The bracket polynomial of the Celtic link shadow

$CK\_4^{2n}$

Franck Ramaharo

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Franck Ramaharo. The bracket polynomial of the Celtic link shadow $CK\_4^{2n}$. 2025. hal-05209475

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# The bracket polynomial of the Celtic link shadow $CK_4^{2n}$

Franck Ramaharo

Mention Mathématiques et Informatique
Université d'Antananarivo
101 Antananarivo, Madagascar

franck.ramaharo@gmail.com

August 12, 2025

## Abstract

We derive the Kauffman bracket polynomial for the shadow of the Celtic link $CK_4^{2n}$ using two complementary approaches. The first approach uses a recursive relation within the Celtic framework of Gross and Tucker, based on diagrammatic identities. The second approach makes use of a 4-tangle algebraic framework: a fundamental tangle is concatenated with itself $n$ times to form an iterated composite tangle, and the Kauffman bracket polynomial is computed by decomposing the state space with respect to the basis elements of the 4-strand diagram monoid.

Keywords: celtic knot, 4-tangle, knot shadow, Kauffman state.

## 1 Introduction

The aim of this paper is to compute the Kauffman bracket polynomial of the Celtic link shadow $CK_4^{2n}$. An example of such a shadow diagram is shown in Figure 1.

<figure><img src="image_2.png" /><figcaption>Figure 1: Celtic knot shadow $CK_4^{20}$</figcaption></figure>

In this paper, we work exclusively with the shadow diagram [4, p. 11–12], that is the 4-regular plane graph underlying the link projection with no information about over- or under-crossings. For simplicity, the term “knot” will be used throughout this paper to refer to either a knot or a link, and specifically to its shadow diagram. Moreover, all diagrams are considered up to isotopy on the sphere $S^2$.

We present two complementary approaches to compute the Kauffman bracket polynomial of the Celtic knot $CK_4^{2n}$. The first approach uses a recursive relation within the Celtic framework of Gross and Tucker [3] by applying diagrammatic identities. Since we
work with the shadow diagram, the recursive relation simplifies significantly and allows us to derive an explicit formula for $\langle CK_4^{2n} \rangle$.

The second approach makes use of a 4-tangle algebraic framework. A fundamental tangle is concatenated with itself $n$ times to form an iterated composite tangle. The Kauffman bracket polynomial is then computed via a recursive process. The strategy consists of decomposing the states of the fundamental tangle with respect to a basis of 14 elements from the 4-strand diagram monoid $\mathcal{K}_4$ [6].

The remainder of the paper is organized as follows. In Section 2, we recall the definition of the bracket polynomial for a knot shadow. In Section 3, we compute the bracket polynomial of the Celtic knot $CK_4^{2n}$ within the Celtic framework. Finally, in Section 4, we compute the bracket polynomial of $CK_4^{2n}$ using the 4-tangle algebraic framework.

## 2 Kauffman bracket polynomials

The Kauffman bracket polynomial for a knot assigns to each shadow diagram $D$ an element $\langle D \rangle \in \mathbb{Z}[x]$. It is defined recursively by the following three axioms:

$$ (\mathbf{K1}): \langle \mathcal{O} \rangle = x; $$

$$ (\mathbf{K2}): \langle \mathcal{O} \cup D' \rangle = x \langle D' \rangle; $$

$$ (\mathbf{K3}): \langle \times \rangle = \langle \mathcal{X} \rangle + \langle \rangle \langle \cdot \rangle. $$

Here, $\mathcal{O}$ denotes a simple closed curves or a circle, $\cup$ denotes disjoint union, and the local diagrams in (K3) represent the two possible smoothings of a crossing. The latter are called the states of that crossing. A state of the diagram refers to the complete configuration obtained by choosing a smoothing at every crossing, resulting in a collection of disjoint circles.

In this setting, the bracket polynomial has a purely combinatorial interpretation. Specifically, $\langle D \rangle$ is a state-sum over all possible ways of smoothing every crossing in $D$:

$$ \langle D \rangle = \sum_{S} x^{|S|}, $$

where the sum is taken over all Kauffman states $S$ of $D$, and $|S|$ denotes the number of disjoint circles in the state $S$. Thus, the coefficient $[x^k] \langle D \rangle$ represents the number of states in which the resolved diagram consists of exactly $k$ circles.

## 3 Within the Celtic framework

Following the Celtic framework of Gross and Tucker [3], the knot $CK_4^{2n}$ is the barrier-free Celtic knot constructed on a $4 \times 2n$ rectangular grid of squares, using the standard Celtic design rules [2, 9]. The Celtic knots $CK_4^2, CK_4^4, CK_4^6$ and $CK_4^{2n}$ are shown in Figure 2.
<figure><img src="image_3.png" /><figcaption>Figure 2: Celtic knot diagrams</figcaption></figure>

Remark 1. In the Celtic framework, axiom (K3) can be written as

$$ \langle CK_4^{2n} \rangle = \langle H_i^j CK_4^{2n} \rangle + \langle V_i^j CK_4^{2n} \rangle, \qquad (3.1) $$

where the notation $H_i^j$ means replace the crossing of a Celtic knot at row $i$, column $j$ by a horizontal pair. The notation $V_i^j$ means replace the crossing at row $i$, column $j$ by a vertical pair [3]. See example in Figure 3.

<figure><img src="image_4.png" /><figcaption>Figure 3: Illustration of $\langle CK_4^{2n} \rangle = \langle H_2^{2n-1} CK_4^{2n} \rangle + \langle V_2^{2n-1} CK_4^{2n} \rangle$</figcaption></figure>

Having introduced the necessary notation, we derive the following Lemma and Theorem from the recursive relations for the Kauffman bracket polynomial given by Gross and Tucker [3].

