On Fibonacci Knots

TL;DR

This paper explores the properties of Fibonacci knots, revealing their Conway polynomials relate to Fibonacci polynomials modulo 2 and identifying conditions under which these knots are not Lissajous knots.

Contribution

It establishes a connection between Fibonacci polynomials and Conway polynomials of Fibonacci links, and determines when Fibonacci knots are not Lissajous knots.

Findings

Conway polynomials of Fibonacci links are Fibonacci polynomials modulo 2.

Fibonacci knots are not Lissajous knots for certain parameters.

Conditions involving n and j determine the Lissajous knot status of Fibonacci knots.

Abstract

We show that the Conway polynomials of Fibonacci links are Fibonacci polynomials modulo 2. We deduce that, when $ n \not\equiv 0 \Mod 4$ and $(n,j) \neq (3,3),$ the Fibonacci knot $ \cF_j^{(n)} $ is not a Lissajous knot.

Figures

Equations

$$\alpha \text{overwithdelims}..\beta =a_1+1\text{overwithdelims}..a_2+1\text{overwithdelims}..a_3+1\text{overwithdelims}..\cdots +1\text{overwithdelims}..a_m=[a_1,\dots ,a_m],\alpha >0.$$

$$\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr\alpha\cr\beta\crcr}}\right)=P^{j}\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1\cr 0\crcr}}\right).$$

$$\nabla_{K}(z)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&0\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr-a_{1}z&1\cr 1&0\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr a_{2}z&1\cr 1&0\crcr}}\right)\cdots\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr(-1)^{m}a_{m}z&1\cr 1&0\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1\cr 0\crcr}}\right)$$

$$\Delta_{K}(t)=\nabla_{K}\bigl{(}t^{1/2}-t^{-1/2}\bigr{)}.$$

$$m\text{overwithdelims}..1-m=[-2,2,\dots ,(-1{)}^{m-1}\cdot 2].$$

$$\nabla(z)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&0\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr z&1\cr 1&0\crcr}}\right)^{m-1}\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1\cr 0\crcr}}\right).$$

$$\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr z&1\cr 1&0\crcr}}\right)^{m}=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr f_{m+1}(z)&f_{m}(z)\cr f_{m}(z)&f_{m-1}(z)\crcr}}\right)$$

$$\Delta(t)=f_{2k+1}\bigl{(}t^{1/2}-t^{-1/2}\bigr{)}=(t^{k}+t^{-k})-(t^{k-1}+t^{k-1})+\cdots+(-1)^{k}.$$

$$[n,n,x]=[n+1,\underset{2k}{\underbrace{-2,2,\dots ,-2,2}},-(1+x)]$$

$$[n,n,n,z]=[n+1,\underset{2k}{\underbrace{-2,2,\dots ,-2,2}},-(n+1),-z].$$

$$\displaystyle G^{k}=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1+2k&2k\cr-2k&1-2k\crcr}}\right)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n&n-1\cr 1-n&2-n\crcr}}\right).$$

$$M(x)=[n+1,-2,2,\dots ,-2,2,-(1+x)],L(u)=[n+1,u],T(x)=-x-1.$$

$$\displaystyle L=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n+1&1\cr 1&0\crcr}}\right),\ T=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&1\cr 0&-1\crcr}}\right),\ M=LG^{k}T.$$

$$\displaystyle M=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n+1&1\cr 1&0\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n&n-1\cr 1-n&2-n\crcr}}\right)\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&1\cr 0&-1\crcr}}\right)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n^{2}+1&n\cr n&1\crcr}}\right)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n&1\cr 1&0\crcr}}\right)^{2},$$

$$[n,n]=[n+1,\underset{2k}{\underbrace{-2,2,\dots ,-2,2}}],[n,n,n]=[n+1,\underset{2k}{\underbrace{-2,2,\dots ,-2,2}},-(n+1)].$$

$$[n{]}_{j+3}=[n+1,\underset{2k}{\underbrace{-2,2,\dots ,-2,2}},-(n+1),-[n{]}_j].$$

