# Tag Archives: linear recurrence

## Finite automata 3: Non-automatic sets

The key ingredients in our proof of the special case of Cobham’s theorem were the pumping lemma and the following fact:

$(*)\qquad [ab^{n+1}c]_q-q^{|b|}[ab^nc]_q=\text{constant}$

We can formally put them together into the following lemma:

Lemma 1. If $S\subseteq\mathbb N$ is an infinite automatic set, then there exist distinct $s_0,s_1,\ldots\in S$ and integers $A>1$ and $B$ such that $s_{n+1}=As_n+B$ for all $n$.

Proof. Say $S$ is $q$-automatic. By the pumping lemma we can take $s_n=[ab^nc]_q$ for some $a,b,c\in\Sigma_q^*$ with $|b|\ge 1$. Note that these are distinct as we don’t allow leading zeros. The conclusion now follows from $(*)$. $\square$

Lemma 1 can be a powerful tool in proving that sets are not automatic, because it transforms a question from automata theory into the language of simple unary recurrences. For instance, we can use it to easily prove the following theorem of Minsky and Papert:

Theorem 1 (Minsky and Papert, 1966). The set of prime numbers is not automatic.

We are actually able to prove the following stronger result:

Theorem 2 (Schützenberger, 1968). An infinite set of prime numbers is not automatic.

Proof. Assume the contrary. Then by lemma 1 there exist distinct prime numbers $p_1,p_2,\dots$ and integers $A>1$ and $B$ such that $p_{n+1}=Ap_n+B$ for all $n\ge 1$. Then

$p_{n+j}\equiv B(A^{j-1}+\dots +A+1)\pmod{p_n}$

for all $j,n\ge 1$. Now pick $n$ large enough so that $p_n\nmid A(A-1)$. Then

$\displaystyle p_{n+p_n-1}\equiv\frac{B(A^{p_n-1}-1)}{A-1}\equiv 0\pmod{p_n}$

by Fermat’s little theorem, a contradiction. $\square$

Note that lemma 1 implies that any set whose elements increase very rapidly cannot be automatic. To put it formally:

Theorem 3. Let $f:\mathbb N\to\mathbb N$ such that $f(n+1)/f(n)\to\infty$ as $n\to\infty$. Then the set $\{f(1),f(2),f(3),\dots\}$ is not automatic.

Proof. Assume the contrary. Then there exist integers $0 such that $f(s_{n+1})=Af(s_n)+B$ for all $n$, where $A>1$ and $B$ are fixed integers. Then $f(s_{n+1})/f(s_n)\to A$ as $n\to\infty$, a contradiction. $\square$

Corollary 1. The set $\{1!,2!,3!,\dots\}$ of factorials is not automatic.

Proof. Take $f(n)=n!$ in theorem 3. $\square$

Corollary 2. The set $\{2^{2^0}+1,2^{2^1}+1,2^{2^2}+1,\dots\}$ of Fermat numbers is not automatic.

Proof. Take $f(n)=2^{2^{n-1}}+1$ in theorem 3. $\square$

It is possible to use lemma 1 in other ways to show that sets are not automatic. For example:

Theorem 4. Let $f:\mathbb N\to\mathbb N$ such that $f(n+1)/f(n)\to\alpha$ as $n\to\infty$, where $\alpha$ is a real number such that $\alpha^m\not\in\mathbb N$ for any $m\in\mathbb N$. Then the set $\{f(1),f(2),f(3),\dots\}$ is not automatic.

Proof. Assume the contrary. Then by lemma 1 there exist distinct positive integers $s_0,s_1,\dots$ such that $f(s_{n+1})=Af(s_n)+B$ for all $n$, where $A>1$ and $B$ are fixed integers. Now

$\displaystyle\lim_{n\to\infty}\frac{f(n+m)}{f(n)}=\alpha^{m}$

for all $m\in\mathbb N$. Therefore, as $f(s_{n+1})/f(s_n)\to A$, it follows that $A=\alpha^m$ for some $m\in\mathbb N$. But this is impossible, as $\alpha^m\not\in\mathbb N$ for any $m\in\mathbb N$. $\square$

Corollary 3. The set $\{0,1,1,2,3,5,\dots\}$ of Fibonacci numbers is not automatic.

Proof. Let $F_n$ denote the $n$-th Fibonacci number. Note that $F_{n+1}/F_n\to\varphi=(1+\sqrt 5)/2$ as $n\to\infty$, and that $\varphi^m=F_m\varphi+F_{m-1}\not\in\mathbb N$ for all $m\in\mathbb N$. So the result follows by theorem 4. $\square$

In the exact same manner we also get:

Corollary4. The set $\{2,1,3,4,7,11,\dots\}$ of Lucas numbers is not automatic.