**Lemma 3.1.** The following four relations hold for bracket polynomials:

$$ \langle H_2^{2n-1} CK_4^{2n} \rangle = (x+1)^2 \langle CK_4^{2n-2} \rangle; \qquad (3.2) $$

$$ \langle V_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle = (x+1) \langle CK_4^{2n-2} \rangle; \qquad (3.3) $$

$$ \langle H_3^{2n-2} H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle = (x+1) \langle V_2^{2n-3} CK_4^{2n-2} \rangle; \qquad (3.4) $$

$$ \langle V_3^{2n-2} H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle = \langle CK_4^{2n-2} \rangle. \qquad (3.5) $$
Proof. Formulas (3.2)–(3.5) follow directly from the diagrammatic relations depicted in Figure 4(a)–(d), respectively.

<figure><img src="image_5.png" /><figcaption>Figure 4: Celtic shadow diagrams for bracket polynomial relations</figcaption></figure>

<figure><img src="image_6.png" /></figure>

**Lemma 3.2.** The bracket polynomial for the sequence $(CK_4^{2n})_n$ is given by the following recursion

$$ \langle CK_4^2 \rangle = x^2 + x; \qquad (3.6) $$

$$ \langle V_2^1 CK_4^2 \rangle = x; \qquad (3.7) $$

$$ \langle CK_4^{2n} \rangle = (x+1)^2 \langle CK_4^{2n-2} \rangle + \langle V_2^{2n-1} CK_4^{2n} \rangle \quad \text{for } n \ge 2; \qquad (3.8) $$

$$ \langle V_2^{2n-1} CK_4^{2n} \rangle = (x+1) \langle V_2^{2n-3} CK_4^{2n-2} \rangle + (x+2) \langle CK_4^{2n-2} \rangle \quad \text{for } n \ge 2. \qquad (3.9) $$

*Proof.* Formulas (3.6) and (3.7) are easily verified. For (3.8), we have

$$ \begin{aligned} \langle CK_4^{2n} \rangle &= \langle H_2^{2n-1} CK_4^{2n} \rangle + \langle V_2^{2n-1} CK_4^{2n} \rangle \quad (\text{by axiom (3.1)}) \\ &= (x+1)^2 \langle CK_4^{2n-2} \rangle + \langle V_2^{2n-1} CK_4^{2n} \rangle \quad (\text{by (3.2)}). \end{aligned} $$

For (3.9), we have

$$ \begin{aligned} \langle V_2^{2n-1} CK_4^{2n} \rangle &= \langle H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle + \langle V_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle \quad (\text{by axiom (3.1)}) \\ &= \langle H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle + (x+1) \langle CK_4^{2n-2} \rangle \quad (\text{by (3.3)}) \\ &= \langle H_3^{2n-2} H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle + \langle V_3^{2n-2} H_1^{2n-2} V_2^{2n-1} CK_4^{2n} \rangle \quad (\text{by axiom (3.1)}) \\ &\quad + (x+1) \langle CK_4^{2n-2} \rangle \\ &= (x+1) \langle V_2^{2n-3} CK_4^{2n-2} \rangle + \langle CK_4^{2n-2} \rangle + (x+1) \langle CK_4^{2n-2} \rangle \quad (\text{by (3.4) and (3.5)}) \\ &= (x+1) \langle V_2^{2n-3} CK_4^{2n-2} \rangle + (x+2) \langle CK_4^{2n-2} \rangle. \end{aligned} $$

<figure><img src="image_7.png" /></figure>
These recursions enable us to derive a closed-form expression for $\langle CK_4^{2n} \rangle$.

**Theorem 3.3.** The bracket polynomial of the Celtic knot $CK_4^{2n}$ is given by

$$ \langle CK_4^{2n} \rangle = \frac{1}{2q} (x^2 + x) \left( (x^2 + 2x + 4 + q) \left(\frac{p+q}{2}\right)^{n-1} - (x^2 + 2x + 4 - q) \left(\frac{p-q}{2}\right)^{n-1} \right), \quad (3.10) $$

where

$$ p := x^2 + 4x + 4 \quad \text{and} \quad q := \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12}. $$

*Proof.* Formulas (3.8) and (3.9) can be expressed in the following matrix form:

$$ \begin{aligned} \left( \begin{array}{c} \langle CK_4^{2n} \rangle \\ \langle V_2^{2n-1} CK_4^{2n} \rangle \end{array} \right) &= \left( \begin{array}{cc} x^2 + 3x + 3 & x+1 \\ x+2 & x+1 \end{array} \right) \left( \begin{array}{c} \langle CK_4^{2n-2} \rangle \\ \langle V_2^{2n-3} CK_4^{2n-2} \rangle \end{array} \right) \\ &= \left( \begin{array}{cc} x^2 + 3x + 3 & x+1 \\ x+2 & x+1 \end{array} \right)^{n-1} \left( \begin{array}{c} \langle CK_4^2 \rangle \\ \langle V_2^1 CK_4^2 \rangle \end{array} \right) \quad (3.11) \\ &= \left( \begin{array}{cc} x^2 + 3x + 3 & x+1 \\ x+2 & x+1 \end{array} \right)^{n-1} \left( \begin{array}{c} x^2 + x \\ x \end{array} \right) \quad \text{by ((3.6) and (3.7))}. \end{aligned} $$