$$s([n{]}_1)=s(n)=n\text{overwithdelims}..1-n=[\underset{2k}{\underbrace{-2,2,\dots -2,2}}],n\ne 1,$$

$$s([n{]}_{j+3})=[\underset{2k}{\underbrace{-2,2,\dots ,-2,2}},-(n+1),-(n+1),-s([n{]}_j)]$$

$$G=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 3&2\cr-2&-1\crcr}}\right),\ Q=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr-(n+1)&1\cr 1&0\crcr}}\right),\ R=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&0\cr 0&-1\crcr}}\right).$$

$$\displaystyle H=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n^{3}+n^{2}+2n+1&n^{2}+1\cr-n^{3}-n&n-n^{2}-1\crcr}}\right).$$

$$S^{-1}HS=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&0\cr 1&1\crcr}}\right)H\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr 1&0\cr-1&1\crcr}}\right)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n^{3}+2n&n^{2}+1\cr n^{2}+1&n\crcr}}\right)=\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n&1\cr 1&0\crcr}}\right)^{3},$$

$$S\left(\hbox{}\,\vbox{\halign{ \hfil${#}\ $\hfil& \hfil$\ {#}\ $\hfil&& \hfil$\ {#}\ $\hfil& \hfil$\ {#}$\hfil\cr n&1\cr 1&0\crcr}}\right)^{3}=HS.$$

$$F_{3m+2}\text{overwithdelims}..F_{3m}=[2,2,-2,-2,\dots ,(-1{)}^{m-1}\cdot 2,(-1{)}^{m-1}\cdot 2]$$

$$F_{3m+1}\text{overwithdelims}..F_{3m}=[2,-2,-2,2,2,\dots ,(-1{)}^{m-1}\cdot 2,(-1{)}^{m-1}\cdot 2,(-1{)}^m\cdot 2]$$

$$\displaystyle\left\{\begin{array}[]{l}\hbox{If }n\equiv 1\,(\hbox{\rm mod}\,4),\ N=\lfloor{\textstyle{{j+2}\overwithdelims. .{3}}}\rfloor(n-2)+j+1,\\ \hbox{If }n\equiv 3\,(\hbox{\rm mod}\,4),\ N=\lfloor{\textstyle{{j+2}\overwithdelims. .{3}}}\rfloor(n+2)-(j+1).\end{array}\right.$$

Full text

keywords: Fibonacci polynomials, Fibonacci knots, continued fractions

1 Introduction

Fibonacci knots (or links) were defined by J. C. Turner ([11]) as rational knots with Conway notation ${\cal C}(1,1,\ldots,1).$ He also considered the generalized Fibonacci knots ${\cal F}_{j}^{(n)}={{\cal C}}(n,n,\ldots,n),$ where $n$ is a fixed integer and the sequence $(n,\ldots,n)$ has length $j$.

In this paper we determine the Conway and Alexander polynomials modulo 2 of Fibonacci knots. We show that the Conway polynomial of a generalized Fibonacci knot is a Fibonacci polynomial modulo $2$.

As an application, we show that if $n\not\equiv 0\,(\hbox{\rm mod}\,4)$ and $(n,j)\neq(3,3)$ the Fibonacci knot ${{\cal F}}_{j}^{(n)}$ is not a Lissajous knot.

Our results are obtained by continued fraction expansions.

2 Conway notation and Fibonacci knots

The Conway notation (J. H. Conway, [3]) is particularly convenient for the important class of rational (or two-bridge) knots. The Conway normal form ${\cal C}(a_{1},a_{2},...,a_{m})$ of a rational knot (or link), is best explained by the following figure.

The number of twists is denoted by the integer $\left|a_{i}\right|$, and the sign of $a_{i}$ is defined as follows: if $i$ is odd, then the right twist is positive, if $i$ is even, then the right twist is negative. On Fig. 1 the $a_{i}$ are positive (the $a_{1}$ first twists are right twists).

