Optimal cube factors of Fibonacci and matchable Lucas cubes

TL;DR

This paper introduces the concept of optimal cube factors in graphs and investigates their properties specifically in Fibonacci and matchable Lucas cubes, linking these findings to Padovan sequences and binomial coefficients.

Contribution

It is the first to study optimal cube factors in Fibonacci and matchable Lucas cubes, providing new insights and results related to these graph classes.

Findings

Derived results connecting optimal cube factors to Padovan sequence

Established relationships between cube factors and binomial coefficients

Provided new bounds or formulas for the cube factors of specific graph families

Abstract

The optimal cube factor of a graph, a special kind of component factor, is first introduced. Furthermore, the optimal cube factors of Fibonacci and matchable Lucas cubes are studied; and some results on the Padovan sequence and binomial coefficients are obtained.

Tables

Table 1. Table 1 : The coefficients of first few Q F ​ ( Γ n , x ) superscript 𝑄 𝐹 subscript Γ 𝑛 𝑥 Q^{F}(\Gamma_{n},x)

Table 2. Table 2 : The first six rows of Lucas triangle Y 𝑌 Y

Table 3. Table 3 : The coefficients of first few Q F ​ ( Ω n , x ) superscript 𝑄 𝐹 subscript Ω 𝑛 𝑥 Q^{F}(\Omega_{n},x)

Equations

$$Q^F(G,x)=\sum _{k=0}^m{q}_k^F(G)x^k$$

$${\Gamma }_{j+1}\square P_2\subseteq {\Gamma }_{j+3},$$

$${\Gamma }_j\subseteq {\Gamma }_{j+1},$$

$$q_k^F({\Gamma }_n)=q_{k-1}^F({\Gamma }_{n-2})+q_k^F({\Gamma }_{n-3}).$$

$$Q^F({\Gamma }_0,x)$$

$$Q^F({\Gamma }_1,x)$$

$$Q^F({\Gamma }_2,x)$$

$$Q^F({\Gamma }_3,x)$$

$$Q^F({\Gamma }_4,x)$$

$$Q^F({\Gamma }_5,x)$$

$$Q^F({\Gamma }_n,x)=x{Q}^F({\Gamma }_{n-2},x)+Q^F({\Gamma }_{n-3},x)$$

$$Q^F({\Gamma }_n,x)=\sum _{k=0}^m{q}_k^F({\Gamma }_n)x^k=\sum _{k=0}^m{q}_{k-1}^F({\Gamma }_{n-2})x^k+\sum _{k=0}^m{q}_k^F({\Gamma }_{n-3})x^k=x\sum _{k=0}^m{q}_k^F({\Gamma }_{n-2})x^k+\sum _{k=0}^m{q}_k^F({\Gamma }_{n-3})x^k.$$

$$Q^F({\Gamma }_n,1)=p_{n+1}.$$

$$Q^F({\Gamma }_n,2)=F_{n+2}.$$

$$\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n=\frac{1+y(x+y)}{1-y^2(x+y)}.$$

$$\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n$$

$$=\sum _{n=3}^{\infty }x{Q}^F({\Gamma }_{n-2},x)y^n+\sum _{n=3}^{\infty }{Q}^F({\Gamma }_{n-3},x)y^n+\sum _{n=0}^2{Q}^F({\Gamma }_n,x)y^n$$

$$=x{y}^2\sum _{n=1}^{\infty }{Q}^F({\Gamma }_n,x)y^n+x{y}^3\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n+\sum _{n=0}^2{Q}^F({\Gamma }_n,x)y^n$$

$$=y^2(x+y)\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n-x{y}^2{Q}^F({\Gamma }_0,x)+\sum _{n=0}^2{Q}^F({\Gamma }_n,x)y^n$$

$$=y^2(x+y)\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n+1+y(x+y).$$

$$(1-y^2(x+y))\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,x)y^n=1+y(x+y).$$

$$\sum _{n=0}^{\infty }{p}_n{y}^n=\frac{1+y}{1-y^2(1+y)}.$$

$$\sum _{n=0}^{\infty }{p}_n{y}^n=\sum _{n=1}^{\infty }{p}_n{y}^n+p_0=y\sum _{n=0}^{\infty }{p}_{n+1}{y}^n+p_0=y\sum _{n=0}^{\infty }{Q}^F({\Gamma }_n,1)y^n+p_0=\frac{y(1+y(1+y))}{1-y^2(1+y)}+1,$$

$$q_k^F({\Gamma }_n)=\left(\genfrac{}{}{0ex}{}{\frac{n+k}{3}}{k}\right)+\left(\genfrac{}{}{0ex}{}{\frac{n+k+1}{3}}{k}\right).$$

