Extended Vuza Canons

Abstract

Starting from well-known constructions of aperiodic tiling rhythmic canons by G. Hajós, N.G. de Bruijn and D.T. Vuza, several generalisations are given. In this way, it is possible to find new aperiodic canons, that we call extended Vuza canons.

Supported by Italian Ministry of Education, University and Research (MIUR), Dipartimenti di Eccellenza Program (2018–2022).

1 Prelude

Canons in music have a very long tradition; among these, a few cases of tiling rhythmic canons (i.e. canons such that, given a fixed tempo, at every beat exactly one voice is playing) have emerged. Only in the last century, stemming from the analogous problem of factorizing finite abelian groups, aperiodic tiling rhythmic canons have been studied: these are canons that tile a certain interval of time in which each voice (inner voice) plays at an aperiodic sequence of beats, and the sequence of starting beats of every voice (outer voice) is also aperiodic. From the musical point of view the seminal paper was probably the four-parts article written by D.T. Vuza between 1991 and 1993 [14,15,16,17], while the mathematical counterpart of the problem was studied also before, e.g. by de Bruijn [5], Sands [13], etc., and after, e.g. by Coven and Meyerowitz [4], Jedrzejewski [9], Amiot [1], Andreatta [3], etc.

A thorough theory of the conditions of existence and the structure of aperiodic tiling rhythmic canons has not been established yet. In this paper we try to give a contribution to this fascinating field.

2 Aperiodic Tiling Canons

We begin fixing some notations and giving the main definitions. In the following, we conventionally denote the cyclic group of remainder classes modulo n by $\mathbb Z_n$ and its elements with the integers $\left\{ 0 , 1 , \ldots , n - 1 \right\}$, i.e., identifying each class with its least non-negative member.

Definition 1

Let $A , B \subset \mathbb Z_n$. Let us define the application

$$ \sigma : A \times B \rightarrow \mathbb Z_n , \left( a , b \right) \mapsto a + b. $$

We set $A + B \doteqdot \text{ Im }(\sigma )$; if $\sigma$ is bijective, we say that A and B are in direct sum, and we write

$$ A \oplus B \doteqdot \text{ Im }(\sigma ). $$

If $\mathbb {Z}_n = A \oplus B$, we call $\left( A , B \right)$ a tiling rhythmic canon of period n; A is called the inner voice and B the outer voice of the canon.

Remark 1

It is easy to see that the tiling property is invariant under translations, i.e., if A is a tiling complement of some set B, also any translate $A + z$ of A is a tiling complement of B (and any translate of B is a tiling complement of A). Thus, without loss of generality, we shall limit our investigation to rhythms containing 0 and consider equivalence classes under translation.

Definition 2

A rhythm $A \subset \mathbb Z_n$ is periodic (of period z) if and only if there exists an element $z \in \mathbb Z_n$, $z \ne 0$, such that $z + A = A$. In this case, A is also called periodic modulo $z \in \mathbb Z_n$. A rhythm $A \subset \mathbb Z_n$ is aperiodic if and only if it is not periodic.

Denote by $\varPhi _d(x)$ the cyclotomic polynomial of index d. Then, tiling rhythmic canons can be characterised as follows.

Lemma 1

Let A be a rhythm in $\mathbb Z_n$ and let A(x) be the characteristic polynomial of A, that is, $A(x) = \sum _{ k \in A }x^k$. Given $B \subset \mathbb Z_n$ and its characteristic polynomial B(x) , we have that

$$\begin{aligned} A(x) \cdot B(x) \equiv \sum _{ k = 0 }^{ n - 1 } x^k = \frac{ x^n - 1 }{ x - 1 } = \prod _{ d \,|\, n , d \ne 1 } \varPhi _d(x) \qquad \mod \left( x^n - 1 \right) \end{aligned}$$

(1)

if and only if A(x) and B(x) are polynomials with coefficients in $\left\{ 0 , 1 \right\}$ and $A \oplus B = \mathbb Z_n$.

As a consequence, for each $d | n$, with $d > 1$, we have

$$ \varPhi _d(x) | A(x) \,\,{\text {or}}\,\, \varPhi _d(x) | B(x). $$

Definition 3

A tiling rhythmic canon $\left( A , B \right)$ in $\mathbb Z_n$ is an aperiodic tiling rhythmic canon if both A and B are aperiodic.

For an extensive discussion on tiling problems, we refer the reader to Amiot [2]. If we indicate the set $\left\{ d \in \mathbb N : d | n \right\}$ by $\mathsf {div}(n)$, the following proposition establishes a polynomial criterion for the aperiodicity of a given rhythm.