The rational links are classified by their Schubert fractions

[TABLE]

Two rational links of fractions ${\displaystyle{{\alpha}\overwithdelims. .{\beta}}}$ and ${\displaystyle{{\alpha^{\prime}}\overwithdelims. .{\beta^{\prime}}}}$ are equivalent if and only if $\alpha=\alpha^{\prime}$ and $\beta^{\prime}\equiv\beta^{\pm 1}({\rm mod}\ \alpha).$ The integer $\alpha$ is the determinant of the link, it is odd for a knot, and even for a two-component link.

The following result is a useful consequence of the continued fraction description of rational links (see [4] p. 207).

Theorem 1

Any rational link has a Conway normal form ${\cal C}(2a_{1},2a_{2},\ldots,2a_{m}).$

The Fibonacci knots (or links) are defined by their Conway notation ${\cal F}_{j}={\cal C}(1,1,\ldots,1),$ where $j$ is the number of crossings. The Schubert fraction of ${\cal F}_{j}$ is ${\displaystyle{{F_{j+1}}\overwithdelims. .{F_{j}}}},$ and its determinant is the Fibonacci number $F_{j+1}.$ It is the reason why J. C. Turner named these knots Fibonacci knots. He also introduced the generalized Fibonacci knots ${\cal F}_{j}^{(n)}={\cal C}(n,n,\ldots n),$ where $n$ is a fixed integer.

We first observe

Proposition 2

${\cal F}_{j}^{(n)}$* is a knot if and only if $n\equiv 0\,(\hbox{\rm mod}\,2)$ and $j\equiv 0\,(\hbox{\rm mod}\,2)$ or $n\not\equiv 0\,(\hbox{\rm mod}\,2)$ and $j\not\equiv 2\,(\hbox{\rm mod}\,3)$.*

Proof. Let us consider the Möbius transformation $P(z)=[n,z]=n+{\displaystyle{{1}\overwithdelims. .{z}}}$. It is convenient to consider its matrix notation P=\left(\hbox{}\,\vbox{\halign{ \hfil{#}\ \hfil& \hfil\ {#}\ \hfil&& \hfil\ {#}\ \hfil& \hfil\ {#}\hfil\cr n&1\cr 1&0\crcr}}\right). Let ${\displaystyle{{\alpha}\overwithdelims. .{\beta}}}=[n,\ldots,n]=P^{j}(\infty)$, it is also

[TABLE]

If $n\equiv 1\,(\hbox{\rm mod}\,2)$ then P\equiv\left(\hbox{}\,\vbox{\halign{ \hfil{#}\ \hfil& \hfil\ {#}\ \hfil&& \hfil\ {#}\ \hfil& \hfil\ {#}\hfil\cr 1&1\cr 1&0\crcr}}\right)\,(\hbox{\rm mod}\,2),\ P^{2}\equiv\left(\hbox{}\,\vbox{\halign{ \hfil{#}\ \hfil& \hfil\ {#}\ \hfil&& \hfil\ {#}\ \hfil& \hfil\ {#}\hfil\cr 0&1\cr 1&1\crcr}}\right)\,(\hbox{\rm mod}\,2),\ P^{3}\equiv{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}\,(\hbox{\rm mod}\,2). We deduce that $\alpha\equiv\beta\equiv 1\,(\hbox{\rm mod}\,2)$ when $j\equiv 1\,(\hbox{\rm mod}\,3)$, $\alpha\equiv 0,\,\beta\equiv 1\,(\hbox{\rm mod}\,2)$ when $j\equiv 2\,(\hbox{\rm mod}\,3)$ and $\alpha\equiv 1,\,\beta\equiv 0\,(\hbox{\rm mod}\,2)$ when $j\equiv 0\,(\hbox{\rm mod}\,3)$. The case $n\equiv 0\,(\hbox{\rm mod}\,2)$ is similar. $\Box$

3 The Conway and Alexander polynomials

The Alexander polynomial, discovered in 1928, is one of the most famous invariant of knots. J. H. Conway discovered an easy way to calculate it. He introduced the “Skein relations” which relate the polynomial of a link $K$ to the polynomials of links obtained by changing one crossing of $K$.