Note that (3.11) remains valid for $n = 1$. Let $M := \begin{pmatrix} x^2 + 3x + 3 & x+1 \\ x+2 & x+1 \end{pmatrix}$ be referred to as the *states matrix*. The characteristic polynomial of $M$ is given by

$$ \chi(M, \lambda) = \left( \lambda - \frac{p-q}{2} \right) \left( \lambda - \frac{p+q}{2} \right), $$

where

$$ p := x^2 + 4x + 4 \quad \text{and} \quad q := \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12}. $$

The *n*-th power of the state matrix is computed using the standard eigenvalue method. We omit the details here, but the computation yields the following results:

$$ \langle CK_4^{2n} \rangle = \frac{1}{2q} (x^2 + x) \left( (x^2 + 2x + 4 + q) \left(\frac{p+q}{2}\right)^{n-1} - (x^2 + 2x + 4 - q) \left(\frac{p-q}{2}\right)^{n-1} \right) $$

and

$$ \langle V_2^{2n-1} CK_4^{2n} \rangle = \frac{1}{2q} x \left( (x^2 + 4x + 2 + q) \left(\frac{p+q}{2}\right)^{n-1} - (x^2 + 4x + 2 - q) \left(\frac{p-q}{2}\right)^{n-1} \right). $$

□

Let $CK(x, y) := \sum_{n \ge 1} \langle CK_4^{2n} \rangle y^n$ denote the generating function of $(\langle CK_4^{2n} \rangle)_n$. By (3.10), we have

$$ CK(x, y) = \frac{2xy(x+1)((x^2 + 2x + 4 - p)y + 2)}{(2 - (p-q)y)(2 - (p+q)y)}. \quad (3.12) $$
Substituting the expressions for $p$ and $q$ in terms of $x$, (3.12) simplifies to

$$CK(x, y) = \frac{xy(x+1)(1-xy)}{1-(x+2)^2y+(x+1)^3y^2}. \qquad (3.13)$$

Table 1 displays the coefficients of the Kauffman bracket polynomial for the Celtic knot $CK_4^{2n}$ for small values of $n$. Recall that the coefficient $[x^k] \langle CK_4^{2n} \rangle$ represents the number of Kaufman states of $CK_4^{2n}$ that consist of exactly $k$ circles.

<table><caption>Table 1: Coefficients in the expansion of $\langle CK_4^{2n} \rangle$ for small values of $n$</caption><thead><tr><th>n \ k</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th><th>11</th><th>12</th><th>13</th><th>14</th></tr></thead><tbody><tr><th>1</th><td>0</td><td>1</td><td>1</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>2</th><td>0</td><td>4</td><td>7</td><td>4</td><td>1</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>3</th><td>0</td><td>15</td><td>40</td><td>42</td><td>23</td><td>7</td><td>1</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>4</th><td>0</td><td>56</td><td>201</td><td>306</td><td>262</td><td>140</td><td>48</td><td>10</td><td>1</td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><th>5</th><td>0</td><td>209</td><td>943</td><td>1877</td><td>2189</td><td>1672</td><td>881</td><td>325</td><td>82</td><td>13</td><td>1</td><td></td><td></td><td></td><td></td></tr><tr><th>6</th><td>0</td><td>780</td><td>4239</td><td>10412</td><td>15368</td><td>15276</td><td>10841</td><td>5660</td><td>2194</td><td>624</td><td>125</td><td>16</td><td>1</td><td></td><td></td></tr><tr><th>7</th><td>0</td><td>2911</td><td>18506</td><td>54051</td><td>96501</td><td>118175</td><td>105495</td><td>71107</td><td>36885</td><td>14817</td><td>4579</td><td>1064</td><td>177</td><td>19</td><td>1</td></tr><tr><th>...</th><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table>

Remark 2.

* Column 1 in Table 1 matches sequence A001353 in the OEIS [8], which is related to the determinant sequence for the weaving knots $S(4, n, (1, -1, 1))$ [5].

* For $n = 2$, $CK_4^4$ is a “4-foil”, the shadow of the link $4_1^2$. The bracket polynomial of an $n$-foil, denoted $F_n$, that is the shadow of the $(2, n)$-torus link [7], is

$$\langle F_n \rangle = (x+1)^n + x^2 - 1. \qquad (3.14)$$

* For $n = 3$, we have the following decomposition.

First, we write $\langle CK_4^6 \rangle = \langle H_2^3 CK_4^6 \rangle + \langle V_2^3 CK_4^6 \rangle$ by axiom (3.1) (see Figure 5).

<figure><img src="image_8.png" /><figcaption>Figure 5: The states of the crossing at the construction dot in line 2 and column 3</figcaption></figure>

Knot $H_2^3 CK_4^6$ is a 6-foil (denoted $F_6$), while knot $V_2^3 CK_4^6$ is the connected sum of two trefoils (denoted $F_3 \# F_3$). Recall that the bracket polynomial of the connected sum of two shadow diagrams $K$ and $K'$ satisfies [7]

$$\langle K \# K' \rangle = x^{-1} \langle K \rangle \langle K' \rangle. \qquad (3.15)$$
Hence, $\langle CK_4^6 \rangle$ can be expressed as

$$ 
\begin{aligned}
\langle CK_4^6 \rangle &= \langle F_6 \rangle + \langle F_3 \# F_3 \rangle \\
&= \langle F_6 \rangle + x^{-1} \langle F_3 \rangle^2 \quad \text{(by (3.15))} \\
&= (x+1)^6 + x^2 - 1 + x^{-1} ((x+1)^3 + x^2 - 1)^2 \quad \text{(by (3.14))} \\
&= x^6 + 7x^5 + 23x^4 + 42x^3 + 40x^2 + 15x.
\end{aligned}
$$

Remark 3. In the context of Celtic design, the coefficient $[x^k] \langle CK_4^{2n} \rangle$ denotes the number of distinct barrier configurations [3] or equivalently, the number of break assignments [1] that result in exactly $k$ connected components in the final design. Specifically, each such configuration corresponds to a choice of either a horizontal barrier (type-0 break) or a vertical barrier (type-1 break) at every interior construction dot (i.e., each crossing in the shadow diagram) which fully determine the state of the plaitwork.

