A matrix approach to the Yang multiplication theorem

TL;DR

This paper introduces a matrix-based method using Laurent polynomials to simplify and clarify the proof of the Yang multiplication theorem, connecting sequence compositions with algebraic identities.

Contribution

It presents a novel matrix approach employing Laurent polynomials to provide a concise proof of the Yang multiplication theorem.

Findings

Matrix approach effectively encodes sequence compositions.

Lagrange identity simplifies proof of Yang multiplication.

Provides transparent algebraic framework for the theorem.

Abstract

In this paper, we use two-variable Laurent polynomials attached to matrices to encode properties of compositions of sequences. The Lagrange identity in the ring of Laurent polynomials is then used to give a short and transparent proof of a theorem about the Yang multiplication.

Equations

$$N_{\mathbf{a}}(j)+N_{\mathbf{b}}(j)+N_{\mathbf{c}}(j)+N_{\mathbf{d}}(j)=0$$

$$N_{\mathbf{s}}(j)=\{\begin{array}{ll}{\sum }_{i=0}^{l-j-1}{s}_i{s}_{i+j}& \text{if 0≤j<l},\\ 0& \text{otherwise},\end{array}$$

$${\varphi }_{\mathbf{a}}(x)$$

$$({\varphi }_{\mathbf{a}}{\varphi }_{\mathbf{a}}^{\ast }+{\varphi }_{\mathbf{b}}{\varphi }_{\mathbf{b}}^{\ast }+{\varphi }_{\mathbf{c}}{\varphi }_{\mathbf{c}}^{\ast }+{\varphi }_{\mathbf{d}}{\varphi }_{\mathbf{d}}^{\ast })(x)=2(m+n).$$

$$({\varphi }_{\mathbf{q}}{\varphi }_{\mathbf{q}}^{\ast }+{\varphi }_{\mathbf{r}}{\varphi }_{\mathbf{r}}^{\ast }+{\varphi }_{\mathbf{s}}{\varphi }_{\mathbf{s}}^{\ast }+{\varphi }_{\mathbf{t}}{\varphi }_{\mathbf{t}}^{\ast })(x)$$

$$=({\varphi }_{\mathbf{a}}{\varphi }_{\mathbf{a}}^{\ast }+{\varphi }_{\mathbf{b}}{\varphi }_{\mathbf{b}}^{\ast }+{\varphi }_{\mathbf{c}}{\varphi }_{\mathbf{c}}^{\ast }+{\varphi }_{\mathbf{d}}{\varphi }_{\mathbf{d}}^{\ast })(x^2)({\varphi }_{\mathbf{e}}{\varphi }_{\mathbf{e}}^{\ast }+{\varphi }_{\mathbf{f}}{\varphi }_{\mathbf{f}}^{\ast }+{\varphi }_{\mathbf{g}}{\varphi }_{\mathbf{g}}^{\ast }+{\varphi }_{\mathbf{h}}{\varphi }_{\mathbf{h}}^{\ast })(x^{2(2m+1)}),$$

$${\varphi }_{\mathbf{q}}(x)$$

$$-x^{2(2m+1)}{\varphi }_{{\mathbf{b}}^{\ast }}(x^2){\varphi }_{\mathbf{e}}(x^{2(2m+1)})+x^{2(2m+1)+1}{\varphi }_{\mathbf{d}}(x^2){\varphi }_{\mathbf{h}}(x^{2(2m+1)}).$$

$$q$$

$$r$$

$$s$$

$$t$$

$$q{q}^{\ast }+r{r}^{\ast }+s{s}^{\ast }+t{t}^{\ast }=(a{a}^{\ast }+b{b}^{\ast }+c{c}^{\ast }+d{d}^{\ast })(e{e}^{\ast }+f{f}^{\ast }+g{g}^{\ast }+h{h}^{\ast }).$$

$${\varphi }_{\mathbf{a}}(x^2),{\varphi }_{\mathbf{b}}(x^2),x{\varphi }_{\mathbf{c}}(x^2),x{\varphi }_{\mathbf{d}}(x^2),$$

$$x^{2m+(1-n)(2m+1)}{\varphi }_{\mathbf{e}}(x^{2(2m+1)}),x^{-n(2m+1)}{\varphi }_{\mathbf{f}}(x^{2(2m+1)}),$$