Proposition 1

A set $A \subset \mathbb Z_n$ is aperiodic if and only if for all $k | n$, with $k \ne n$, we have

that is, if and only if for all $k \in \mathsf {div}(n) \setminus \left\{ n \right\}$ there exists $d \in \mathsf {div}(n) \setminus \mathsf {div}(k)$ such that .

The following result, in conjunction with Theorem 2, identifies which are the periods of aperiodic tiling rhythmic canons.

Theorem 1 (Vuza)

Let

• $\mathcal V \doteqdot \{ n \in \mathbb N : n = p_1 n_1 p_2 n_2 n_3 \text{ with } \gcd \left( p_1 n_1 , p_2 n_2 \right) = 1 \text{ and } p_1 , n_1 , p_2 , n_2$, $n_3 > 1 \}$, and

• $\mathcal {H} \doteqdot \left\{ p^\alpha , p^\alpha q , p^2 q^2 , p q r , p^2 q r , p q r s : \alpha \in \mathbb N, \ p , q , r , s \text{ distinct } \text{ primes } \right\}$,

then $\mathbb N^* = \mathcal V \sqcup \mathcal H$.

The minimum period necessary for an aperiodic canon is 72, and the corresponding $p_i$ and $n_i$ are:

$$ \left( p_1 , n_1 , p_2 , n_2 , n_3 \right) = \left( 2 , 2 , 3 , 3 , 2 \right) . $$

3 Extended Vuza Canons

The canons with periods 72, 108, 120, 144 and 168 have been completely enumerated by Vuza [14], Fripertinger [7], Amiot [1], Kolountzakis and Matolcsi [11].

An exhaustive construction method for aperiodic tiling rhythmic canons is not known to date; the first method to find some of them was provided by the following result (see [8] by Hajós, Theorem 1 in [5] by de Bruijn, and Proposition 2.2 in [14] by Vuza).

Theorem 2

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$ and

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$.

Then $\mathbb Z_n$ admits an aperiodic tiling rhythmic canon.

Example 1

In the hypotheses of Theorem 2, an example of tiling canon of $\mathbb Z_n$ with two aperiodic subsets is given by the following construction by F. Jedrzejewski (see Theorem 227 in [9]). Indicating with $\mathbb I_k$ the set $\left\{ 0 , 1 , \ldots , k - 1 \right\}$, let us call:

$$ \begin{aligned} A_1&= n_3 p_1 n_1 \mathbb I_{ n_2 } \\ U_1&= n_3 p_1 n_1 n_2 \mathbb I_{ p_2 } \\ V_1&= n_3 n_2 \mathbb I_{ p_2 } \\ K_1&= \left\{ 0 \right\} \end{aligned} \qquad \begin{aligned} A_2&= n_3 p_2 n_2 \mathbb I_{ n_1 } \\ U_2&= n_3 p_2 n_2 n_1 \mathbb I_{ p_1 } \\ V_2&= n_3 n_1 \mathbb I_{ p_1 } \\ K_2&= \left\{ 1 , 2 , \ldots , n_3 - 1 \right\} . \end{aligned} $$

Then taking

$$ \begin{aligned} A&= A_1 \oplus A_2 \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) , \end{aligned} $$

we have the canon $\mathbb Z_n = A \oplus B$.

Remark 2

From now on, given $p_1$, $n_1$, $p_2$, $n_2$, and $n_3$, we will denote by $A_1$, $A_2$, $U_1$, $U_2$, $V_1$, and $V_2$ the sets so called in Example 1.

Many other ways of constructing aperiodic tiling canons are possible, see for example de Bruijn [5], Vuza [14], Fidanza [6], and Jedrzejewski [9]. These methods fall into a category treated by F. Jedrzejewski (Theorem 14 in [10]). We refine his result lifting the hypothesis that $p_1$ and $p_2$ are prime and proving that B is aperiodic if $n_3$ satisfies a simple arithmetic constraint.

Theorem 3

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that:

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$;

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$;

• 3. there is no prime q such that $q | n_3$, but .

Let H be the subgroup $H = n_3 \mathbb I_{ p_1 n_1 p_2 n_2 }$ of $\mathbb Z_n$ and let K be a complete set of cosets representatives for $\mathbb Z_n$ modulo H such that K is the disjoint union $K = K_1 \sqcup K_2$. Then the pair $\left( A , B \right)$ defined by

$$ \begin{aligned} A&= A_1 \oplus A_2 \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) \end{aligned} $$

is an aperiodic tiling rhythmic canon of $\mathbb Z_n$.