Let us consider the case $n = 3$ and focus on the two extreme values $k = 1$ and $k = 6$.

For $k = 1$, there are 15 distinct barrier configurations that result in a single connected component. These 15 configurations fall into six equivalence classes under the action of the dihedral group $D_2$, which includes reflection across the vertical axis, reflection across the horizontal axis and 180° rotation [1]. The representatives of these equivalence classes are shown in Figure 6. The size of equivalence classes are for Figure 6(a): 2, 6(b): 2, 6(c): 2, 6(d): 1, 6(e): 4, and 6(b): 4.

<figure><img src="image_9.png" /><figcaption>Figure 6: The 6 equivalence classes for the single-component barrier configurations on $CK_4^6$</figcaption></figure>

For $k = 6$, there is only one such configuration that produces exactly 6 disjoint simple closed curves. This unique configuration is illustrated in Figure 7.
<figure><img src="image_10.png" /><figcaption>Figure 7: The unique 6-component barrier configuration for $CK_4^6$</figcaption></figure>

# 4 Within the 4-tangle framework

In this section, we make use of a 4-tangle algebraic framework to analyze the Kauffman bracket of Celtic knot shadows.

Kauffman's state model for tangles shows that any state $S$ of a 4-tangle diagram $T := \begin{tikzpicture}[baseline=(current bounding box.center)] \node[draw, circle, minimum size=1.5pt] (T) at (0,0) {};
\draw[thick] (T.north) -- (T.south);
\draw[thick] (T.east) -- (T.west);
\draw[thick] (T.north) -- (T.east);
\draw[thick] (T.south) -- (T.west);
\node[above] at (T.north) {$g_1$};
\node[below] at (T.south) {$g_2$};
\node[left] at (T.east) {$g_3$};
\node[right] at (T.west) {$g_4$};
\draw[thick] (T.north) -- (T.south);
\draw[thick] (T.north) -- (T.east);
\draw[thick] (T.south) -- (T.west);
\draw[thick] (T.south) -- (T.east);
\node[left] at (T.north) {$g_5$};
\node[right] at (T.south) {$g_6$};
\node[left] at (T.east) {$g_7$};
\node[right] at (T.west) {$g_8$};
\node[left] at (T.north) {$g_9$};
\node[right] at (T.south) {$g_{10}$};
\node[left] at (T.east) {$g_{11}$};
\node[right] at (T.west) {$g_{12}$};
\draw[thick] (T.north) -- (T.south);
\draw[thick] (T.north) -- (T.east);
\draw[thick] (T.south) -- (T.west);
\draw[thick] (T.south) -- (T.east);
\node[left] at (T.north) {$g_{13}$};
\node[right] at (T.south) {$g_{14}$};
\end{tikzpicture}$ lies in the module generated by the disjoint union of loops and elements of the 4-strand diagram monoid $K_4$ which consists of all planar isotopy classes of 4-endpoint diagrams with no closed components [6].

The monoid $K_4$ has 14 distinct basis elements, here denoted $g_1, g_2, \dots, g_{14}$, corresponding to the 14 possible ways to connect four boundary points in pairs without crossings. These are depicted in Figure 8.

<figure><img src="image_11.png" /><figcaption>Figure 8: The states basis element of $K_4$</figcaption></figure>

Every state $S$ arising from a bracket expansion of $T$ can be expressed as a disjoint union

$$S = \bigcirc^k \sqcup g,$$

where $k \ge 0$, $\bigcirc^k = \bigcirc \sqcup \bigcirc \sqcup \dots \sqcup \bigcirc$ denotes the disjoint union of $k$ circles, and $g \in K_4$.

Thus, by the bracket axioms (K1) and (K2), the bracket evaluation of the state $S$ is [4, p. 98]

$$\langle S \rangle = x^k \langle g \rangle.$$

Consequently, the full bracket polynomial of the 4-tangle $T$ is a linear combination of the form
$$ \langle T \rangle = \sum_S \langle S \rangle = \sum_{i=1}^{14} a_i \langle g_i \rangle, $$

where each coefficient $a_i \in \mathbb{Z}[x]$ counts (with powers of $x$) the number of states that reduce to the basis element $g_i$, possibly with additional circles.

