.

For every couple of integers $(l,b)$ satisfying $l\in\mathbf{N},0\leq b<q^{l}$, let us define a relation $R_{l,b}$ over the sequence space: we say $R_{l,b}((a_{n})_{n\in\mathbf{N}},(b_{n})_{n\in\mathbf{N}})$ if and only if

[TABLE]

• Definition

Let $\sum_{i=0}^{\infty}a_{i}x^{i}$ be a power series, we say it is $q$-automatic if the sequence of coefficients $(a_{n})_{n\in\mathbf{N}}$ is $q$-automatic.

Similarly we define operators $O_{l,b}$ over the space of power series:

[TABLE]

Now let us consider a detailed version of a well-known theorem, see, for example, [AS92].

Proposition 1

let $f\in F((x))$ be a $k$-automatic power series, then there exist polynomials $a_{0}(x),a_{1}(x),...,a_{m}(x)\in F[x]$ with $a_{0}(x)a_{m}(x)\neq 0$ such that

[TABLE]

Furthermore, the coefficients of $a_{0}(x),a_{1}(x),...,a_{x}(t)$ depend only on $R_{l,b}$ relations over the $q$-kernel of the sequence of the coefficients of $f$.

• Proof

Let $B$ denote the $k$-kernel of the sequence of coefficients of $f$, and $N$ denote the cardinal of $B$. We can then associate each element in $B$ with a power series by

[TABLE]

Let $B^{\prime}$ denote the image of $B$ by the previous map. For each power series in $B^{\prime}$, we have

[TABLE]

Remarking that if the sequence $(a_{n})_{n\in\mathbf{N}}$ is in $B$, then $(a_{kn+j})_{n\in\mathbf{N}}$ is also in $B$, for $j=0,1,...,k-1$. If we write

[TABLE]

Then

[TABLE]

Particularly, we can do the same thing for $f(x),f(x^{k}),...,f(x^{k^{N}})$:

[TABLE]

with $c_{b}^{j}$ defined only by $R_{l,b}$ relations. But as the cardinal of $B^{\prime}$ is $N$, the linear forms at the right-hand side of above equalities are linearly dependent. As a result, if we neglect the linear dependence between elements in $B^{\prime}$, we can have a linear dependence between $f(x),f(x^{k}),...,f(x^{k^{N}})$ such that the coefficients depend only on $c_{b}^{j}$. So these coefficients depend only on $R_{l,b}$ relations.

Here we make this proposition precise by some examples:

• Example

Let us consider a periodic sequence

[TABLE]

which is $2$-automatic.

Now let us write down the associated power sequence $F(x)=a+bx+ax^{2}+bx^{3}+...$ and two other sequences $A(x)=a+ax+ax^{2}+ax^{3}...$, $B(x)=b+bx+bx^{2}+bx^{3}...$ with constant coefficients.

So

[TABLE]

[TABLE]

[TABLE]

so we have the following dependence:

[TABLE]

[TABLE]

[TABLE]

$F(x)$ satisfies the functional equation

[TABLE]

This functional equation does not depend on the values of $a$ and $b$.

• Example

Let us consider the Thue-Morse sequence

[TABLE]

which is $2$-automatic.

Now let us write down the associated power sequence $F(x)=a+bx+bx^{2}+ax^{3}+...$ and another sequence $G(x)=b+ax+ax^{2}+bx^{3}...$, by changing $a$ to $b$ and $b$ to $a$:

So

[TABLE]

and

[TABLE]

so we have the following dependence:

[TABLE]

[TABLE]

[TABLE]

$F(x)$ satisfies the functional equation

[TABLE]

This functional equation does not depend on the values of $a$ and $b$.

Proposition 2

For a given functional equation $F:\sum_{s=0}^{m}a_{s}(t)f(t^{k^{s}})=0$, there exist finitely many polynomials $p_{1},p_{2},...,p_{r}$ with $p_{i}(0)=1,\forall i\in[0,r]$, such that the associated theta functions $G_{r}(x)=\prod_{s=0}^{\infty}p_{r}(x^{q^{s}})$ satisfying equation $F$.