The following result is a beautiful application of his algorithm.

Theorem 3** ([4])**

*Let $K={{\cal C}}(2a_{1},2a_{2},\ldots,2a_{m})$ be a rational knot (or link).

The Conway polynomial of $K$ is*

[TABLE]

The Alexander polynomial of $K$ is

[TABLE]

Let us consider a simple example.

Example 4** (The torus links)**

The torus link ${\mathrm{T}}(2,m)$ has Conway normal form ${\cal C}(m)={\cal F}_{1}^{(m)}$. It is the link of fraction ${\displaystyle{{m}\overwithdelims. .{1}}}$ or ${\displaystyle{{m}\overwithdelims. .{1-m}}}.$ We have the continued fraction (of length $m-1$)

[TABLE]

Hence, the Conway polynomial is

[TABLE]

It is well known that

[TABLE]

where $f_{m}(z)$ are the Fibonacci polynomials defined by $f_{0}(z)=0,\ f_{1}(z)=1,\ f_{m+1}(z)=zf_{m}(z)+f_{m-1}(z)$ ([12]). We conclude that the Conway polynomial of ${\mathrm{T}}(2,m)$ is the Fibonacci polynomial $f_{m}(z)$ (see also [6]). If $m=2k+1$ (i.e. ${\mathrm{T}}(2,m)$ is a knot) we obtain the Alexander polynomial

[TABLE]

The recently introduced Lissajous knots ([2, 5, 10, 4]) are non singular Lissajous space curves. We will show that in many cases, Fibonacci knots are not Lissajous knots. Let us first recall the following

Theorem 5** ([5, 10])**

If $K$ is a rational Lissajous knot then $\Delta_{K}(t)\equiv 1\,(\hbox{\rm mod}\,2).$

Consequently, we see that a non trivial torus knot is never a Lissajous knot.

Moreover, Theorem 3 provides many examples of knots which are not Lissajous knots.

Corollary 6

Let $b_{i}\equiv 2\,(\hbox{\rm mod}\,4),m>1.$ The Conway polynomial of ${{\cal C}}(b_{1},\ldots,b_{m})$ is equivalent to $f_{m+1}(z)\,(\hbox{\rm mod}\,2).$

Corollary 7

If $n\equiv 2\,(\hbox{\rm mod}\,4),$ the modulo $2$ Conway polynomial of ${\cal F}_{j}^{(n)}$ is $f_{j+1}(z).$

Hence these knots are not Lissajous knots by Theorem 5.

The following result is an immediate consequence of Theorem 3.

Corollary 8

If $n\equiv 0\,(\hbox{\rm mod}\,4),$ the modulo $2$ Conway polynomial of ${\cal F}_{j}^{(n)}$ is [math] if $j$ is odd, and $1$ if $j$ is even.

It is not known whether the knot ${\cal F}_{2}^{(4)}={\cal C}(4,4)$ is Lissajous or not (see [1]).

4 The modulo 2 Conway polynomial of Fibonacci knots

We shall now study the knots ${\cal F}_{j}^{(n)},$ where $n=2k+1$ is an odd integer.

Lemma 9

Let $n=2k+1.$ We have the identities

[TABLE]

[TABLE]

Proof. Let us prove the first formula. We shall use matrix notations for Möbius transformations. Let $G(u)=[-2,2,u]={\displaystyle{{3u+2}\overwithdelims. .{-2u-1}}}.$ Its matrix is G=\left(\hbox{}\,\vbox{\halign{ \hfil{#}\ \hfil& \hfil\ {#}\ \hfil&& \hfil\ {#}\ \hfil& \hfil\ {#}\hfil\cr 3&2\cr-2&-1\crcr}}\right), and consequently we get by induction

[TABLE]

Let

[TABLE]

The corresponding matrices are

[TABLE]