$$q_k^F({\Gamma }_n)=\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{m}{k}\right),& \text{if }n+k=3m,3m-1;\\ 0,& \text{if }n+k=3m+1.\end{array}$$

$$\frac{1}{1-y^2(x+y)}$$

$$=\sum _{j=0}^{\infty }\sum _{n-2j=0}^j\left(\genfrac{}{}{0ex}{}{j}{n-2j}\right)x^{3j-n}{y}^n$$

$$=\sum _{n=0}^{\infty }\sum _{j=\lfloor \frac{n+2}{3}\rfloor }^{\lfloor \frac{n}{2}\rfloor }\left(\genfrac{}{}{0ex}{}{j}{n-2j}\right)x^{3j-n}{y}^n;$$

$$[x^k][y^n]\frac{1}{1-y^2(x+y)}=\left(\genfrac{}{}{0ex}{}{\frac{n+k}{3}}{k}\right).$$

$$[x^k][y^n]\frac{y(x+y)}{1-y^2(x+y)}$$

$$=[x^{k-1}][y^{n-1}]\frac{1}{1-y^2(x+y)}+[x^k][y^{n-2}]\frac{1}{1-y^2(x+y)}$$

$$=\left(\genfrac{}{}{0ex}{}{\frac{n+k-2}{3}}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{\frac{n+k-2}{3}}{k}\right)$$

$$=\left(\genfrac{}{}{0ex}{}{\frac{n+k+1}{3}}{k}\right).$$

$$q_k^F({\Gamma }_{n-k})=\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{m}{k}\right),& \text{if }n=3m-1\text{ or }3m;\\ 0,& \text{if }n=3m-2;\end{array}$$

$$q_k^F({\Gamma }_{n+2k})=\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{k+m}{k}\right),& \text{if }n=3m-1,3m.\\ 0,& \text{otherwise.}\end{array}$$

$$p_n=\sum _{j=\lfloor \frac{n+1}{3}\rfloor }^{\lfloor \frac{n}{2}\rfloor }\left(\genfrac{}{}{0ex}{}{j+1}{n-2j}\right).$$

$$p_{n+1}=Q^{F}(\Gamma_{n},1)=\sum_{j=\lfloor\frac{n+2}{3}\rfloor}^{\lfloor\frac{n+1}{2}\rfloor}\Bigg{(}\binom{j}{n-2j}+\binom{j}{n-2j+1}\Bigg{)}=\sum_{j=\lfloor\frac{n+2}{3}\rfloor}^{\lfloor\frac{n+1}{2}\rfloor}\binom{j+1}{n-2j+1}.$$

Full text

Key words: optimal cube factor; Fibonacci cube; matchable Lucas cube; Padovan sequence; Yang Hui triangle; Lucas triangle

2010 AMS Subj. Class.: 05C30; 05C70; 05A10; 11B83

1 Introduction

The factorizations of graph were been studied by Akiyama and Kano [1, 2], Yu and Liu [12], etc. For a set $\{A_{1},A_{2},A_{3},\ldots\}$ of graphs, an $\{A_{1},A_{2},A_{3},\ldots\}$-factor of a graph $G$ is a spanning subgraph of $G$ such that each of its components is isomorphic to one of $\{A_{1},A_{2},A_{3},\ldots\}$ [12, Chapter 4]. An $\{A_{1},A_{2},A_{3},\ldots\}$-factor is called a cube factor if each $A_{i}$ is isomorphic to a $k$-cube, where $i=1,2,\ldots$ and $k$ is a non-negative integer.

Let $G$ be a graph, let $\mathbf{F}$ be a cube factor of $G$; let $q^{F}_{k}(G)$ denote the number of subgraphs that are isomorphic to $k$-cube in $\mathbf{F}$ and let $q^{F}_{k}(G)=0$ if no $k$-cube in $\mathbf{F}$ exists. A cube factor $\mathbf{F}$ of $G$ is called as optimal, if there is no cube factor $\mathbf{F}^{\prime}$, such that $\sum_{k\geq 0}q^{F^{\prime}}_{k}(G)<\sum_{k\geq 0}q^{F}_{k}(G)$. Since the optimal cube factor is always denoted by $\mathbf{F}$, we will only write the optimal cube factor. Moreover, let

[TABLE]

be optimal cube factor polynomial of $G$. Hence $Q^{F}(G,1)$ is the smallest for a graph $G$.

The Padovan sequence $\{p_{n}\}$ is introduced by Stewart [9, Chapter 8], where $p_{0}=p_{1}=p_{2}=1$, and $p_{n}=p_{n-2}+p_{n-3}$ for $n\geq 3$.

We first consider the optimal cube factor polynomial of Fibonacci cubes [3], obtain the generating function and coefficients of which, and the relation between Padovan sequence and the polynomials, moreover give formulae of Padovan numbers and the coefficients; in what follows, we explore matchable Lucas cubes [10] in the same way.