$$x^{-n(2m+1)}{\varphi }_{\mathbf{g}}(x^{2(2m+1)}),x^{(1-n)(2m+1)}{\varphi }_{\mathbf{h}}(x^{2(2m+1)}),$$

$${\varphi }_{\mathbf{a}}(x^2),{\varphi }_{\mathbf{b}}(x^2),{\varphi }_{\mathbf{c}}(x^2),{\varphi }_{\mathbf{d}}(x^2),{\varphi }_{\mathbf{e}}(x^{2(2m+1)}),{\varphi }_{\mathbf{f}}(x^{2(2m+1)}),{\varphi }_{\mathbf{g}}(x^{2(2m+1)}),{\varphi }_{\mathbf{h}}(x^{2(2m+1)}),$$

$${\psi }_Q(x,y)={\psi }_{\mathbf{a}}(x){\psi }_{\mathbf{f}}(y)+{\psi }_{\mathbf{c}}(x){\psi }_{\mathbf{g}}(y)+{\psi }_{\mathbf{b}}(x){\psi }_{\mathbf{e}}(y)+{\psi }_{\mathbf{d}}(x){\psi }_{\mathbf{h}}(y).$$

$$({\psi }_Q{\psi }_Q^{\ast }+{\psi }_R{\psi }_R^{\ast }+{\psi }_S{\psi }_S^{\ast }+{\psi }_T{\psi }_T^{\ast })(x,y)$$

$$=({\psi }_{\mathbf{a}}{\psi }_{\mathbf{a}}^{\ast }+{\psi }_{\mathbf{b}}{\psi }_{\mathbf{b}}^{\ast }+{\psi }_{\mathbf{c}}{\psi }_{\mathbf{c}}^{\ast }+{\psi }_{\mathbf{d}}{\psi }_{\mathbf{d}}^{\ast })(x)({\psi }_{\mathbf{e}}{\psi }_{\mathbf{e}}^{\ast }+{\psi }_{\mathbf{f}}{\psi }_{\mathbf{f}}^{\ast }+{\psi }_{\mathbf{g}}{\psi }_{\mathbf{g}}^{\ast }+{\psi }_{\mathbf{h}}{\psi }_{\mathbf{h}}^{\ast })(y).$$

$${\varphi }_{\mathbf{a}}(x)$$

$${\psi }_{\mathbf{a}}(x)$$

$$N_{\mathbf{a}}(j)=\{\begin{array}{ll}{\sum }_{i=0}^{l-j-1}{a}_i{a}_{i+j}^{\ast }& \text{if 0≤j<l},\\ 0& \text{otherwise}.\end{array}$$

$$\sum _{i=1}^n{N}_{{\mathbf{a}}_i}(j)=\{\begin{array}{ll}w& \text{if j=0},\\ 0& \text{otherwise}.\end{array}$$

$${\psi }_{{\mathbf{a}}^{\ast }}(x)={\psi }_{\mathbf{a}}^{\ast }(x).$$

$$\mathbf{a}/0=(a_0,0,a_1,\dots ,0,a_{l-1})\in {\mathcal{R}}^{2l-1},0/\mathbf{a}=(0,a_0,0,\dots ,a_{l-1},0)\in {\mathcal{R}}^{2l+1}.$$

$${\psi }_{\mathbf{a}/0}(x)={\psi }_{0/\mathbf{a}}(x)={\psi }_{\mathbf{a}}(x^2).$$

$${\psi }_{\mathbf{a}/0}(x)$$

$${\psi }_{0/\mathbf{a}}(x)$$

$$\mathrm{seq}(A)=({\mathbf{a}}_0\mid {\mathbf{a}}_1\mid \cdots \mid {\mathbf{a}}_{m-1})\in {\mathcal{R}}^{mn},$$

$${\psi }_A(x,y)=\sum _{i=0}^{m-1}{\psi }_{{\mathbf{a}}_i}(x)y^{2i+1-m}.$$

$${\psi }_{{\mathbf{f}}^t\mathbf{a}}(x,y)={\psi }_{\mathbf{a}}(x){\psi }_{\mathbf{f}}(y).$$