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## Finite automata 1: Automatic sets

In one of my recent courses I had to learn about finite automata theory. The main purpose was to understand a generalisation of the celebrated Skolem-Mahler-Lech theorem in positive characteristic. Let $\mathbb N_0$ denote the set of non-negative integers and $K$ be a field. For a linear recurrence $f:\mathbb N_0\to K$, define its zero set to be $\{n\in\mathbb N_0 : f(n)=0\}$. Then the Skolem-Mahler-Lech theorem states the following:

Theorem (Skolem-Mahler-Lech, 1953). The zero set of a linear recurrence $f:\mathbb N_0\to\mathbb K$, where $K$ is a field of characteristic $0$, is a finite union of arithmetic progressions along with a finite set.

Derksen in his 2005 paper was able to provide the correct analogue of this result in positive characteristic for the first time:

Theorem (Derksen, 2007). The zero set of a linear recurrence $f:\mathbb N_0\to K$, where $K$ is a field of characteristic $p>0$, is a $p$-automatic set.

To understand this theorem we need to know about finite automata. In this and the next few posts I want to write about some interesting elementary results about automatic sets. Since my knowledge of computer science is epsilon(!), I will mostly touch on some very basic mathematical aspects.

A deterministic finite-state automaton (DFA) is a directed multigraph $\Gamma$ satisfying certain properties. Firstly, its vertex-set $Q$ (which we call the set of states) contains a special vertex (or state) $q_0$, called the initial state. We also have a finite non-empty set $\Sigma$ of symbols. Each edge of $\Gamma$ is labelled with a symbol from $\Sigma$. As a result, we have a transition function $\delta:Q\times\Sigma\to Q$ such that if an edge from $q_1$ to $q_2$ is labelled with the symbol $a\in\Sigma$, then we write $\delta(q_1,a)=q_2$. Finally, we have a set $F\subseteq Q$ (possibly empty) called the set of accepting or final states.

So $\Gamma$ can be formally written as a quintuple $(\Sigma,Q,q_0,\delta,F)$.

Let $\Sigma^*$ be the free monoid on $\Sigma$. We can extend the transition function $\delta$ to a map $\delta:Q\times\Sigma^*\to Q$ by setting $\delta(q_0,wa)=\delta(\delta(q_0,a),w)$ for all $a\in\Sigma$ and $w\in\Sigma^*$. Thus, to $\Gamma$ we can associate a language

$\mathcal L_\Gamma:=\{w\in\Sigma^*:\delta(q_0,w)\in F\}\subseteq\Sigma^*$.

We call such a language produced from a DFA a regular language. If $w\in\mathcal L_\Gamma$ for some $w\in\Sigma^*$, we say that $w$ is accepted by $\Gamma$.

Let $\mathbb N$ denote the set of the positive integers, and let $k>1$ be an integer. Define $\Sigma_k:=\{0,1,\dots,k-1\}$. A subset $S\subset\mathbb N$ is said to be $k$-automatic if there exists a DFA $\Gamma=(\Sigma_k,Q,q_0,\delta,F)$ such that $S$ is precisely the set of those $j\in\mathbb N$ whose base-$k$ representation, when regarded as a string in $\Sigma^*$, is accepted by $\Gamma$. If $S\subseteq\mathbb N$ is $k$-automatic for some $k>1$, we say $S$ is an automatic set.

For $w=w_nw_{n-1}\cdots w_0\in\Sigma_k^*$ with each $w_i\in\Sigma_k$ we define

$[w]_k:=w_nk^n+w_{n-1}k^{n-1}+\cdots+w_0.$

In this case we also define $|w|:=n+1$ to be the length of $w$.

So $S\subseteq\mathbb N$ is $k$-automatic precisely when $S=\{[w]_k:w\in\mathcal L_\Gamma\}$ for some DFA $\Gamma$.

Note that we can set $\delta(q,0)$ to be in $Q\backslash F$ for every $q\in Q$ (if $F=Q$, just create a new state). Thus we may without loss of generality assume that all our DFAs only accept words without leading zeros.

A basic result about regular languages is the pumping lemma:

Lemma (Pumping lemma). If $\mathcal L$ is an infinite regular language, then there exist words $a,b,c$ with $|b|\ge 1$ such that $ab^nc\in\mathcal L$ for all $n\ge 0$.