Proof

The proof that $A \oplus B = \mathbb Z_n$ and that the set A is aperiodic is the same as in Vuza (Proposition 2.2 in [14]). We are left to prove that B is aperiodic. Consider the characteristic polynomial B(x) :

$$ B(x) = \frac{ x^{ n_3 p_1 n_1 } - 1 }{ x^{ n_3 n_1 } - 1 } \frac{ x^n - 1 }{ x^{ n_3 p_1 n_1 n_2 } - 1 } K_1(x) + \frac{ x^{ n_3 p_2 n_2 } - 1 }{ x^{ n_3 n_2 } - 1 } \frac{ x^n - 1 }{ x^{ n_3 p_2 n_2 n_1 } - 1 } K_2(x). $$

Given any $h \in \mathbf {\mathrm {div}}(n) \setminus \left\{ n \right\}$, we look for a $d \in \mathbf {\mathrm {div}}(n) \setminus \mathbf {\mathrm {div}}(h)$ such that . Let us consider the cases:

• 1. if , then since

  $$ \varPhi _{ n_3 p_2 n_2 n_1 }(x) | \frac{ x^n - 1 }{ x^{ n_3 p_1 n_1 n_2 } - 1 } $$

  but

  

  In particular, by Lemma 4 of Rédei’s paper [12].

• 2. if , then $\varPhi _{ n_3 p_1 n_1 n_2 }(x) | B(x)$ (symmetrically to the previous case).

There are no other possibilities: in fact, if we had $n_3 p_2 n_2 n_1 | h$ and $n_3 p_1 n_1 n_2 | h$, then $h = \alpha n_3 p_2 n_2 n_1 = \beta n_3 p_1 n_1 n_2$ and therefore $\alpha p_2 = \beta p_1$. Since $\gcd \left( p_1 , p_2 \right) = 1$, it would follow $\alpha = p_1$ and $\beta = p_2$ and so $h = n$, which is a contradiction.    $\square$

Example 2

Consider $n = 216$; let $p_1 = 2$, $n_1 = 2$, $p_2 = 3$, $n_2 = 3$, and $n_3 = 6$. Theorem 3 ensures that, defining

$$ \begin{aligned} A&= 24 \mathbb I_3 \oplus 54 \mathbb I_2 \\ B&= \left( 72 \mathbb I_3 \oplus 12 \mathbb I_2 \oplus \left\{ 0 , 106 \right\} \right) \sqcup \left( 108 \mathbb I_2 \oplus 18 \mathbb I_3 \oplus \left\{ 21 , 43 , 122 , 167 \right\} \right) , \end{aligned} $$

$A \oplus B = \mathbb Z_{216}$ and $\left( A , B \right)$ is an aperiodic tiling rhythmic canon.

In a first generalization of Theorem 3, rhythm B is the disjoint union of three sets, one being periodic both modulo $n/p_1$ and modulo $n/p_2$.

Theorem 4

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that:

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$;

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$;

• 3. there is no prime q such that $q | n_3$, but .

Let H be the subgroup $H = n_3 \mathbb I_{ p_1 n_1 p_2 n_2 }$ of $\mathbb Z_n$ with $n = p_1 n_1 p_2 n_2 n_3$, K be a complete set of cosets representatives for $\mathbb Z_n$ modulo H such that K is the disjoint union $K = K_1 \sqcup K_2 \sqcup K_3$ with $K_1 , K_2 \ne \emptyset$, and $W = n_3 n_1 n_2 \mathbb I_{ p_1 p_2 }$. Then the pair $\left( A , B \right)$ defined by

$$ \begin{aligned} A&= A_1 \oplus A_2 \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) \sqcup \left( W \oplus K_3 \right) \end{aligned} $$

is an aperiodic tiling rhythmic canon of $\mathbb Z_n$.

Proof

The only case we need to consider is $K_3 \ne \emptyset$ (notice that this is possible only if $n_3 > 2$). We already know, from Theorem 2 that A is aperiodic; B is aperiodic too, since

$$ B(x) = U_1(x) V_2(x) K_1(x) + U_2(x) V_1(x) K_2(x) + W(x) K_3(x) $$

and the cyclotomic polynomials $\varPhi _{ n_3 p_2 n_2 n_1 }$ and $\varPhi _{ n_3 p_1 n_1 n_2 }$ divide exactly 2 of the summands on the right hand side.

We now prove that $A \oplus B = \mathbb Z_n$: to this aim we make use of the following facts, proven by F. Jedrzejewski (Theorem 14 in [10]):

$$ \begin{aligned} A_1 + U_1 + V_2&= A_1 + U_1 + U_2 \\ A_2 + U_2 + V_1&= A_2 + U_2 + U_1. \end{aligned} $$