$${\psi }_{{\mathbf{f}}^t\mathbf{a}}(x,y)$$

$$=\sum _{i=0}^{m-1}{f}_i{\psi }_{\mathbf{a}}(x)y^{2i+1-m}$$

$$={\psi }_{\mathbf{a}}(x)\sum _{i=0}^{m-1}{f}_i{y}^{2i+1-m}$$

$$={\psi }_{\mathbf{a}}(x){\psi }_{\mathbf{f}}(y).$$

Full text

1. Introduction

Many classes of complementary sequences have been investigated in the literature (see [1]). A quadruple of $(\pm 1)$-sequences $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})$ of length $m,m,n,n$, respectively, is called base sequences if

[TABLE]

for all positive integers $j$, where

[TABLE]

for $\boldsymbol{s}=(s_{0},\ldots,s_{l-1})\in\{\pm 1\}^{l}$. We denote by $BS(m,n)$ the set of base sequences of length $m$, $m$, $n$, $n$. If $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})\in BS(m,n)$, then it is complementary with weight $2(m+n)$. In [11], Yang proved the following theorem, which is known as one version of the Yang multiplication theorem:

Theorem 1.1** ([11, Theorem 4]).**

If $BS(m+1,m)\neq\emptyset$ and $BS(n+1,n)\neq\emptyset$, then $BS(m^{\prime},m^{\prime})\neq\emptyset$ with $m^{\prime}=(2m+1)(2n+1)$.

The well-known Hadamard conjecture states that Hadamard matrices of order $4n$ exist for every positive integer $n$. A consequence of Theorem 1.1 is the existence of a Hadamard matrix of order $8m^{\prime}$ for a positive integer $m^{\prime}$ satisfying the hypotheses. Indeed, a class of sequences called $T$-sequences with length $2m^{\prime}$ can be obtained from $BS(m^{\prime},m^{\prime})$ [8], and Hadamard matrices of order $8m^{\prime}$ can be produced from $T$-sequences with length $2m^{\prime}$ by using Goethals–Seidel arrays [12]. For more information on $T$-sequences, we refer the reader to [1, 2, 3, 4].

In order to prove Theorem 1.1, Yang used the Lagrange identity for polynomial rings. Let $\mathbb{Z}[x^{\pm 1}]$ be the ring of Laurent polynomials over $\mathbb{Z}$ and $*:\mathbb{Z}[x^{\pm 1}]\rightarrow\mathbb{Z}[x^{\pm 1}]$ be the involutive automorphism defined by $x\mapsto x^{-1}$. Let $\boldsymbol{a}=(a_{0},\dots,a_{l-1})\in\mathbb{Z}^{l}$. We define the Hall polynomial $\phi_{\boldsymbol{a}}(x)\in\mathbb{Z}[x^{\pm 1}]$ of $\boldsymbol{a}$ by

[TABLE]

It is easy to see that a quadraple $(\pm 1)$-sequences $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})$ of length $m,m,n,n$, respectively, is a base sequences if and only if

[TABLE]

Suppose $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})\in BS(n+1,n)$ and $(\boldsymbol{f},\boldsymbol{g},\boldsymbol{h},\boldsymbol{e})\in BS(m+1,m)$. The proof of Theorem 1.1 in [11] is by establishing the identity

[TABLE]

after defining the sequences $\boldsymbol{q}$, $\boldsymbol{r}$, $\boldsymbol{s}$, $\boldsymbol{t}$ appropriately such that, in particular,

[TABLE]

A key to the proof is the Lagrange identity (see [11, Theorem L]): given $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ in a commutative ring with an involutive automorphism $*$, set

[TABLE]

Then

[TABLE]

However, the derivation of (1) from (3) is not so immediate since one has to define $a,b,c,d,e,f,g,h,$ as

[TABLE]

rather than

[TABLE]

respectively. We note that Đoković and Zhao [7] observed some connection between the Yang multiplication theorem and the octonion algebra. More information on the Yang multiplication theorem and constructions of complementary sequences can be found in [5].