**Definition 4.1.** The concatenation (or multiplication) of two 4-tangles $T$ and $T'$, denoted $TT'$, is the 4-tangle formed by placing $T$ directly above $T'$ and connecting the bottom endpoints of $T$ to the corresponding top endpoints of $T'$ with non-crossing planar arcs:

<figure><img src="image_12.png" /></figure>

Let $T_n := TT \cdots T$ denote the 4-tangle obtained by concatenating the 4-tangle $T$ with itself $n$ times, with $T_0 := g_1$. By construction, for each $n \ge 0$, the Kauffman bracket polynomial $\langle T_n \rangle$ lies in the linear span of a fixed basis of tangle states $\{\langle g_1 \rangle, \dots, \langle g_{14} \rangle\}$. That is, there exist coefficients $a_1^{(n)}, \dots, a_{14}^{(n)}$ such that

$$ \langle T_n \rangle = \sum_{i=1}^{14} a_i^{(n)} \langle g_i \rangle \qquad (4.1) $$

In the following, we compute the Kauffman bracket polynomial for $CK_4^{2n}$ using a recursive approach based on the concatenation of a fundamental 4-tangle $G$, defined as

<figure><img src="image_13.png" /></figure>

The following result can easily be established

<figure><img src="image_14.png" /></figure>

or more formally,

$$ \langle G \rangle = \langle g_1 \rangle + \langle g_2 \rangle + \langle g_3 \rangle + \langle g_4 \rangle + \langle g_5 \rangle + \langle g_6 \rangle + \langle g_7 \rangle + \langle g_8 \rangle + \langle g_9 \rangle + \langle g_{11} \rangle. \qquad (4.2) $$
**Lemma 4.1.** For $n \ge 1$, $G_n$ satisfies the following decomposition:

$$
\begin{align}
\langle G_n \rangle ={}& a_1^{(n-1)} \langle g_1 \rangle + (a_1^{(n-1)} + (x+2)a_2^{(n-1)} + (x+1)a_6^{(n-1)} + a_{14}^{(n-1)}) \langle g_2 \rangle \nonumber \\
& + (a_1^{(n-1)} + (x+1)a_3^{(n-1)} + a_5^{(n-1)} + a_8^{(n-1)}) \langle g_3 \rangle \nonumber \\
& + (a_1^{(n-1)} + (x+2)a_4^{(n-1)} + (x+1)a_7^{(n-1)} + a_{13}^{(n-1)}) \langle g_4 \rangle \nonumber \\
& + (a_1^{(n-1)} + (x+1)a_3^{(n-1)} + (x+2)a_5^{(n-1)} + a_8^{(n-1)}) \langle g_5 \rangle \nonumber \\
& + (a_2^{(n-1)} + (x+1)a_6^{(n-1)} + a_{14}^{(n-1)}) \langle g_6 \rangle \nonumber \\
& + (a_4^{(n-1)} + (x+1)a_7^{(n-1)} + a_{13}^{(n-1)}) \langle g_7 \rangle \tag{4.3} \\
& + (a_1^{(n-1)} + (x+1)a_3^{(n-1)} + a_5^{(n-1)} + (x+2)a_8^{(n-1)}) \langle g_8 \rangle \nonumber \\
& + (a_1^{(n-1)} + (x+1)a_3^{(n-1)} + (x+2)a_5^{(n-1)} + (x+2)a_8^{(n-1)} \nonumber \\
& \qquad + (x^2+3x+3)a_9^{(n-1)} + (x^2+3x+2)a_{10}^{(n-1)}) \langle g_9 \rangle \nonumber \\
& + (a_9^{(n-1)} + (x+1)a_{10}^{(n-1)}) \langle g_{10} \rangle \nonumber \\
& + (a_1^{(n-1)} + (x+2)a_2^{(n-1)} + (x+2)a_4^{(n-1)} + (x+1)a_6^{(n-1)} + (x+1)a_7^{(n-1)} \nonumber \\
& \qquad + (x^2+3x+3)a_{11}^{(n-1)} + (x^2+3x+2)a_{12}^{(n-1)} + (x+2)a_{13}^{(n-1)} + (x+2)a_{14}^{(n-1)}) \langle g_{11} \rangle \nonumber \\
& + (a_{11}^{(n-1)} + (x+1)a_{12}^{(n-1)}) \langle g_{12} \rangle \nonumber \\
& + (a_4^{(n-1)} + (x+1)a_7^{(n-1)} + (x+2)a_{13}^{(n-1)}) \langle g_{13} \rangle \nonumber \\
& + (a_2^{(n-1)} + (x+1)a_6^{(n-1)} + (x+2)a_{14}^{(n-1)}) \langle g_{14} \rangle . \nonumber
\end{align}
$$

*Proof.* If $\langle G \rangle = \sum_{j=1}^{8} \langle g_{i_j} \rangle$, where the $g_{i_j}$ represent the basis states contributing to the 4-tangle $G$ as determined in (4.2), then for the iterated 4-tangle $G_n$, we have

$$
\langle G_n \rangle = \langle G_{n-1} G \rangle = \sum_{i=1}^{14} a_i^{(n-1)} \sum_{j=1}^{8} \langle g_i g_{i_j} \rangle,
$$

where $a_i^{(n-1)}$ denotes the coefficient of the bracket $\langle g_i \rangle$ in the expansion of $\langle G_{n-1} \rangle$.

Each bracket $\langle g_i g_{i_j} \rangle$ corresponds to the Kauffman state evaluation of the concatenated
configuration formed by combining state $g_i$ from the $G_{n-1}$ factor with state $g_{i_j}$ from $G$.
The result of this product is determined by a state multiplication rule which is tabulated
in Table 2. In this table, the entry in row $j$ and column $i$ gives the resulting state from
the formal product $g_i g_{i_j}$, with $i = 1, \dots, 14$ indexing the column labels (top row) and
$j = 1, \dots, 8$ indexing the column labels (leftmost column). The conclusion follows by
expressing the result in terms of the $\langle g_i \rangle$'s.
$\square$