By an easy check, we see that

$$ U_1 + U_2 = n_3 n_1 n_2 \left( p_1 \mathbb I_{ p_2 } + p_2 \mathbb I_{ p_1 } \right) = n_3 n_1 n_2 \mathbb Z_{ p_1 p_2 } = W, $$

and $| U_1 | | U_2 | = p_2 p_1 = | W |$. This means that

$$ U_1 \oplus U_2 = W. $$

We obtain that

$$\begin{aligned}&A + B = \left( A_1 + A_2 \right) + \left( \left( U_1 + V_2 + K_1 \right) \sqcup \left( U_2 + V_1 + K_2 \right) \sqcup \left( W + K_3 \right) \right) \\&\qquad \quad = \left( A_1 + A_2 + U_1 + V_2 + K_1 \right) \sqcup \left( A_1 + A_2 + U_2 + V_1 + K_2 \right) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \sqcup \left( A_1 + A_2 + W + K_3 \right) \\&\qquad \quad =\left( A_1 + A_2 + U_1 + U_2 + K_1 \right) \sqcup \left( A_1 + A_2 + U_2 + U_1 + K_2 \right) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \;\;\; \sqcup \left( A_1 + A_2 + U_1 + U_2 + K_3 \right) \\&\qquad \quad = A_1 + A_2 + U_1 + U_2 + \left( K_1 \sqcup K_2 \sqcup K_3 \right) \\&\qquad \quad = A_1 + U_1 + A_2 + U_2 + K. \end{aligned}$$

Again, an easy computation shows that

$$ \begin{aligned} \left( A_1 + U_1 \right) + \left( A_2 + U_2 \right)&= n_3 p_1 n_1 \mathbb I_{ p_2 n_2 } + n_3 p_2 n_2 \mathbb I_{ p_1 n_1 } \\&= n_3 \mathbb I_{ p_1 n_1 p_2 n_2 } \\&= H \end{aligned} $$

and so

$$ A + B = H + K = \mathbb Z_n. $$

Moreover, since $| A | | B | = n = | H | | K |$, the sum $A + B$ is direct.    $\square$

Example 3

Let us go back to $n = 216$ with the same choices of $p_1$, $n_1$, $p_2$, $n_2$, and $n_3$. By Theorem 4, we find a new aperiodic tiling rhythmic canon $\left( A , B \right)$ defining

$$ \begin{aligned} A&= 24 \mathbb I_3 \oplus 54 \mathbb I_2 \\ B&= \left( 72 \mathbb I_3 \oplus 12 \mathbb I_2 \oplus \left\{ 0 , 106 \right\} \right) \sqcup \left( 108 \mathbb I_2 \oplus 18 \mathbb I_3 \oplus \left\{ 21 , 43 \right\} \right) \sqcup \left( 36 \mathbb I_6 \oplus \left\{ 122 , 167 \right\} \right) . \end{aligned} $$

The second generalization of Theorem 3 widens the definitions of sets $A_1$, $A_2$, $V_1$, and $V_2$. We precede it with a useful lemma.

Lemma 2

Suppose that a subset $S \subseteq \mathbb Z_n$ is periodic of period $m | n$, i.e. $S + m = S$, and for $i = 0 , \ldots , k - 1$ let $S_i = \left\{ a \in S : a \equiv i \mod k \right\}$ where k is a divisor of m. Then also the sets $S_i$ are periodic of period m for every i.

Proof

It is sufficient to observe that since m is a multiple of k the remainder classes modulo k are invariant by the translation by m, hence also $S_i + m = S_i$.

   $\square$

Theorem 5

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that:

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$;

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$;

• 3. there is no prime q such that $q | n_3$, but .

Let H be the subgroup $H = n_3 \mathbb I_{ p_1 n_1 p_2 n_2 }$ of $\mathbb Z_n$, and $K = K_1 \sqcup K_2$ (with $K_1 , K_2 \ne \emptyset$) be a complete set of cosets representatives for $\mathbb Z_n$ modulo H. Take

• $\tilde{A}_1$ as a complete aperiodic set of coset representatives for $\mathbb Z_{ p_2 n_2 }$ modulo $n_2 \mathbb I_{ p_2 }$;

• $\tilde{A}_2$ as a complete aperiodic set of coset representatives for $\mathbb Z_{ p_1 n_1 }$ modulo $n_1 \mathbb I_{ p_1 }$;

• $\tilde{V}_1^1 , \ldots , \tilde{V}_1^j$ as complete aperiodic sets of coset representatives for $\mathbb Z_{ p_2 n_1 }$ modulo $p_2 \mathbb I_{ n_1 }$;

• $\tilde{V}_2^1 , \ldots , \tilde{V}_2^h$ as complete aperiodic sets of coset representatives for $\mathbb Z_{ p_1 n_2 }$ modulo $p_1 \mathbb I_{ n_2 }$.