• Proof

If $p(x)$ is a such polynomial satisfying $p(0)=1$. Let us denote by $G(x)$ the associated power series. By hypothesis, it satisfies the functional equation $F$:

[TABLE]

On the other hand, the power series $G$ satisfies another functional equation:

[TABLE]

Plugging the second equation into the first one, we get

[TABLE]

An observation is that all terms in the sum contain a factor $p(x^{q^{m-1}})$ except the last one. So we have

[TABLE]

with $p(0)=1$, so there are finitely many choices for $p(x)$.

Proposition 3

For a fixed number $k$, there are finitely many polynomials $p_{1},p_{2},...,p_{r}$ such that the theta functions $G_{j}(x)=\prod_{s=0}^{\infty}p_{j}(x^{q^{s}})$ are $q$-automatic and the sizes of their $q$-kernels are bounded by $k$.

• Proof

Fixing the size of the $q$-kernel, we fix the number of possibilities of $R_{l,b}$ relations, so the possible functional equations, and we conclude by Proposition 4.2.

3 Infinite product of polynomials

Let $p=\sum_{i=1}^{n}a_{i}x^{i}$ be a polynomial with coefficients in $\mathbf{C}$ and $q$ be an integer larger than $1$. It is known that the coefficients of the power series

[TABLE]

form a $q$-regular sequence [Dum93], here we want to study when this sequence is $q$-automatic.

Firstly, let us suppose that the degree of $p$, noted $deg(p)$, satisfies $q^{k-1}<deg(p)\leq q^{k}$ for some $k\in\mathbf{N}$ and write

[TABLE]

Then the coefficients $c_{i}$ satisfy a recurrence relation:

[TABLE]

for all $r$ such that $0\leq r\leq q-1$ and $c_{n}=0$ for all negative indices.

Lemma 1

The sequences $(c_{qn+i-j})_{n\in\mathbf{N}}$, for all $i$ and $j$ such that $0\leq i\leq q-1$ and $0\leq j\leq 2q^{k}$, can be represented as linear combinations of sequences $\left\{(c_{n-i})_{n\in\mathbf{N}}|0\leq i<2q^{k}\right\}$.

• Proof

Because of the previous equality, we have

[TABLE]

for all $n,i,j$ defined as above. Now let us check that all sequences appearing on the right-hand side of these equalities are in the set defined in the statement. It is enough to calculate the shifting indices and we have the bounds as follows,

[TABLE]

which proves the statement.

• Example

Let us consider the case where $p(x)=1+x+x^{2}+x^{3}+x^{4}$ and $q=2$, the sequence of coefficients of the power series $F(x)=\prod_{s=0}^{\infty}p(x^{q^{s}})$ is denoted by $(c_{n})_{n\in\mathbf{N}}$, so we have

[TABLE]

from which we can deduce

[TABLE]

[TABLE]

Using the above lemma, we get

[TABLE]

and

[TABLE]

Because of the previous fact, we can introduce some transition matrices: for all integers $r$ such that $0\leq r\leq q-1$ let us define $\Gamma_{r}$ as a square matrix of size $2q^{k}+1$ satisfying

[TABLE]

for all $n\in\mathbf{N}$.

Let us denote by $G$ the semi-group generated by all $\Gamma_{r}$ and multiplication.

Proposition 4

$a\in\left\{c_{n}|n\in\mathbf{N}\right\}$* if and only if there exists a matrix $g\in G$ such that $a$ is the first element in the first row of the matrix $g$, in other words, $a=g(1,1)$. Furthermore, $(c_{n})_{n\in\mathbf{N}}$ is automatic if and only if $G$ is a finite semi-group.*

• Proof

The first part of this proposition is trivial, for any $r\in\mathbf{N}$, let us consider its $q$-ary expansion $r=\overline{s_{k_{1}}s_{k_{1}-1}...s_{0}}$. Using Lemma 4.1, we have

[TABLE]

which proves the first part of the statement.

For the second part, let us define maps $\gamma_{r}$ for all integers $r$ by $\gamma_{r}(n)=q(q(...q(q(n)+s_{0})...)+s_{k_{1}-1})+s_{k_{1}}$ for all $n\in\mathbf{N}$ if $r=\overline{s_{k_{1}}s_{k_{1}-1}...s_{0}}$. Then there is an equality for all $r$:

[TABLE]

But the the last matrix in the above equality is constant and invertible, so each element of a matrix $g\in G$ is a finite linear composition of elements in the sequence $(c_{n})_{n\in\mathbf{N}}$, so the finiteness of elements in $(c_{n})_{n\in\mathbf{N}}$ is equivalent to the finiteness of elements in $G$. And using the fact that $(c_{n})_{n\in\mathbf{N}}$ is an automatic sequence, we conclude the statement.