In this paper, we give a more straightforward proof of Theorem 1.1. Our approach is by constructing a matrix $Q$ from the eight sequences $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d},\boldsymbol{e},\boldsymbol{f},\boldsymbol{g},\boldsymbol{h}$ and produce Laurent polynomials $\psi_{\boldsymbol{s}}(x)$ for $\boldsymbol{s}\in\{\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d},\boldsymbol{e},\boldsymbol{f},\boldsymbol{g},\boldsymbol{h}\}$ of single variable and a Laurent polynomial $\psi_{Q}(x,y)$ of two variables for a matrix $Q$, such that

[TABLE]

This gives an interpretation of the Lagrange identity in term of sequences and matrices, i.e. there exist matrices $Q,R,S,T$ such that

[TABLE]

Then (1) follows immediately by noticing $\psi_{Q}(x,x^{(2n+1)})=\psi_{\boldsymbol{q}}(x)$ and $(\psi_{\boldsymbol{a}}\psi^{*}_{\boldsymbol{a}})(x)=(\phi_{\boldsymbol{a}}\phi^{*}_{\boldsymbol{a}})(x^{2})$.

The paper is organized as follows. In Section 2, we will define a Laurent polynomial $\psi_{\boldsymbol{a}}(x)$ for a sequence $\boldsymbol{a}$ and introduce basic properties of $\psi_{\boldsymbol{a}}(x)$. We will also show how to combine sequences and matrices to produce new sequences and matrices, eventually leading to a construction of a matrix from a given set of eight sequences. Finally, in Section 3, we will prove Theorem 1.1 as a consequence of the Lagrange identity in the ring of Laurent polynomials of two variables. We note here that Theorem 1.1 [11, Theorem 4] is known as one of the Yang multiplication theorem. Other versions of the Yang multiplication theorem will be investigated in subsequent papers.

2. Preliminary Results

Let $\mathcal{R}$ be a commutative ring with identity and let $*$ be an involutive automorphism of $\mathcal{R}$. Moreover, let $\mathcal{R}[x^{\pm 1}]$ be the ring of Laurent polynomials over $\mathcal{R}$ and $*:\mathcal{R}[x^{\pm 1}]\rightarrow\mathcal{R}[x^{\pm 1}]$ be the extension of the involutive automorphism $*$ of $\mathcal{R}$ defined by $x\mapsto x^{-1}$.

Definition 2.1**.**

Let $\boldsymbol{a}=(a_{0},\dots a_{l-1})\in\mathcal{R}^{l}$. We define the Hall polynomial $\phi_{\boldsymbol{a}}(x)\in\mathcal{R}[x^{\pm 1}]$ of $\boldsymbol{a}$ by

[TABLE]

We define a Laurent polynomial $\psi_{\boldsymbol{a}}(x)\in\mathcal{R}[x^{\pm 1}]$ by

[TABLE]

Hall polynomials have been used not only by Yang, but also others. See [6] and references therein. For a sequence $\boldsymbol{a}=(a_{0},\ldots,a_{l-1})\in\mathcal{R}^{l}$ of length $l$ we define $\boldsymbol{a}^{*}\in\mathcal{R}^{l}$ by $(a_{l-1}^{*},\ldots,a_{0}^{*})$. It follows immediately that $\boldsymbol{a}^{**}=\boldsymbol{a}$ for every $\boldsymbol{a}\in\mathcal{R}^{l}$.

Definition 2.2**.**

For a sequence $\boldsymbol{a}=(a_{0},\ldots,a_{l-1})$ of length $l$ with entries in $\mathcal{R}$, we define the non-periodic autocorrelation $N_{\boldsymbol{a}}$ of $\boldsymbol{a}$ by

[TABLE]

We say that a set of sequences $\{\boldsymbol{a}_{1},\ldots,\boldsymbol{a}_{n}\}$ not necessarily all of the same length, is complementary with weight $w$ if

[TABLE]

By Definition 2.2 with $\mathcal{R}=\mathbb{Z}$, we see that $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})\in BS(m,n)$ if and only if $\{\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d}\}$ is complementary with weight $2(m+n)$.

Lemma 2.3**.**

Let $l$ be a positive integer and $\boldsymbol{a}\in\mathcal{R}^{l}$. Then

[TABLE]

Proof.