Set $K_1 = K_1^1 \sqcup \cdots \sqcup K_1^j$ and $K_2 = K_2^1 \sqcup \cdots \sqcup K_2^h$, where $K_\alpha ^s = \left\{ k_\alpha ^{ j_{ s - 1 } + 1 }, \ldots , k_\alpha ^{j_s} \right\}$ are non-empty subsets of $K_\alpha$ for $\alpha = 1 , 2$. Then the pair $\left( A , B \right)$ defined by

$$\begin{aligned}&\; A = n_3 p_1 n_1 \tilde{A}_1 \oplus n_3 p_2 n_2 \tilde{A}_2 \\&B = \Bigl ( \left( U_1 \oplus n_3 n_1 \tilde{V}_2^1 \oplus \left\{ k_1^1 , \ldots , k_1^{l_1} \right\} \right) \sqcup \cdots \\&\quad \quad \;\; \cdots \sqcup \left( U_1 \oplus n_3 n_1 \tilde{V}_2^j \oplus \left\{ k_1^{ l_{j-1} + 1} , \ldots , k_1^{ | K_1 | } \right\} \right) \Bigr )\\&\qquad \quad \;\;\; \sqcup \Bigl ( \left( U_2 \oplus n_3 n_2 \tilde{V}_1^1 \oplus \left\{ k_2^1 , \ldots , k_2^{m_1} \right\} \right) \sqcup \cdots \\&\qquad \quad \cdots \sqcup \left( U_2 \oplus n_3 n_2 \tilde{V}_1^h \oplus \left\{ k_2^{ m_{ h - 1 } + 1 } , \ldots , k_2^{ | K_2 | } \right\} \right) \Bigr ) \end{aligned}$$

is an aperiodic tiling rhythmic canon of $\mathbb Z_n$.

Proof

We have

• $n_3 p_1 n_1 \tilde{A}_1 + U_1 = n_3 p_1 n_1 \left( \tilde{A}_1 \oplus n_2 \mathbb I_{ p_2 } \right) = n_3 p_1 n_1 \mathbb I_{ p_2 n_2 } = A_1 + U_1$;

• $n_3 p_2 n_2 \tilde{A}_2 + U_2 = n_3 p_2 n_2 \left( \tilde{A}_2 \oplus n_1 \mathbb I_{ p_1 } \right) = n_3 p_2 n_2 \mathbb I_{ p_1 n_1 } = A_2 + U_2$;

• $A_1 + n_3 n_1 \tilde{V}_2 = n_3 n_1 \left( p_1 \mathbb I_{ n_2 } + \tilde{V}_2 \right) = n_3 n_1 \mathbb I_{ p_1 n_2 } = A_1 + V_2$;

• $A_2 + n_3 n_2 \tilde{V}_1 = n_3 n_2 \left( p_2 \mathbb I_{ n_1 } + \tilde{V}_1 \right) = n_3 n_2 \mathbb I_{ p_2 n_1 } = A_2 + V_1$.

For the sake of simplicity, we now give the proof in the case $j = 1$ and $h = 1$. The general case is completely analogous. We compute

$$\begin{aligned}&A + B = \left( n_3 p_1 n_1 \tilde{A}_1 + n_3 p_2 n_2 \tilde{A}_2 \right) \\&\qquad \qquad \quad \; +\left( \left( U_1 + n_3 n_1 \tilde{V}_2 + K_1 \right) \sqcup \left( U_2 + n_3 n_2 \tilde{V}_1 + K_2 \right) \right) \\&\qquad \quad = \left( n_3 p_1 n_1 \tilde{A}_1 + n_3 p_2 n_2 \tilde{A}_2 + U_1 + n_3 n_1 \tilde{V}_2 + K_1 \right) \\&\qquad \qquad \quad \; \sqcup \left( n_3 p_1 n_1 \tilde{A}_1 + n_3 p_2 n_2 \tilde{A}_2 + U_2 + n_3 n_2 \tilde{V}_1 + K_2 \right) \\&\qquad \quad = \left( A_1 + n_3 p_2 n_2 \tilde{A}_2 + U_1 + n_3 n_1 \tilde{V}_2 + K_1 \right) \\&\qquad \qquad \quad \; \sqcup \left( n_3 p_1 n_1 \tilde{A}_1 + A_2 + U_2 + n_3 n_2 \tilde{V}_1 + K_2 \right) \\&\qquad \quad = \left( A_1 + n_3 p_2 n_2 \tilde{A}_2 + U_1 + V_2 + K_1\right) \\&\qquad \qquad \quad \; \sqcup \left( n_3 p_1 n_1 \tilde{A}_1 + A_2 + U_2 + V_1 + K_2\right) \\&\qquad \quad = \left( A_1 + n_3 p_2 n_2 \tilde{A}_2 + U_1 + U_2 + K_1\right) \\&\qquad \qquad \quad \; \sqcup \left( n_3 p_1 n_1 \tilde{A}_1 + A_2 + U_2 + U_1 + K_2\right) \\&\qquad \quad = A_1 + A_2 + U_1 + U_2 + \left( K_1 \sqcup K_2 \right) \\&\qquad \quad = A_1 + U_1 + A_2 + U_2 + K \\&\qquad \quad = \mathbb Z_n. \end{aligned}$$

A cardinality argument analogous to that used in Theorem 4 shows that the sum is direct.