2 Optimal cube factors of Fibonacci cubes

The Fibonacci sequence $\{F_{n}\}$ is defined as follows: $F_{0}=0$, $F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for $n\geq 2$. The Fibonacci cubes were introduced by Hsu [3], and lots of conclusions on Fibonacci cubes, include structure and enumerative properties, ware obtained [4, 5, 7, 13, 14]. Let $\Gamma_{n}$ be the $n$-th Fibonacci cube, the dimension of maximum cube of $\Gamma_{n}$ is $\lfloor n/2\rfloor$ [5], and the structure of $\Gamma_{n}$ $(n\geq 3)$ is shown as in Figure 1.

An easy induction gives the following theorem.

Theorem 2.1**.**

An optimal cube factor of $\Gamma_{n}$ could be obtained as follows:

*For every $k\geq 0$, select the most $k$-dimensional cubes in $\Gamma_{n}$ under selecting the most $k+1$-dimensional cubes and deleting them from $\Gamma_{n}$. *

Proof**.**

We proceed by induction on the $n$ of the Fibonacci cube. Verification for small values of $0\leq n<3$ is trivial. Assume that the optimal cube factors of $j$ and $j+1$ can be obtained as described above, by the structure of $\Gamma_{j+3}$ as shown in Figure 1, we have

[TABLE]

and

[TABLE]

where $G_{1}\mathbin{\square}G_{2}$ denotes the Cartesian product of graphs $G_{1}$ and $G_{2}$, and $P_{2}$ is a path with two vertices.

In addition, it is easy to see that the optimal cube factor of $\Gamma_{j+3}$ consists of the optimal cube factors of both $\Gamma_{j+1}\mathbin{\square}P_{2}$ and $\Gamma_{j}$. By the definition of Cartesian product, the optimal cube factor of $\Gamma_{j+1}\mathbin{\square}P_{2}$ corresponds to $\Gamma_{j+1}$. Therefore, the optimal cube factor of $\Gamma_{j+3}$ could be determined by $\Gamma_{j+1}$ and $\Gamma_{j}$. From the induction hypothesis, an optimal cube factor of $\Gamma_{j+3}$ can be obtained in the same way.

This complete the induction. $\square$

In fact, we just need to consider $0\leq k\leq\lfloor\frac{n}{2}\rfloor$ for $\Gamma_{n}$. Moreover, as a consequence of Theorem 2.1, we have the recurrence relation of $q^{F}_{k}(\Gamma_{n})$.

Corollary 2.2**.**

For $n\geq 3$ and $0\leq k\leq\lfloor\frac{n}{2}\rfloor$,

[TABLE]

The first few of $Q^{F}(\Gamma_{n},x)$ is listed as following.

[TABLE]

The recurrence relation of $Q^{F}(\Gamma_{n},x)$ is obtained easily.

Proposition 2.3**.**

For $n\geq 3$,

[TABLE]

Proof**.**

By Theorem 2.2,

[TABLE]

The proof is completed. $\square$

By recursion formula of Padovan sequences and Proposition 2.3, it is not difficult to verify the relation between $Q^{F}(\Gamma_{n},1)$ and Padovan sequences.

Corollary 2.4**.**

The sum of all coefficients of $Q^{F}(\Gamma_{n},x)$ is the $(n+1)$-th Padovan number, that is

[TABLE]

And

[TABLE]

Furthermore, some tedious computation yields the generation function of $Q^{F}(\Gamma_{n},x)$.

Theorem 2.5**.**

The generation function of $Q^{F}(\Gamma_{n},x)$ is given by

[TABLE]

Proof**.**

By the recurrence relation of $Q^{F}(\Gamma_{n},x)$, i.e. Proposition 2.3,

[TABLE]

Thus,

[TABLE]

$\square$

By the relation between $Q^{F}(\Gamma_{n},1)$ and Padovan sequence, we have a consequence on Padovan sequence.

Corollary 2.6**.**

The generating function of Padovan sequence $\{p_{n}\}$ is

[TABLE]

Proof**.**

Combining Corollary 2.4 and Theorem 2.5,

[TABLE]

which completes the proof. $\square$

Let $[x^{n}]$ denote the coefficient extraction operator, that is, $[x^{n}]g(x)$ denotes the coefficient of $x^{n}$ in the power series expansion of $g(x)$ [11]. A perfectly obvious property of this symbol, which we will use repeatedly, is $[x^{n}]\{x^{m}g(x)\}=[x^{n-m}]g(x)$.

Expanding the generating function of $Q^{F}(\Gamma_{n},x)$ into power series, we have the formula of $q^{F}_{k}(\Gamma_{n})$.