<figure><img src="image_15.png" /></figure>
<figure><img src="image_16.png" /><figcaption>Table 2: Multiplication table of $g_i$, $i = 1, \dots, 14$ and $g_{i,j}$, $j = 1, \dots, 8$</figcaption></figure>
**Notation 4.2.** For convenience, we identify the bracket polynomial expression in $\langle G \rangle = \sum_{i=1}^{14} a_i \langle g_i \rangle$ with the coefficient vector

$$A := A(G) = [a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, a_{11}, a_{12}, a_{13}, a_{14}]^T.$$

Similarly, we identify the bracket polynomial $\langle G_n \rangle$ with the vector

$$A_n := A(G_n) = [a_1^{(n)}, a_2^{(n)}, a_3^{(n)}, a_4^{(n)}, a_5^{(n)}, a_6^{(n)}, a_7^{(n)}, a_8^{(n)}, a_9^{(n)}, a_{10}^{(n)}, a_{11}^{(n)}, a_{12}^{(n)}, a_{13}^{(n)}, a_{14}^{(n)}]^T.$$

**Lemma 4.3.** For $n \ge 1$, the bracket vector of $G_n$ satisfies the following recursion:

$$ \begin{pmatrix} a_1^{(n)} \\ a_2^{(n)} \\ a_3^{(n)} \\ a_4^{(n)} \\ a_5^{(n)} \\ a_6^{(n)} \\ a_7^{(n)} \\ a_8^{(n)} \\ a_9^{(n)} \\ a_{10}^{(n)} \\ a_{11}^{(n)} \\ a_{12}^{(n)} \\ a_{13}^{(n)} \\ a_{14}^{(n)} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & t & 0 & 0 & 0 & s & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & s & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & t & 0 & 0 & s & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & s & 0 & t & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & s & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & s & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & s & 0 & 1 & 0 & 0 & t & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & s & 0 & t & 0 & 0 & t & v & u & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & s & 0 & 0 & 0 \\ 1 & t & 0 & t & 0 & s & s & 0 & 0 & 0 & v & u & t \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & s & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & s & 0 & 0 & 0 & 0 & t & 0 \\ 0 & 1 & 0 & 0 & 0 & s & 0 & 0 & 0 & 0 & 0 & 0 & t \end{pmatrix} \begin{pmatrix} a_1^{(n-1)} \\ a_2^{(n-1)} \\ a_3^{(n-1)} \\ a_4^{(n-1)} \\ a_5^{(n-1)} \\ a_6^{(n-1)} \\ a_7^{(n-1)} \\ a_8^{(n-1)} \\ a_9^{(n-1)} \\ a_{10}^{(n-1)} \\ a_{11}^{(n-1)} \\ a_{12}^{(n-1)} \\ a_{13}^{(n-1)} \\ a_{14}^{(n-1)} \end{pmatrix}, \quad (4.4) $$

where

$$s := x+1, \quad t := x+2, \quad u := x^2+3x+2 \quad \text{and} \quad v := x^2+3x+3.$$

*Proof.* Recursion (4.4) is directly derived from (4.3). □

<figure><img src="image_17.png" /></figure>

Note that (4.4) can also be written as

$$A_n = M A_{n-1} = M^n A_0, \quad (4.5)$$

with

$$A_0 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]^T,$$

where $M$ denotes the $14 \times 14$ matrix which is referred to as *states matrix*.

The $n$-th power of the state matrix is computed using the standard eigenvalue method. Hence, the characteristic polynomial of $M$ is given by

$$\chi(M, \lambda) = (\lambda-1)(\lambda-x-1)^3 (\lambda-x-2-r)^3 (\lambda-x-2+r)^3 \left(\lambda - \frac{p-q}{2}\right)^2 \left(\lambda - \frac{p+q}{2}\right)^2,$$
where

$$r := \sqrt{2x+3}, \quad p := x^2+4x+4 \quad \text{and} \quad q := \sqrt{x^4+4x^3+12x^2+20x+12}.$$

Then, the values for the $a_i^n$'s are given as follows:

$$a_1^{(n)} = 1$$

$$a_2^{(n)} = 2wqx(x^2 - 1)^2 \left\{ \lambda_1^n (2rx^2 - 4r) + \lambda_2^n ((r+1)x^2 + 2x) + \lambda_3^n ((r-1)x^2 - 2x) - 4rx^2 + 4r \right\}$$

$$a_3^{(n)} = 4wqx^2(x^2 - 1)^2 \left\{ \lambda_2^n((r+1)x + 2) + \lambda_3^n((r-1)x - 2) - 2rx \right\}$$

$$a_4^{(n)} = 2wqx(x^2 - 1)^2 \left\{ \lambda_1^n (2rx^2 - 4r) + \lambda_2^n ((r+1)x^2 + 2x) + \lambda_3^n ((r-1)x^2 - 2x) - 4rx^2 + 4r \right\}$$

$$a_5^{(n)} = 2wqx^2(x^2 - 1)^2 \left\{ \lambda_2^n (-2x^2 - 2x + 2 - 2r) + \lambda_3^n (2x^2 + 2x - 2 - 2r) + 4r \right\}$$

$$a_6^{(n)} = 2wqx^2(x^2 - 1) \left\{ \lambda_2^n (-2x^3 - (2r+2)x^2 + 2x + 2 + 2r) + \lambda_3^n (2x^3 + (2-2r)x^2 - 2x - 2 + 2r) + 4rx^2 - 4 \right\}$$

$$a_7^{(n)} = 2wqx^2(x^2 - 1) \left\{ \lambda_2^n (-2x^3 - (2r+2)x^2 + 2x + 2 + 2r) + \lambda_3^n (2x^3 + (2-2r)x^2 - 2x - 2 + 2r) + 4rx^2 - 4 \right\}$$