The proof that A is aperiodic follows from Vuza’s argument (Proposition 2.2 in [14]), as above. Assume now that B is periodic of period a: we can assume without loss of generality that $a = n/p$ where p is a prime number. Hypothesis now implies that a is a multiple of $n_3$: but then by Lemma 2 also the sets $B_i = B \cap \left( \left\{ i \right\} + n_3 \mathbb Z_n \right)$ must be periodic of period a. However, the sets $B_i$ are simply translates of $U_1 \oplus n_3 n_1 \tilde{V}_2$ by elements of $K_1$ or of $U_2 \oplus n_3 n_2 \tilde{V}_1$ by elements of $K_2$ (remember that also the elements of $U_1$ and $U_2$ are multiple of $n_3$): on their turn, $U_1 \oplus n_3 n_1 \tilde{V}_2$ and $U_2 \oplus n_3 n_2 \tilde{V}_1$ are indeed periodic resp. of period $n/p_1$ and $n/p_2$, but since $p_1$ and $p_2$ are coprime no common period smaller than n is possible. A contradiction follows since we assumed both $K_1$ and $K_2$ to be non-empty.    $\square$

Remark 3

Note that Theorems 3–5 hold trivially if hypothesis is replaced by the condition that $n_3$ is prime.

Example 4

This time we choose $n = 252$; let $p_1 = 2$, $n_1 = 7$, $p_2 = 3$, $n_2 = 3$, and $n_3 = 2$. We can take e.g.

$$ \begin{aligned} \tilde{A}_1&= \left\{ 0 , 2 , 7 \right\} \\ \tilde{V}_1&= \left\{ 0 , 10 , 17 \right\} \\ K_1&= \left\{ 0 \right\} \end{aligned} \qquad \begin{aligned} \tilde{A}_2&= \left\{ 0 , 1 , 3 , 4 , 9 , 12 , 13 \right\} \\ \tilde{V}_2&= \left\{ 0 , 1 \right\} = \mathbb I_{ p_1 } \\ K_2&= \left\{ 1 \right\} \end{aligned} $$

obtaining a new canon $\left( A , B \right)$ where

$$ \begin{aligned} A&= 28 \tilde{A}_1 \oplus 18 \tilde{A}_2 \\&= \left\{ 0 , 56 , 196 \right\} \oplus \left\{ 0 , 18 , 54 , 72 , 162 , 216 , 234 \right\} \\ B&= \left( U_1 \oplus 14 \tilde{V}_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus 6 \tilde{V}_1 \oplus K_2 \right) \\&= \left( \left\{ 0 , 84 , 168 \right\} \oplus \left\{ 0 , 14 \right\} \oplus \left\{ 0 \right\} \right) \sqcup \left( \left\{ 0 , 126 \right\} \oplus \left\{ 0 , 60 , 102 \right\} \oplus \left\{ 1 \right\} \right) . \end{aligned} $$

Definition 4

We call Vuza canons all the canons obtained using the constructions described in Theorems 2, 3, 4, 5.

It is possible to stretch this type of constructions even further. With the following theorem, we improve the result of Jedrzejewski (Theorem 21 in [10]).

Theorem 6

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that:

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$;

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$;

• 3. there is no prime q such that $q | n_3$, but .

Let H be the subgroup $H = n_3 \mathbb I_{ p_1 n_1 p_2 n_2 }$ of $\mathbb Z_n$. Suppose that L and K are proper subsets of $\mathbb Z_{ n_3 }$ such that $L \oplus K = \mathbb Z_{ n_3 }$ and $K = K_1 \sqcup K_2$, with $K_1 , K_2 \ne \emptyset$. Then the pair $\left( A , B \right)$ defined by

$$ \begin{aligned} A&= A_1 \oplus A_2 \oplus L \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) \end{aligned} $$

is an aperiodic tiling rhythmic canon of $\mathbb Z_n$.