Proof. Since $\mathcal L$ is infinite, there are words of arbitrarily large length in $\mathcal L$. Let $w=w_nw_{n-1}\cdots w_0\in\mathcal L$ be a word of sufficient length. Then there exist $0\le i such that $w_i\cdots w_0=w_j\cdots w_0\in\mathcal L$. Taking $a=w_n\cdots w_j$, $b=w_{j-1}\cdots w_i$, $c=w_{i-1}\cdots w_0$ shows that $ab^nc\in\mathcal L$ for all $n\ge 0$. Moreover, since $i, it follows that $|b|\ge 1$. $\square$

In the following posts we are going to use the pumping lemma, together with other machinery, to derive some interesting results on the automaticity of certain sets.

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## Some divisibility properties of Fibonacci-like sequences

Here we shall aim to generalise the corollary from the last post, and on the way generalise some other important properties of the Fibonacci sequence, to general Fibonacci-like sequences.

Consider a sequence $\{a_n\}$ with $a_0=0, a_1=1$ and $a_n=ua_{n-1}+va_{n-2}$ for $n\ge 2$, where $u,v\in\mathbb Z$ such that $u^2+4v\neq 0$. Let $\alpha,\beta$ be the roots of $x^2-ux-v=0$. Then

$\displaystyle a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$.

Let $p$ be an odd prime. Suppose $(\alpha)$ and $(p)$ have a common factor. By Dedekind’s factorisation criterion, $(p)$ is either prime or it factors into two (not necessarily distinct) prime ideals each with norm $p$. In the first case, we must have $(p)\mid (\alpha)$ i.e. $p^2\mid v$, taking ideal norms. In the second case, if one of the prime factors of $(p)$ divides $(\alpha)$, then (again taking norms) $p$ must divide $v$.

So we take $(p,v)=1$ which ensures that $(\alpha)$ and $(p)$ are coprime. Thus we deduce from the general exponent lifting result,

Lemma 1. If $p$ is a prime factor of $a_k$ such that $p\nmid v$, then $p^r\parallel a_k\Rightarrow p^{n+r}\parallel a_{kp^n}\ \forall n\ge 0$.

Next, recalling the proof of the corollary from last time, we seem to require the condition $(a_m,a_n)=a_{(m,n)}$.

Lemma 2. If $(u,v)=1$, then $(a_m,a_n)=a_{(m,n)}$ for all $m,n$.

Proof. First, it is clear that $(a_m,a_n)/a_{(m,n)}\in\mathcal O_K$, where $K=\mathbb Q(\alpha)$, and that $(a_m,a_n)/a_{(m,n)}\in\mathbb Q$. Thus $(a_m,a_n)/a_{(m,n)}\in\mathbb Z$.

Let $p$ be a prime factor of $(a_m,a_n)$. Then for $m\ge n$,

$\displaystyle p\mid\frac{\alpha^m-\beta^m}{\alpha-\beta}-\frac{\alpha^{m-n}(\alpha^n-\beta^n)}{\alpha-\beta}=\frac{\beta^n(\alpha^{m-n}-\beta^{m-n})}{\alpha-\beta}=\beta^na_{m-n}$.

If $(p)$ and $(\beta^n)$ share a common factor, then (taking ideal norms,) $p\mid v$. Hence $a_n\equiv ua_{n-1}\equiv u^{n-1}\equiv 0\pmod p$, a contradiction as $(u,p)=1$.

Hence $(a_m,a_n)\mid a_{m-n}$ and by the Euclidean algorithm we get $(a_m,a_n)\mid a_{(m,n)}$. Thus $(a_m,a_n)=a_{(m,n)}$, as desired. $\square$

Now we define $k(n)$ to be the least $k$ such that $n\mid a_k$, and we are ready for our generalisation!

Theorem. Let $\{a_n\}$ be given by $a_0=0,a_1=1$ and $a_n=ua_{n-1}+va_{n-2}$ for $n\ge 2$, where $u,v$ are coprime integers such that $u^2+4v\neq 0$. Then for an odd prime $p$ for which $k(p)$ exists,

$p^r\parallel a_{k(p)}\Rightarrow k(p^n)=\begin{cases}p^{n-r}k(p)&\text{ if }n\ge r,\\ k(p)&\text{ if }n\le r.\end{cases}$