Straightforward. ∎

Lemma 2.4**.**

For sequences $\boldsymbol{a}_{1},\ldots,\boldsymbol{a}_{n}$ with entries in $\mathcal{R}$, the following are equivalent.

1. (i)

$\boldsymbol{a}_{1},\ldots,\boldsymbol{a}_{n}$* are complementary with weight $w$,* 2. (ii)

$\sum_{i=1}^{n}(\phi_{\boldsymbol{a}_{i}}\phi^{*}_{\boldsymbol{a}_{i}})(x)=w$, 3. (iii)

$\sum_{i=1}^{n}(\psi_{\boldsymbol{a}_{i}}\psi^{*}_{\boldsymbol{a}_{i}})(x)=w$.

Proof.

It is straightforward to check that (i) is equivalent to (ii). Equivalence of (ii) and (iii) is clear since for any sequence $\boldsymbol{a}$, $\phi_{\boldsymbol{a}}(x^{2})\phi^{*}_{\boldsymbol{a}}(x^{2})=\psi_{\boldsymbol{a}}(x)\psi^{*}_{\boldsymbol{a}}(x)$ from Definition 2.1. ∎

Definition 2.5**.**

Let $\boldsymbol{a}=(a_{0},\ldots,a_{l-1})\in\mathcal{R}^{l}$. Define

[TABLE]

Lemma 2.6**.**

For every $\boldsymbol{a}\in\mathcal{R}^{l}$,

[TABLE]

Proof.

By Definition 2.1 and Definition 2.5 , we have

[TABLE]

∎

Now, we will define a Laurent polynomial of two variables for arbitrary matrices. Let $\mathcal{R}[x^{\pm 1},y^{\pm 1}]$ be the ring of Laurent polynomials in two variables $x,y$. We define an involutive ring automorphism $*:\mathcal{R}[x^{\pm 1},y^{\pm 1}]\rightarrow\mathcal{R}[x^{\pm 1},y^{\pm 1}]$ by $x\mapsto x^{-1}$, $y\mapsto y^{-1}$ and $a\mapsto a^{*}$ for $a\in\mathcal{R}$.

Definition 2.7**.**

For $A\in\mathcal{R}^{m\times n}$, we denote the row vectors of a matrix $A$ by $\boldsymbol{a}_{0},\ldots,\boldsymbol{a}_{m-1}$. Define

[TABLE]

where $\mid$ denotes concatenation, and

[TABLE]

Clearly, we have $\psi_{A\pm B}(x,y)=\psi_{A}(x,y)\pm\psi_{B}(x,y)$ for every $A,B\in\mathcal{R}^{m\times n}$. Note that we may regard $\mathcal{R}^{n}$ as $\mathcal{R}^{1\times n}$. So, for every $\boldsymbol{a}\in\mathcal{R}^{n}$, we have $\boldsymbol{a}^{t}\in\mathcal{R}^{n\times 1}$ where $t$ denotes the transpose of a matrix.

Lemma 2.8**.**

Let $\boldsymbol{f}\in\mathcal{R}^{m}$ and $\boldsymbol{a}\in\mathcal{R}^{n}$. Then

[TABLE]

Proof.

Let $\boldsymbol{f}=(f_{0},\ldots,f_{m-1})$. Then

[TABLE]

∎

Lemma 2.9**.**

If $A\in\mathcal{R}^{m\times n}$, then

[TABLE]

Proof.

Let $\boldsymbol{a}_{0},\ldots,\boldsymbol{a}_{m-1}$ be the row vectors of $A$. Since $\phi_{\operatorname{seq}(A)}(x)=\sum_{i=0}^{m-1}x^{ni}\phi_{\boldsymbol{a}_{i}}(x)$, we have

[TABLE]

∎

3. Main Result

We will present our result by three steps. The following lemma is essential to describe the Yang multiplication theorem by using matrix approach.

Lemma 3.1**.**

Let

[TABLE]

Set

[TABLE]

Then

[TABLE]

Proof.