Proof

$$ \begin{aligned} A + B&= \left( A_1 + A_2 + L \right) + \left( \left( U_1 + V_2 + K_1 \right) \sqcup \left( U_2 + V_1 + K_2 \right) \right) \\&= \left( A_1 + A_2 + L + U_1 + V_2 + K_1 \right) \sqcup \left( A_1 + A_2 + L + U_2 + V_1 + K_2 \right) \\&= \left( A_1 + A_2 + L + U_1 + U_2 + K_1 \right) \sqcup \left( A_1 + A_2 + L + U_2 + U_1 + K_2 \right) \\&= A_1 + A_2 + L + U_1 + U_2 + \left( K_1 \sqcup K_2 \right) \\&= A_1 + U_1 + A_2 + U_2 + L + K. \end{aligned} $$

The sum is direct because the computation of the cardinality leads to

$$ | A_1 | | A_2 | | U_1 | | U_2 | | L \oplus K | = n. $$

Aperiodicity of A is immediate from Lemma 2, since $A_1 + A_2$ is aperiodic, and B is the union of the subsets $B_i$ contained in different remainder classes modulo $n_3$, some of which have a period coprime with the period of the other ones (exactly as in the previous theorem).

Example 5

Choosing again $n = 216$ and the same values for $p_1$, $n_1$, $p_2$, $n_2$, and $n_3$ as in Example 3, we set $L = \left\{ 0 , 1 \right\}$, $K_1 = \left\{ 2 \right\}$, and $K_2 = \left\{ 0 , 4 \right\}$. By Theorem 6, we get that

$$ \begin{aligned} A&= 24 \mathbb I_3 \oplus 54 \mathbb I_2 \oplus L \\ B&= \left( 72 \mathbb I_3 \oplus 12 \mathbb I_2 \oplus K_1 \right) \sqcup \left( 108 \mathbb I_2 \oplus 18 \mathbb I_3 \oplus K_2 \right) \end{aligned} $$

define an aperiodic tiling rhytmic canon.

To prove our next result we take advantage of the equivalent polynomial formulation of tilings. Using it, in [4] E.M. Coven, and A. Meyerowitz introduced two sufficient conditions for a rhythm A to be a factor of a tiling rhythmic canon. To state them we need the following definitions.

Definition 5

$R_A \doteqdot \left\{ d : \varPhi _d(x) | A(x) \right\}$ and $S_A \doteqdot \left\{ p^\alpha \in R_A : p \text{ prime } \right\}$.

The Coven-Meyerowitz conditions are the following:

• T1 $| A | = \prod _{ p^\alpha \in S_A }p$;

• T2 for all $p^\alpha , q^\beta , r^\gamma , \ldots \in S_A$, $p^\alpha q^\beta r^\gamma \cdots \in R_A$, where $p^\alpha , q^\beta , r^\gamma , \ldots$ are powers of distinct primes.

The polynomial approach provides a few new important properties.

Lemma 3

Let $A(x), B(x) \in \mathbb N[x]$ and $n \in \mathbb N^*$. Then

$$\begin{aligned} A(x) B(x) \equiv \sum _{ k=0 }^{ n - 1 } x^k \mod \left( x^n - 1 \right) \end{aligned}$$

(T0)

if and only if

• 1. $A(x) , B(x) \in \left\{ 0 , 1 \right\} [x]$, so they are the characteristic polynomials of sets A and B, and

• 2. $A \oplus B = \left\{ r_1 , \ldots , r_n \right\} \subset \mathbb Z$, with $r_i \ne r_j \mod n$ for all $i , j \in \left\{ 1 , \ldots , n \right\}$ with $i \ne j$.

Lemma 4

Let $f(x) \in \mathbb Z[x]$ and $n \in \mathbb N^*$. The following are equivalent:

• 1. $f(x) \equiv \sum _{ k = 0 }^{ n - 1 } x^k \mod \left( x^n - 1 \right)$;

• 2.

  • (a) $f(1) = n$ and

  • (b) for every $d | n$, with $d > 1$, we have $\varPhi _d(x) | f(x)$.

Definition 6

Let A be a subset of $\mathbb Z_n$ and let $S_A = \left\{ p^\alpha , q^\beta , \ldots , r^\gamma \right\}$. We call the extension of A any rhythm $\overline{A}$ whose characteristic polynomial is

$$ \overline{A}(x) = \varPhi _{ p^\alpha } \left( x^{ \frac{ n }{ p^\alpha k_p } } \right) \varPhi _{ q^\beta } \left( x^{ \frac{ n }{ q^\beta k_q } } \right) \cdots \varPhi _{ r^\gamma } \left( x^{ \frac{ n }{ r^\gamma k_r } } \right) . $$

where $k_p , k_q , \ldots , k_r$ are divisors of n such that .

Note that by definition clearly $S_A = S_{ \overline{A} }$.

Proposition 2

Let $A \oplus B = \mathbb Z_n$ and let B satisfy condition (T2). Then $\overline{A} \oplus B = \mathbb Z_n$, too.

Proof

Since $p^\alpha$ is a prime power, then

$$ \varPhi _{ p^\alpha } \left( x^{ \frac{ n }{p^{ \alpha }k_p} } \right) \in \left\{ 0 , 1 \right\} [x], $$

and so $\overline{A}(x) \in \mathbb N[x]$. Moreover,

• $\overline{A}(1) B(1) = n$ and

• $\varPhi _d(x) | \overline{A}(x) B(x)$ for all $d | n$, with $d > 1$.