$$a_8^{(n)} = 2wqx^2(x^2 - 1)^2 \left\{ \lambda_2^n (-2x^2 - 2x + 2 - 2r) + \lambda_3^n (2x^2 + 2x - 2 - 2r) + 4r \right\}$$

$$a_9^{(n)} = 2wx(x^2 - 1) \left\{ \lambda_2^n q (4x^4 + 4x^3 + (4r-8)x^2 - 4x + 4 - 4r) + \lambda_3^n q (-4x^4 - 4x^3 + (4r+8)x^2 + 4x - 4 - 4r) + \lambda_4^n r (x^7 + 4x^6 + 6x^5 + 2x^4 - (q^2+8)x^3 - (2q+16)x^2 + (2q^2-16)x + 4q - 8) + \lambda_5^n r (-x^7 - 4x^6 - 6x^5 - 2x^4 + (q^2+8)x^3 + (16-2q)x^2 + (16-2q^2)x + 4q + 8) - 4qx^2 \right\}$$

$$a_{10}^{(n)} = 4wx(x^2 - 1) \left\{ \lambda_2^n q((-r-1)x^3 - 2x^2 + (r+1)x + 2) + \lambda_3^n q((1-r)x^3 + 2x^2 + (r-1)x - 2) + \lambda_4^n r(x^5 + 2x^4 + qx^3 - 2x^2 - (2q+4)x - 4) + \lambda_5^n r(-x^5 - 2x^4 + qx^3 + 2x^2 + (4-2q)x + 4) + 2qrx \right\}$$

$$a_{11}^{(n)} = 2wx(x^2 - 1) \left\{ \lambda_2^n q((-4r-4)x^3 - 8x^2 + (4r+4)x + 8) + \lambda_3^n q((4-4r)x^3 + 8x^2 + (4r-4)x - 8) + \lambda_4^n r(-x^6 - 6x^5 - 10x^4 + 2qx^3 + (q^2+20)x^2 + (24-4q)x + 8 - 2q^2) + \lambda_5^n r(x^6 + 6x^5 + 10x^4 + 2qx^3 - (q^2+20)x^2 - (4q+24)x - 8 + 2q^2) + 4qrx^3 \right\}$$

$$a_{12}^{(n)} = 4wx(x^2 - 1) \left\{ \lambda_2^n q (2x^3 + (2r+2)x^2 - 2x - 2 - 2r) + \lambda_3^n q (-2x^3 + (2r-2)x^2 + 2x + 2 - 2r) + \lambda_4^n r (-x^4 - 4x^3 - qx^2 + 8x + 4 + 2q) \right\}$$
$$ + \lambda_5^n r (x^4 + 4x^3 - qx^2 - 8x - 4 + 2q) - 2qrx^2 \} $$

$$ a_{13}^{(n)} = 2wqx(x^2 - 1)^2 \{ \lambda_1^n (-2rx^2 + 4r) + \lambda_2^n ((r+1)x^2 + 2x) + \lambda_3^n ((r-1)x^2 - 2x) - 4r \} $$

$$ a_{14}^{(n)} = 2wqx(x^2 - 1)^2 \{ \lambda_1^n (-2rx^2 + 4r) + \lambda_2^n ((r+1)x^2 + 2x) + \lambda_3^n ((r-1)x^2 - 2x) - 4r \}, $$

where

$$ r := \sqrt{2x+3}, \quad q := \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12}, \quad w := \frac{1}{8qrx^2(x^2-2)(x^2-1)^2}, $$

$$ \lambda_1 := x+1, \quad \lambda_2 := x+2-\sqrt{2x+3}, \quad \lambda_3 := x+2+\sqrt{2x+3}, $$

$$ \lambda_4 := \frac{1}{2} \left( x^2 + 4x + 4 + \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12} \right), $$

$$ \text{and} \quad \lambda_5 := \frac{1}{2} \left( x^2 + 4x + 4 + \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12} \right). $$

Next, we introduce a closure for a 4-tangle $T$, denoted $\bar{T}$, which is obtained by connecting the two top endpoints to each other and the two bottom endpoints to each other without introducing any crossings as displayed in Figure 9(a). In the same fashion, the closure for $G_n$, denoted $\bar{G}_n$, is illustrated in Figure 9(b).

<figure><img src="image_18.png" /><figcaption>Figure 9: The closure operation</figcaption></figure>

**Lemma 4.4.** *The bracket polynomial for the closure of $G_n$ satisfies*

$$ \langle \overline{G_n} \rangle = a_{11}^{(n)} x^4 + (a_2^{(n)} + a_4^{(n)} + a_9^{(n)} + a_{12}^{(n)} + a_{13}^{(n)} + a_{14}^{(n)}) x^3 \\ + (a_1^{(n)} + a_5^{(n)} + a_6^{(n)} + a_7^{(n)} + a_8^{(n)} + a_{10}^{(n)}) x^2 + a_3^{(n)} x. \quad (4.6) $$

*Proof.* The crossings are smoothed without affecting the closure, hence

$$ \langle \overline{G_n} \rangle = \sum_{i=1}^{14} a_i^{(n)} \langle \overline{g_i} \rangle, \quad (4.7) $$

and the only point remaining concerns the evaluation of the brackets to the closure of the elements of $\mathcal{K}_4$ (see Figure 10). $\square$
<figure><img src="image_19.png" /><figcaption>Figure 10: Closures of the elements of $\mathcal{K}_4$</figcaption></figure>