Proposition 5

For given integers $q\geq 2$ and $d\geq 0$, there exist finitely many polynomials of degree $d$ defined over the field of rational numbers $\mathbf{Q}$, such that $\prod_{s=0}^{\infty}p(x^{q^{s}})=\sum_{i=1}^{\infty}c_{i}x^{i}$ is a $q$-automatic power series.

• Proof

Suppose that the sequence $(c_{n})_{n\in\mathbf{N}}$ generated by $\prod_{s=0}^{\infty}p(x^{q^{s}})=\sum_{i=1}^{\infty}c_{i}x^{i}$ is automatic. Let us consider a sequence of matrices $(\Gamma_{n})_{n\in\mathbf{N}}$, such that $\Gamma_{i}$ are defined as above for $i=0,1,..,q-1$ and $\Gamma_{qi+j}=\Gamma_{i}\Gamma_{j}$ for all $i\geq 1$ and $j=0,1,...,q-1$.

It is easy to see that this matrix sequence is automatic because $G$ is finite. And also the automata of this matrix sequence is the same as the one of $(c_{n})_{n\in\mathbf{N}}$, because $c_{n}$ is exactly the element at the position $(1,1)$ of the matrix $\Gamma_{n}$. To conclude the statement, we have to prove two things: firstly the number of automata generating the sequences $(\Gamma_{n})_{n\in\mathbf{N}}$ is finite, secondly, the output functions for each automaton are also finite.

For the first point, it is enough to show that $|G|$ is bounded by a function depending only on $d$ and $q$, which is proved by Theorem 1.3 of [MS77]. It says that given naturals $n$ and $k$, there exist, up to semi-group isomorphism, only a finite number of finite sub-semi-groups of $M_{n}(F)$ generated by at most $k$ elements.

For the second point, it is a consequence of Proposition 4.3.

Proposition 6

Let $f$ be a polynomial satisfying the hypothesis in Proposition 4.5, then all its coefficients belong to $\mathbf{Z}$.

• Proof

Let us denote by $d$ the degree of $f$ and write down all coefficients of $f$ in the form $a_{i}=\frac{p_{i}}{q_{i}}$ such that $(p_{i},q_{i})=1$, and similarly for all coefficient of $F$, let us write down $c_{i}=\frac{r_{i}}{t_{i}}$ with $(r_{i},t_{i})=1$. If there are some coefficients of $f$ which are rational numbers but not integers, then there exist a prime $p$ and two integers $d_{1}$ and $d_{2}$ satisfying :

[TABLE]

and

[TABLE]

with $d_{1}>0,d_{2}>0$. In fact, because of the hypothesis, there exists $a_{i}=\frac{p_{i}}{q_{i}}$ with $q_{i}\neq 1$. So there exists a prime $p$ such that $p|q_{i}$, thus $d_{1}\neq 0$. Let us suppose $a_{j}=\frac{p_{j}}{q_{j}}$ with the smallest index such that $p^{d_{1}}|q_{j}$. Now let us check

[TABLE]

If $c_{j}=\frac{r_{j}}{t_{j}}$ with $p|t_{j}$ then $d_{2}\geq 1$; otherwise, there are some $a_{k},c_{j}$ such that $p^{d_{1}}|q_{k}t_{j}$, but with the assumption of smallest index, $p^{d_{1}}\nmid q_{k}$, so $p|t_{j}$ thus $d_{2}\geq 1$.

Let $l_{1}$ be the smallest index such that $p^{d_{1}}|q_{l_{1}}$ and similarly let $l_{2}$ be the smallest index such that $p^{d_{2}}|s_{l_{2}}$. Now let us consider the coefficient $c_{l_{2}q+l_{1}}$, which can be calculated as

[TABLE]

Let us consider the sum at the right-hand side, for any couple of $(a_{i},c_{j})$, if $i<t_{1}$, then $p^{d_{1}}\nmid q_{i}$, the maximality of $d_{2}$ leads to $p^{d_{1}+d_{2}}\nmid q_{i}t_{j}$; similarly, if $i>t_{1}$, then $j<t_{2}$ thus $p^{d_{2}}\nmid t_{j}$, so that $p^{d_{1}+d_{2}}\nmid q_{i}t_{j}$; but if $i=t_{1}$, then $j=t_{2}$, so $p^{d_{1}}|q_{i}$ and $p^{d_{2}}|t_{i}$. As a result, $p^{d_{1}+d_{2}}|c_{t_{2}q+t_{1}}$, contradicts the maximality of $d_{2}$.