Consequently

[TABLE]

that is $M(x)=[n,n,x]$ which proves the first identity. If we substitute $x=[n,z]$ in Formula (2), we obtain the second identity (3). $\Box$

Corollary 10

Let $n=2k+1.$ We have the continued fractions

[TABLE]

Let us denote $[n]_{j}=[\underbrace{n,\ldots,n}_{j}].$ If $j\not\equiv 1\,(\hbox{\rm mod}\,3),$ we get the continued fractions

[TABLE]

When $j\equiv 1\,(\hbox{\rm mod}\,3)$, there is no continued fraction expansion of $[n]_{j}$ with even quotients, by Prop. 2. In this case, we shall get a continued fraction expansion for ${\displaystyle{{\alpha}\overwithdelims. .{\beta-\alpha}}},$ which is another fraction of the same knot. Let $s$ be the Möbius transformation defined by $s(x)={\displaystyle{{x}\overwithdelims. .{1-x}}}.$ We have $s({\displaystyle{{\alpha}\overwithdelims. .{\beta}}})={\displaystyle{{\alpha}\overwithdelims. .{\beta-\alpha}}}.$

Proposition 11

Let $n=2k+1.$ We have the continued fractions

[TABLE]

[TABLE]

.

Proof. The first formula has already been proved. Let us prove the second formula. We shall use the Möbius maps $G(x)=[-2,2,x]={\displaystyle{{3x+2}\overwithdelims. .{-2x-1}}},\ Q(x)=[-(n+1),x],\ R(x)=-x$ corresponding to

[TABLE]

Let us define the Möbius transformation $H=G^{k}\cdot Q^{2}\cdot R.$ We obtain, using Form. (8)

[TABLE]

Let $S$ be a matrix corresponding to the Möbius map $s$. We have

[TABLE]

and then

[TABLE]

This means that s\bigl{(}[n,n,n,x]\bigr{)}=h\circ s(x), which proves our formula. $\Box$

Remark 12

By considering the case $n=1$ where $h(x)=[-2,-2,-x]$, we obtain the following interesting continued fractions of length $2m$:

[TABLE]

Using the corollary 10 we obtain similarly

[TABLE]

of length $2m$ and ${\displaystyle{{F_{3m+3}}\overwithdelims. .{F_{3m+2}}}}=[2,-{\displaystyle{{F_{3m+2}}\overwithdelims. .{F_{3m}}}}]$ of length $2m+1.$

Of course, these fractions correspond to Fibonacci knots (or links). They are not Lissajous knots because of corollary 6.

It is straightforward to calculate the Conway polynomials of our Fibonacci knots, using Prop. 11.

Theorem 13

Let us denote by $\nabla_{j}^{(n)}(z)$ the modulo 2 Conway polynomial of the Fibonacci link ${\cal F}_{j}^{(n)}.$ We have $\nabla_{j}^{(n)}(z)=f_{N}(z)$ where

[TABLE]

Corollary 14

If $n\not\equiv 0\,(\hbox{\rm mod}\,4)$ and $(n,j)\neq(3,3),$ the Fibonacci link ${\cal F}_{j}^{(n)}$ is not a Lissajous knot.

It is not known whether the knot ${\cal F}_{3}^{(3)}={\cal C}(3,3,3)$ is a Lissajous knot ([1]).

Question 15

It would be interesting to study the wider classes of knots defined by their Conway notation ${\cal C}(\pm n,\pm n,\ldots\pm n).$

If $n=1$ we obtain all the rational knots ([8, 9]).

If $n=2$ we obtain the important class of rational fibered knots (see [7]).

In general, we obtain knots with fractions ${\displaystyle{{\alpha}\overwithdelims. .{\beta}}}$ such that $(\alpha,\beta)\equiv(0,\pm 1)\ {\rm or}\ (\pm 1,0)\,(\hbox{\rm mod}\,n).$

Acknowledgements: We would like to thank Pr. C. Lamm for having suggested this problem to us.