Replacing the $a_i^{(n)}$'s and the $\langle g_i \rangle$'s by their formal expressions, (4.7) becomes

$$ \langle \overline{G}_n \rangle = \frac{1}{2q} x \left( \lambda_5^n (x^3 + 2x^2 + (q+2)x + 2) - \lambda_4^n (x^3 + 2x^2 - (q-2)x + 2) \right) \\ = \frac{1}{2q} x \left( \left( \frac{p+q}{2} \right)^n (x^3 + 2x^2 + (q+2)x + 2) - \left( \frac{p+q}{2} \right)^n (x^3 + 2x^2 - (q-2)x + 2) \right), $$

where

$$ p := x^2 + 4x + 4 \quad \text{and} \quad q := \sqrt{x^4 + 4x^3 + 12x^2 + 20x + 12}. $$

Finally, the following Theorem establishes the link between $\langle \overline{G}_n \rangle$ and the Kaufman bracket polynomial of the Celtic knot $\langle CK_4^{2n} \rangle$:

**Theorem 4.5.** $\langle \overline{G}_n \rangle$ and $\langle CK_4^{2n} \rangle$ satisfy the following identity:

$$ \langle \overline{G}_n \rangle = (x+1)^2 \langle CK_4^{2n} \rangle \quad \text{for } n \ge 1. \qquad (4.8) $$

*Proof.* Identity (4.8) follows immediately from diagram in Figure 9(b). $\square$

<figure><img src="image_20.png" /></figure>

Theorem 4.5 gives us the closed form for $\langle CK_4^{2n} \rangle$ based on $\langle \overline{G}_n \rangle$:

$$ \langle CK_4^{2n} \rangle = \frac{1}{(x+1)^2} \langle \overline{G}_n \rangle. \qquad (4.9) $$

We have

$$ \begin{aligned} \langle \overline{G}_n \rangle &= \frac{1}{2q} x \left( \left( \frac{p+q}{2} \right)^{n-1} \left( \frac{x^2+4x+4+q}{2} \right) (x^3+2x^2+(q+2)x+2) \right. \\ & \qquad \left. - \left( \frac{p-q}{2} \right)^{n-1} \left( \frac{x^2+4x+4-q}{2} \right) (x^3+2x^2-(q-2)x+2) \right) \\ &= \frac{1}{2q} x \left( \left( \frac{p+q}{2} \right)^{n-1} \left( \frac{x^5}{2} + 3x^4 + (q+7)x^3 + (3q+9)x^2 + \left( \frac{q^2}{2} + 3q+8 \right) x + 4 + q \right) \right) \end{aligned} $$
$$ - \left(\frac{p-q}{2}\right)^{n-1} \left(\frac{x^5}{2} + 3x^4 + (7-q)x^3 + (9-3q)x^2 + \left(\frac{q^2}{2} - 3q + 8\right)x + 4 - q\right) $$

(replacing $q^2$ by $x^4 + 4x^3 + 12x^2 + 20x + 12$)

$$ = \frac{1}{2q}x \left( \left( \frac{p+q}{2} \right)^{n-1} \left( x^5 + 5x^4 + (q+13)x^3 + (3q+19)x^2 + (3q+14)x + 4 + q \right) \right) $$

$$ - \left(\frac{p-q}{2}\right)^{n-1} \left(x^5 + 5x^4 + (13-q)x^3 + (19-3q)x^2 + (14-3q)x + 4 - q\right) $$

$$ = \frac{1}{2q}x \left( \left( \frac{p+q}{2} \right)^{n-1} (x+1)^3 (x^2+2x+4+q) \right) $$

$$ - \left(\frac{p-q}{2}\right)^{n-1} (x+1)^3 (x^2+2x+4-q) $$

$$ \langle \overline{G_n} \rangle = \frac{1}{2q} x (x+1)^3 \left( (x^2 + 2x + 4 + q) \left(\frac{p+q}{2}\right)^{n-1} - (x^2 + 2x + 4 - q) \left(\frac{p-q}{2}\right)^{n-1} \right). $$

Applying (4.9), we obtain the expected result.

## References

[1] R. Antonsen and L. Taalman, “Categorizing Celtic Knot Designs”, In Proceedings of Bridges 2021: Mathematics, Art, Music, Architecture, Culture, 2021, pp. 87–94.

[2] I. Bain, *Celtic Knotwork*, Sterling Publishing Co., Inc., 1992.

[3] J. L. Gross and T. W. Tucker, “A Celtic Framework for Knots and Links”, Discrete Comput. Geom. 46 (2011), 86–99.

[4] L. H. Kauffman, *Knots and Physics*, World Scientific, 1993.

[5] S. J. Kim, R. Stees and L. Taalman, “Sequences of spiral knot determinants”, J. Integer Seq. 19 (2016), 1–14.

[6] N. V. Kitov and M. V. Volkov, “Identities of the Kauffman Monoid $K_4$ and of the Jones Monoid $J_4$”, In A. Blass, P. Cégielski, N. Dershowitz, M. Droste and B. Finkbeiner (eds), *Fields of Logic and Computation III. Lecture Notes in Computer Science* 12180 (2020), Springer, pp. 156–178.

[7] F. Ramaharo, Statistics on some classes of knot shadows, arXiv.org e-Print archive, 2018, https://arxiv.org/abs/1802.07701

[8] N. J. A. Sloane, *The On-Line Encyclopedia of Integer Sequences*, published electronically at http://oeis.org, 2025.

[9] A. Sloss, *How to Draw Celtic Knotwork*, Cassel & Co, 2002.

**2010 Mathematics Subject Classifications:** 57M25; 05A19.