4 Rational functions generated by infinite products

Here we consider the following question: for a given polynomial $p$ and an integer $q$, when does $F(x)=\prod_{s=0}^{\infty}p(x^{q^{s}})$ equal a rational function. This question has already been studies in [DN15] when restricting the polynomial to the cyclotomic case, this section can be considered as a generalization of the previous work.

Proposition 7

Let $p$ be a polynomial taking coefficients over $\mathbf{C}$ and $q$ be an integer larger than $1$, then there is an equivalence between:

(1) $\prod_{s=0}^{\infty}p(x^{q^{s}})$ is a rational function.

(2) there exists a polynomial $Q(x)$ such that $p(x)=\frac{Q(x^{q})}{Q(x)}$ and all roots of $Q(x)$ are roots of unity, if $\delta$ is a root of $Q(x)$ then $\delta^{q^{t}}$ is a root of $Q$ for all $t\in\mathbf{N}$.

• Proof

(2) implies (1) is straightforward, let us check (1) implies (2).

Let $F(x)=\prod_{s=0}^{\infty}p(x^{q^{s}})$ be a rational function, say $F(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are coprime, using the functional equation $F(x)=p(x)F(x^{q})$, we get

[TABLE]

As $deg(p(x))>0$, so that $deg(Q(x))>deg(P(x))$, and $P(x^{q})|P(x)Q(x^{q})$ if $deg(P(x))>0$, then $P(x^{q})$ and $Q(x^{q})$ should have at least one common root, which contradicts that $P(x)$ and $Q(x)$ are coprime, so we have

[TABLE]

and

[TABLE]

Now let us study the roots of $Q(x)$, let us suppose $0\leq|r_{1}|\leq|r_{2}|\leq...\leq|r_{m}|$ where $r_{i}$ are the roots of $Q(x)$ and $|r_{i}|$ is the modulus of $r_{i}$. Firstly $|r_{m}|$ can not be too large, if $|r_{m}|>1$ then each root of $Q(x^{q})$ should have a modulus strictly smaller than $|r_{m}|$, on the other hand $Q(x)|Q(x^{q})$, which is impossible. For the same reason, $|r_{1}|$ can not be a real number between [math] and $1$. So $|r_{i}|$ are either [math] or $1$, but if $x|Q(x)$, the infinite product of $p(x)$ will not converge, so $|r_{i}|=1$ for all roots of $Q(x)$. Using once more $Q(x)|Q(x^{q})$, if $\delta$ is a root of $Q(x)$ then it is a root of $Q(x^{q})$ which implies $\delta^{q}$ is a root of $Q(x)$, we can do it recursively and we obtain $\delta^{q^{t}}$ is a root of $Q$ for all $t\in\mathbf{N}$, as a corollary, $\delta$ can only be a root of unity. So we prove (2) using (1).

5 Infinite product of inverse of polynomials

In this section, we consider the power sequence defined as follows:

[TABLE]

where $q$ is an integer larger than $1$ and $p=\sum_{i=0}^{n}b_{i}x^{i}$ is a polynomial such that $p(0)=1$ defined as before.

Such a sequence satisfies the functional equation

[TABLE]

If we write $\frac{1}{p(x)}=\sum_{i=0}^{\infty}a_{i}x^{i}$, then

[TABLE]

for all $n\in\mathbf{N}$ and $i$ such that $0\leq i\leq q-1$.

Proposition 8

If the coefficients of the power series $F(x)=\prod_{s=0}^{\infty}\frac{1}{p(x^{q^{s}})}=\sum_{i=0}^{\infty}c_{i}x^{i}$ take finitely many values in $\mathbf{C}$, then the roots of $f$ are all of modulus $1$.

• Proof

Firstly, let us prove that the moduli of all roots of $p$ are not smaller than $1$. Otherwise, let us chose one of those which have smallest modulus, say $\alpha$, because of the above definition, we can conclude that

[TABLE]

for all $k$ larger than $1$.