Theorem 2.7**.**

Let $\binom{n}{k}=0$ if $n\notin\mathbb{Z}$. For $n\geq 0$,

[TABLE]

In detail, for $n\geq 0$,

[TABLE]

Proof**.**

By Theorem 2.5, we have

[TABLE]

namely

[TABLE]

Thus

[TABLE]

$\square$

The coefficients of the first few $Q^{F}(\Gamma_{n},x)$ listed are as shown in Table 1. It is not difficult to see that these coefficients are corresponding to the binomial coefficient. Furthermore, we have more general conclusions on $q^{F}_{k}(\Gamma_{n})$ and expressions of the Padovan sequence.

Corollary 2.8**.**

*The number of terms with nonzero coefficients of $Q^{F}(\Gamma_{n},x)$ is $\lfloor\frac{n+4}{3}\rfloor$. *

Corollary 2.9**.**

For $n\geq 0$ and $0\leq k\leq\lfloor\frac{n}{2}\rfloor$,

[TABLE]

and

[TABLE]

In addition, the $n$-th Padovan number is calculated by

[TABLE]

Proof**.**

It is immediate that $q^{F}_{k}(\Gamma_{n-k})$ and $q^{F}_{k}(\Gamma_{n+2k})$ by Theorem 2.7.

And by Corollary 2.4,

[TABLE]

The proof is completed. $\square$

Corollary 2.10**.**

For $n\geq 0$,

[TABLE]

and

[TABLE]

3 Optimal cube factors of matchable Lucas cubes

The Lucas sequence $\{L_{n}\}$ is defined as follows: $L_{0}=2$, $L_{1}=1$, and $L_{n}=L_{n-1}+L_{n-2}$ for $n\geq 2$. The Lucas triangle (see A029635 in OEIS [8]) [6] $Y$ is shown in Table 2, and the entry Lucas triangle is given by

[TABLE]

where $0\leq k\leq n$ and $\binom{-1}{-1}=1$.

The matchable Lucas cubes are introduced by Wang et al. [10] as follows: $|\Omega_{0}|=1$, $\Omega_{1}=P_{2}$, $\Omega_{2}=P_{3}$, $\Omega_{3}=P_{4}$, and the structure of $\Omega_{n}$ for $n\geq 5$ is shown as in Figure 2, where $P_{n}$ is the path with $n$ vertices. Furthermore, the first eight matchable Lucas cubes are shown as in Figure 3 [10].

The proof of those results are quite similar to that given earlier for Fibonacci cubes and so is omitted.

Proposition 3.1**.**

For $n\geq 5$,

[TABLE]

We list the first few of $Q^{F}(\Omega_{n},x)$ as follows.

[TABLE]

Obviously, Proposition 3.1 yields the recurrence relation of $Q^{F}(\Omega_{n},x)$.

Proposition 3.2**.**

For $n\geq 5$,

[TABLE]

Consider the generating function of $Q^{F}(\Omega_{n},x)$ with $x=1,2$, we have two special sequences as follows.

Corollary 3.3**.**

For $n\geq 0$, $Q^{F}(\Omega_{n},1)$ is the $n+1$-th Padovan number too, i.e.

[TABLE]

and for $n\geq 2$,

[TABLE]

In addition, from Proposition 3.2, we obtain two generating functions.

Theorem 3.4**.**

The generating functions of $Q^{F}(\Omega_{n},x)$ and $p_{n}$ are

[TABLE]

and

[TABLE]

*respectively. *

We can obtain again the generating function of Padovan sequence as same as Corollary 2.6, so omit it. In general, a tedious calculation gives coefficients of $Q^{F}(\Omega_{n},x)$ from Theorem 3.4.

Theorem 3.5**.**

Let $\binom{n}{k}=0$ if $n\notin\mathbb{Z}$. For $n\geq 2$,

[TABLE]

In detail, for $n\neq 1$ and $k\neq 0,1$,

[TABLE]

Corollary 3.6**.**

*For $n\geq 4$, the number of terms with nonzero coefficients of $Q^{F}(\Omega_{n},x)$ is $\lfloor\frac{n+5}{3}\rfloor$. *

Corollary 3.7**.**

For $n\neq 1$ and $k\neq 0,1$,

[TABLE]

And for $n\neq 1$ and $k\neq 0$,

[TABLE]

Moreover, for $n\geq 0$, the $n$-th Padovan number can be given again.

Proof**.**

Since $q^{F}_{k}(\Omega_{n-k})$ and $q^{F}_{k}(\Omega_{n+2k})$ are obvious, we need only calculate $p_{n}$. By Corollary 3.3 and Theorem 3.5,

[TABLE]

$\square$

In addition, from Yang Hui and Lucas triangles, we have a corollary.

Corollary 3.8**.**

For $n>2$,

[TABLE]

And for $n\geq 6$,

[TABLE]