Proof. If $p\mid v$, then $p\mid a_{k(p)}-va_{k(p)-2}=ua_{k(p)-1}$. By definition of $k(p)$, $p\nmid a_{k(p)-1}$. Hence $p\mid u$, which is impossible as $(u,v)=1$. Thus $p\nmid v$ and the proof of the corollary in the previous post now applies. $\square$

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## A generalisation of ‘lifting the exponent’

Lifting the exponent is a popular result in elementary number theory. It basically says that if $\alpha,\beta\in\mathbb Z$ and $a_n=(\alpha^n-\beta^n)/(\alpha-\beta)$, then for an odd prime $p$ coprime to $\alpha,\beta$ and for $r\ge 1$,

Theorem 1. (Lifting the exponent) $p^r\parallel a_k\Rightarrow p^{n+r}\parallel a_{kp^n}\ \forall n\ge 0$.

Here, as usual, $p^r\parallel x$ means that $p^r\mid x$ but $p^{r+1}\nmid x$.

We shall generalise this to general rings of algebraic integers. In the following, $K$ will denote an algebraic number field and $\mathcal O_K$ its ring of integers. Let $\alpha,\beta\in\mathcal O_K$ such that the ideals $(\alpha), (\beta)$ are coprime to $(p)$, and $a_n\in\mathbb Z\ \forall n$.

Theorem 2. $p^r\parallel a_k\Rightarrow p^{n+r}\parallel a_{kp^n}\ \forall n\ge 0$.

Proof. We proceed by induction on $n$. The base case $n=0$ is clear. Suppose that the assertion holds for $n-1$. We want to show that $p^{n+r}\parallel a_{kp^n}$, that is,

$\displaystyle p^{n+r}\parallel\frac{\alpha^{kp^n}-\beta^{kp^n}}{\alpha-\beta}=a_{kp^{n-1}}\cdot\frac{\alpha^{kp^n}-\beta^{kp^n}}{\alpha^{kp^{n-1}}-\beta^{kp^{n-1}}}$

By hypothesis, $p^{n+r-1}\parallel a_{kp^{n-1}}$. So we need to show that $\displaystyle p\parallel\frac{\alpha^{kp^n}-\beta^{kp^n}}{\alpha^{kp^{n-1}}-\beta^{kp^{n-1}}}$.

Let $c=\alpha^{kp^{n-1}}, d=\beta^{kp^{n-1}}$. We want to show that $\displaystyle p\parallel\frac{c^p-d^p}{c-d}$.

Since $p\mid a_k$, we have $a_k\in (p)$. Hence $\alpha^k-\beta^k=(\alpha-\beta)a_k\in (p)$, i.e. $\alpha^k\equiv\beta^k\pmod p$. Hence $c\equiv d\pmod p$. Let $c=d+\mu p$ for some $\mu\in\mathcal O_K$. Then

$\displaystyle\frac{c^p-d^p}{c-d}=\frac{(d+\mu p)^p-d^p}{\mu p}=\frac{p\cdot d^{p-1}\mu p+\ldots}{\mu p}\equiv pd^{p-1}\pmod{p^2}.$

If this is $0$, then $(p)\mid (d)^{p-1}$, which is impossible by unique factorisation as $(p)$ and $(d)$ are coprime. $\square$

Corollary. Let $\{F_n\}$ be the Fibonacci sequence. Then

$p^r\parallel F_{k(p)}\Rightarrow k(p^n)=\begin{cases}p^{n-r}k(p)&\text{ if }n\ge r,\\ k(p)&\text{ if }n\le r.\end{cases}$

(Refer to this post for notation.)

Proof. Take $\alpha=\varphi=\frac{1+\sqrt 5}{2}, \beta=\bar\varphi=\frac{1-\sqrt 5}{2}$ (with $K=\mathbb Q(\sqrt 5)$ and $\mathcal O_K=\mathbb Z[\varphi]$) and let $k=k(p)$. We proceed by induction on $n$. Suppose $p^{n+r}\mid F_j$ for some $j. By hypothesis, $k(p^{n+r-1})=kp^{n-1}$ so $p^{n+r-1}\mid p^{n+r}\mid F_j\Rightarrow kp^{n-1}\mid j. Then $1\le \frac{j}{kp^{n-1}}, so $\gcd(\frac{j}{kp^{n-1}},p)=1$, i.e. $\gcd(j,kp^n)=kp^{n-1}$. Hence $\gcd(F_{kp^n},F_j)=F_{kp^{n-1}}$ which implies $p^{n+r}\mid F_{kp^{n-1}}$, a contradiction. $\square$

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## Fibonacci sequence modulo n: Part 2

See here.

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