Let us consider the equality,

[TABLE]

the right-hand side converges when $x$ tends to $\alpha$ while the left-hand side diverges, in fact $\prod_{s=1}^{\infty}\frac{1}{p(\alpha^{q^{s}})}$ converges to a non-zero value because

[TABLE]

which converges, however, $\frac{1}{p(x^{q^{s}})}$ has a pole at $x=\alpha$.

Secondly, let us prove that the moduli of all roots of $p$ are not larger than $1$. Otherwise, let us chose one of them, say $\beta$, and an integer $t$ such that $|\beta|^{q^{t}}>|a|/|b|+1$, where $|a|$ is the largest modulus of the sequence $(c_{i})_{i\in\mathbf{N}}$ and $|b|$ is the smallest non-zero modulus of this sequence. Now consider the following series

[TABLE]

It is easy to see that $\left\{d_{i}|i\in\mathbf{N}\right\}$ is finite, because such a series can be obtained by multiplying a polynomial to $F(x)$, but on the other hand, we have the inequality,

[TABLE]

which diverges. This contradicts the fact that $\left\{d_{i}|i\in\mathbf{N}\right\}$ is finite. In conclusion, the roots of $f$ are all of modulus $1$.

Proposition 9

If the power series $F(x)=\prod_{s=0}^{\infty}\frac{1}{p(x^{q^{s}})}=\sum_{i=0}^{\infty}c_{i}x^{i}$ is a $q$-regular sequence, then the roots of $p$ are all roots of unity, furthermore, the order of each root is multiple of $q$.

• Proof

If $F(x)=\prod_{s=0}^{\infty}\frac{1}{p(x^{q^{s}})}=\sum_{i=0}^{\infty}c_{i}x^{i}$ is a $q$-regular sequence, then $F^{\prime}(x)=\sum_{i=1}^{\infty}c_{i}ix^{i-1}$ is also $q$-regular. On the other hand, we know $\frac{1}{F(x)}=\prod_{s=0}^{\infty}p(x^{q^{s}})$ is $q$-regular, so

[TABLE]

is $q$-regular. In the same way we have $(\log F(x^{q}))^{\prime})$ is $q$-regular so that

[TABLE]

is $q$-regular, then we conclude by Theorem 3.3 [AS92] that all roots are roots of unity.

To prove the second part, we use a method introduced in [Bec94]. We firstly define some notation. Let us denote by $A_{t,i}$ the operator of power series:

[TABLE]

for all $i$ such that $0\leq i\leq q^{t}-1$.

If there exists a root of $p$ which’s order is not a multiple of $q$, say $\alpha$, then for all formal power series $f$, let us define $ord(f(x))$ to be the order of pole of $f$ at point $\alpha$. It is easy to check that there exists a $t\in\mathbf{N}$ such that for all $f\in F[[x]]$, $ord(f(x))=ord(f(x^{q^{t}}))$ so there are some $i$ such that $ord(f(x))\leq ord(A_{t,i}(f(x)))$.

Now let us define a sequence of power series $(s_{i})_{i\in\mathbf{N}}$ and a sequence of integer $(I_{i})_{i\in\mathbf{N}}$ such that $s_{0}=1$, $0\leq I_{i}\leq q^{t}-1,\forall i$ and $ord(A_{I_{i}}(\frac{s_{i}}{p(x)}))\geq ord(\frac{s_{i}}{p(x)})$ and we define $s_{i+1}=A_{I_{i}}(\frac{s_{i}}{p(x)})$, so we can easily check

[TABLE]

and by induction

[TABLE]

However,

[TABLE]

the sequence $s_{i}$ are linearly independent, so $F(x)$ can not be a regular sequence.

Theorem 1

If the power series $F(x)=\prod_{s=0}^{\infty}\frac{1}{p(x^{q^{s}})}=\sum_{i=0}^{\infty}c_{i}x^{i}$ is a $q$-regular sequence, then there exists a polynomial $Q(x)$ such that $p(x)|\frac{Q(x^{q})}{Q(x)}$, so $F(x)$ can be written as

[TABLE]

where $R(x)=\frac{Q(x^{q})}{Q(x)F(x)}$, which is a polynomial.