By Lemma 2.3 and Lemma 2.8, we have

[TABLE]

Thus, by applying the Lagrange identity, the result follows. ∎

For the remainder of this section, we fix a multiplicatively closed subset $\mathcal{T}$ of $\mathcal{R}\setminus\{0\}$ satisfying $-1\in\mathcal{T}=\mathcal{T}^{*}$. Also, we denote $\mathcal{T}_{0}=\mathcal{T}\cup\{0\}$. Denote by $\operatorname{supp}(\boldsymbol{a})$ and $\operatorname{supp}(A)$ the set of indices of nonzero entries of a sequence $\boldsymbol{a}=(a_{0},\ldots,a_{l-1})\in\mathcal{R}^{l}$ and a matrix $A=[a_{ij}]_{0\leq i\leq m-1,0\leq j\leq n-1}\in\mathcal{R}^{m\times n}$, respectively. We say that sequences $\boldsymbol{a},\boldsymbol{b}$ are disjoint if $\operatorname{supp}(\boldsymbol{a})\cap\operatorname{supp}(\boldsymbol{b})=\emptyset$. Matrices $A,B$ are also said to be disjoint if $\operatorname{supp}(A)\cap\operatorname{supp}(B)=\emptyset$.

Lemma 3.2**.**

Let $m$ and $n$ be positive integers,

[TABLE]

Set

[TABLE]

Write

[TABLE]

Then $Q,R,S,T\in\mathcal{T}^{(2m+1)\times(2n+1)}$ satisfy

[TABLE]

Proof.

Notice that $\boldsymbol{a}^{\prime},\boldsymbol{b}^{\prime},\boldsymbol{c}^{\prime},\boldsymbol{d}^{\prime}\in\mathcal{T}_{0}^{2n+1}$ and $\boldsymbol{e}^{\prime},\boldsymbol{f}^{\prime},\boldsymbol{g}^{\prime},\boldsymbol{h}^{\prime}\in\mathcal{T}_{0}^{2m+1}$.

Since $\operatorname{supp}(\boldsymbol{s}^{\prime*})=\operatorname{supp}(\boldsymbol{s}^{\prime})$ for every $\boldsymbol{s}\in\{\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d},\boldsymbol{e},\boldsymbol{f},\boldsymbol{g},\boldsymbol{h}\}$ and $(\boldsymbol{s}^{\prime},\boldsymbol{t}^{\prime})$ is disjoint whenever

[TABLE]

matrices $A$ and $B$ are disjoint whenever $A\neq B$ and

[TABLE]

Also,

[TABLE]

Hence

[TABLE]

By a similar argument, we obtain

[TABLE]

Therefore, $Q,R,S,T\in\mathcal{T}^{(2m+1)\times(2n+1)}$. The claimed identity follows from Lemma 2.6 and Lemma 3.1. ∎

Theorem 3.3**.**

Let $m,n$ be positive integers, and suppose

[TABLE]

satisfy

[TABLE]

Then there exist $\boldsymbol{q},\boldsymbol{r},\boldsymbol{s},\boldsymbol{t}\in\mathcal{T}^{(2m+1)(2n+1)}$ such that

[TABLE]

Proof.

Define $Q,R,S,T$ as in (4), (5), (6), (7), respectively. Write

[TABLE]

By Lemma 3.2, $\boldsymbol{q},\boldsymbol{r},\boldsymbol{s},\boldsymbol{t}\in\mathcal{T}^{(2m+1)(2n+1)}$. Applying Lemma 2.9 and Lemma 3.2, we have

[TABLE]

Hence the proof is complete. ∎

Finally, we see that Theorem 1.1 follows from Theorem 3.3 by setting $\mathcal{T}=\{\pm 1\}\subseteq\mathbb{Z}$. Hence, our method gives a more transparent proof of Theorem 1.1. Indeed, by taking $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c},\boldsymbol{d})\in BS(n+1,n)$ and $(\boldsymbol{f},\boldsymbol{g},\boldsymbol{h},\boldsymbol{e})\in BS(m+1,m)$, the hypotheses in Theorem 3.3 are satisfied by Lemma 2.4. Then the resulting sequences $(\boldsymbol{q},\boldsymbol{r},\boldsymbol{s},\boldsymbol{t})$ belong to $BS(m^{\prime},m^{\prime})$ by Lemma 2.4 where $m^{\prime}=(2m+1)(2n+1)$.

Acknowledgements

We would like to thank Robert Craigen for valuable advice.