By Lemma 4, this means that

$$ \overline{A}(x) B(x) \equiv \sum _{ k = 0 }^{ n - 1 } x^k \mod \left( x^n - 1 \right) , $$

that is, condition (T0) in Lemma 3 holds. Therefore $\overline{A}(x) \in \left\{ 0 , 1 \right\} [x]$ and $\overline{A} \oplus B = \mathbb Z_n$, that is, $\overline{A}$ tiles with B.    $\square$

Combining Theorem 6 and Proposition 2, we are able to find new Vuza canons where L is not a subset of $\mathbb Z_{ n_3 }$.

Theorem 7

Let $n = p_1 n_1 p_2 n_2 n_3 \in \mathbb N$ such that:

• 1. $p_1 , n_1 , p_2 , n_2 , n_3 > 1$;

• 2. $\gcd \left( p_1 n_1 , p_2 n_2 \right) = 1$;

• 3. there is no prime q such that $q | n_3$, but .

Let H be the subgroup $H = n_3 \mathbb I_{ p_1 n_1 p_2 n_2 }$ of $\mathbb Z_n$. Suppose that L and K are proper subsets of $\mathbb Z_{ n_3 }$ such that $L \oplus K = \mathbb Z_{ n_3 }$ and $K = K_1 \sqcup K_2$, with $K_1 , K_2 \ne \emptyset$. Let $\tilde{L}$ be an extension of L; then the pair $\left( A , B \right)$ defined by

$$ \begin{aligned} A&= A_1 \oplus A_2 \oplus \tilde{L} \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) \end{aligned} $$

is an aperiodic tiling rhythmic canon of $\mathbb Z_n$.

Table 1. The number of aperiodic rhythms for non-Hajós values of n from 72 to 280, generated with the constructions described in Theorems 2–7.

Proof

Since, by definition, $A_1$ and $A_2$ coincide with their own extensions, the extension of $A_1 \oplus A_2 \oplus L$ is A. By Theorem 6, $A_1 \oplus A_2 \oplus L \oplus B = \mathbb Z_n$, therefore Proposition 2 implies that $A \oplus B = \mathbb Z_n$.

We already know from Theorem 6 that B is aperiodic. To show that A is aperiodic, consider $\tilde{L}(x)$. By hypothesis 3 $S_{ \tilde{L}}$ does not contain any maximal prime power dividing n, as $S_{ A_1 }$ and $S_{ A_2 }$. As a consequence, $S_A = S_{ A_1 } \cup S_{ A_2 } \cup S_{ \tilde{L}}$ does not contain any such prime power, either. By Proposition 1, A can not be periodic.    $\square$

Definition 7

We call extended Vuza canons all the canons obtained using the constructions of Theorems 6 and 7, possibly combined with those of Theorems 2, 3, 4 and 5.

Example 6

We show now an extended Vuza canon with period $n = 240$ ($p_1 = 2 , n_1 = 2 , p_2 = 5 , n_2 = 3 , n_3 = 4$). Set $L = \mathbb I_2$; then $\tilde{L} = 15 \mathbb I_2$. Choosing $K_1 = \left\{ 2 \right\}$ and $K_2 = \left\{ 0 \right\}$, we obtain the canon

$$ \begin{aligned} A&= A_1 \oplus A_2 \oplus \tilde{L} \\&= 16 \mathbb I_3 \oplus 60 \mathbb I_2 \oplus 15 \mathbb I_2 \\ B&= \left( U_1 \oplus V_2 \oplus K_1 \right) \sqcup \left( U_2 \oplus V_1 \oplus K_2 \right) \\&= \left( 48 \mathbb I_5 \oplus 8 \mathbb I_2 \oplus \left\{ 2 \right\} \right) \sqcup \left( 120 \mathbb I_2 \oplus 12 \mathbb I_5 \oplus \left\{ 0 \right\} \right) . \end{aligned} $$

It is worth noting that it would not be possible to obtain such a canon without applying Theorem 7.

We include below a table showing the number of Vuza canons and extended Vuza canons for all the periods n with values between 72 and 280 (Table 1).

As a final comment, one could say that the recipes by Hajós, de Bruijn and Vuza to generate aperiodic tiling rhythmic canons are deceivingly simple. Their basic mechanism can be (and has indeed been) generalised in several ways; this paper gives a generalisation on its own, but Theorem 7 can certainly still be improved. Further studies should follow, aiming at lifting the hypotheses used in the present results and (hopefully) at establishing a systematic theory of aperiodic tiling rhythmic canons given by all the known constructions, and eventually of all aperiodic tiling rhythmic canons straightaway.