6 Applications

In this section we will consider some examples of automatic power series of type

[TABLE]

where $p$ is a polynomial of degree $d$ with coefficients in $\mathbf{Q}$ and $l\geq 2$. It has been proved by Proposition 4.5 that the number of such polynomials $p$ is fixed once given the degree $d$ of the polynomial and $l$. But when $l$ and $d$ are both large, it will be difficult to compute the semi-group of matrix discussed in Section 4.2. Here we show a method applied on a particular example to generate the couples $(p,l)$ such that $\prod_{s=0}^{\infty}p(x^{l^{s}})=\sum_{i=1}^{\infty}c_{i}x^{i}$ is an automatic power sequence.

Let us consider firstly the power series $F_{1}(x)$ defined by $p_{1}(x)=1+x-x^{3}-x^{4}$ and $l=2$, it is easy to check that

[TABLE]

And it is well known that $\prod_{s=0}^{\infty}(1+x^{2^{s}})=\frac{1}{1-x}=\sum_{i=1}^{\infty}x^{i}$ and $\prod_{s=0}^{\infty}(1-x^{2^{s}})=\sum_{i=1}^{\infty}b_{n}x^{i}$, where $(b_{n})_{n\in\mathbf{N}}$ is the Thue-Morse sequence beginning with $1,-1$. So the coefficient of term $x^{n}$ in $F_{1}(x)$, say $f_{1}(n)$, can be calculated by

[TABLE]

The sequence $(f_{1}(n))_{n\in\mathbf{N}}$ is bounded because of the fact that $b_{2n+1}+b_{2n}=0$, so $F_{1}(x)$ is a $2$-automatic power sequence. Moreover the transition matrices $\Gamma_{1}$ and $\Gamma_{0}$ can be defined by

[TABLE]

[TABLE]

Remarking that

[TABLE]

[TABLE]

let us consider the the power series $F_{2}(x)$ defined by $p_{2}(x)=1+x+x^{2}-x^{4}-x^{5}-2x^{6}-x^{7}-x^{8}+x^{10}+x^{11}+x^{12}=(x^{2}+x+1)(x^{6}-1)(x^{4}-1)$ and $l=4$, the transition matrices of this polynomial are

[TABLE]

[TABLE]

If we define a sequence of matrices $(\alpha_{n})_{n\in\mathbf{N}}$ by $\alpha_{4n+i}=\alpha_{n}\alpha_{i},0\leq i\leq 3$, then the $n$-th coefficient of $F_{2}(x)$ is $f_{2}(n)=\alpha_{n}(1,1)$. However the matrices $\alpha_{i}$ for $i=0,1,2,3$ are all of form $\begin{pmatrix}A_{i}&0\\ B_{i}&C_{i}\\ \end{pmatrix}$ with $A_{i}$ of size $4\times 4$, $B_{i}$ of size $4\times 1$, $C_{i}$ of size $1\times 1$ and [math] the [math]-matrix of size $1\times 4$, so $\alpha_{n}(1,1)$ can be calculated only by the multiplications between $A_{i}$. Remarking that this four matrices are nothing else then $\Gamma_{0}^{2},\Gamma_{1}\Gamma_{0},\Gamma_{0}\Gamma_{1},\Gamma_{0}^{2}$, we conclude that the sequence $(f_{2}(n))_{n\in\mathbf{N}}$ is bounded so $4$-automatic.

By the same method, the power series $F_{3}(x)$ defined by $p_{3}(x)=1+x+x^{2}-x^{4}-x^{5}+x^{7}+x^{8}-x^{10}-x^{11}-x^{12}=(x^{2}+x+1)(x^{6}+1)(1-x^{4})$ and $l=4$ is also $4$-automatic. In fact, its transition matrices are

[TABLE]

[TABLE]

and once more they are of form $\begin{pmatrix}A_{i}&0\\ B_{i}&C_{i}\\ \end{pmatrix}$ with $A_{0}=-\Gamma_{0}\Gamma_{0}\Gamma_{1}\Gamma_{1},A_{1}=-\Gamma_{1}\Gamma_{0}\Gamma_{0}\Gamma_{0},A_{2}=\Gamma_{0}\Gamma_{1}\Gamma_{0}\Gamma_{0},A_{3}=\Gamma_{1}\Gamma_{1}\Gamma_{0}\Gamma_{0}$.

Furthermore, as

[TABLE]

we have

[TABLE]

[TABLE]

Proposition 10

The power series

[TABLE]

and

[TABLE]

are $4$-automatic.