The Number of All m-Independent Sets of K Disjoint 3-Regular Graphs with a Given Vertex

1. Introduction

A 3-regular graph is a special kind of undirected simple graph in which the degree of each vertex (i.e., the number of edges connected to that vertex) is 3. In other words, in such a graph, each vertex is connected to exactly 3 other vertices. Let $Q_3=(V,E)$ be a simple undirected 3-regular graph with $|V|=8$, $|E|=12$. An independent set is a subset S of V, such that any two vertices in S are not adjacent. A maximum independent set is an independent set of maximal size; its size is denoted by $\alpha \left(G\right)$. We say that an m-independent set means that this is an independent set and it has size m.

In 1965, Moon and Moser described in their paper [14] all n-vertex graphs with the maximum number of maximal independent sets. Later, the same problem was also solved for various graph classes, such as trees [2,4,9,13] connected graphs [3,5], forests and graphs with at most one cycle [2], triangle-free graphs [6], graphs with at most r cycles [7,11], and unicyclic connected graph [12]. Graphs with the second [8] and the third [10] largest number of maximal independents sets were also described.

In this paper, we described the graph consisting of K disconnected graph $Q_3$ denoted as $K{Q}_3$, where K can be taken from $\{1,2,\dots ,k\}$ and k is also a positive integer. Under this set of graphs, we consider its number of certain m-independent sets, where m can be taken from $\{1,2,\dots ,4K\}$. However, we will first give formulas to compute the number of m-independent sets of $k{Q}_3$, where m can be obtained from $\{1,2,\dots ,4k\}$, since $K=k$. We will partition the range of m. In the first case, m is taken from $\{1,2,\dots ,k\}$; in the second case, m is taken from $\{k+1,\dots ,4k\}$. We also do the same for the case of $K{Q}_3$, K can only be taken from $\{1,2,\dots ,k-1\}$ obtained.

2. Some Definitions and Notations

Let p is a positive integer, it means that in selecting the independent sets, we obtain points all in these arbitrary p graphs, i.e., take points in these p graphs to m. We know that for graph $Q_3$, the maximum number of points in its independent set is 4, i.e., $\alpha \left(Q_3\right)=4$. Therefore, for K disjoint graphs $Q_3$ (see Figure 1), the value of m can be obtained in $[1,4K]$. When $K=k$, the value of m is obtained in $[1,4k]$, and when K belongs to some number in $[1,k-1]$, then the value of m can be obtained in $[1,4K]$. Therefore, in this paper, we are going to consider all the cases.

For $k{Q}_3$, we use $T_m\left(k{Q}_3\right)$ to denote the number of a certain m-independent set, where the value of m is taken in [1,4k]. When $1\le m\le k$, the range of values of p is $\left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$. We need to compute the number of m-independent for some p selected graphs, denoted by $|A_p|$. Thus, $T_m\left(k{Q}_3\right)={\sum }_{p=⌈{\displaystyle \frac{m}{4}}⌉}^m|A_p|$. When $k<m\le 4k$, the range of values of p is $\left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$, in which case, $T_m\left(k{Q}_3\right)={\sum }_{p=⌈{\displaystyle \frac{m}{4}}⌉}^k|A_p|$. For $K{Q}_3$, we denote by $S_m\left(K{Q}_3\right)$ the number of its certain m-independent sets, and again we consider the range of values of p to obtain the corresponding formula.

3. Main Results

We first consider the number of certain m-independent sets of $k{Q}_3$, where $1\le m\le 4k$.

Theorem 3.1.

Let $m,k,p$ be positive integers with $1\le m\le 4k$, $T_m\left(k{Q}_3\right)$ is the number of a certain m-independent set for graph $k{Q}_3$.

(i) If $1\le m\le k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$, then

$$T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_p|+|A_m|.$$

(ii) If $k<m\le 4k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$, then

$$T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|+|A_k|.$$

where

$$\begin{array}{ccc}\hfill & |A_p|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i\sum _{All{e}_j}{\displaystyle \frac{(p-i)!}{{\prod }_{All\delta \left(e_j\right)}\delta \left(e_j\right)!}}\prod _{j=1}^{p-i}\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{e_j}}\right)+\tau \left(e_j\right)4\right)\right],\hfill & \\ \hfill & |A_m|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^m,\hfill & \\ \hfill \end{array}$$

$$\begin{array}{ccc}\hfill & |A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{m-k}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}\mathrm{for}k<m<2k,\hfill & \\ \hfill & |A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}+{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4\right)}^k\mathrm{for}m=2k,\hfill & \\ \hfill & |A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^k\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}\mathrm{for}2k<m\le 4k,\hfill & \\ \hfill & \zeta =\left[\sum _{All{\ell }_j}{\displaystyle \frac{i!}{{\prod }_{All\delta \left({\ell }_j\right)}\delta \left({\ell }_j\right)!}}\prod _{j=1}^i\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{{\ell }_j}}\right)+\tau \left({\ell }_j\right)4\right)\right].\hfill & \\ \hfill \end{array}$$

We will explain the meaning of each symbol in it.

$\left(i\right)$ℓ is a positive integer, which is the same number of points taken for some i of these arbitrary p graphs, i.e.,ℓ, and $1\le \ell \le ⌊{\displaystyle \frac{m}{p}}⌋\le 4$. (Because for each graph, the number of points selected for each graph cannot be more than 4 if the points are guaranteed to be independent, and this is related to $\alpha \left(Q_3\right)=4$).

$\left(ii\right)$$\tau (\ell )=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}\ell =2\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$

$\left(iii\right)$$e_j$ is the number of points selected for each of the remaining $p-i$ graphs under the selection of graphs in $\left(i\right)$, where $j\in [1,p-i]$, and there are $\ell <e_1\le e_2\le \cdots \le e_{p-i}\le 4$, and ${\sum }_{j=1}^{p-i}{e}_j=m-i\ell$. Thus, the summation term in $\xi$ refers to the selection of all different $e_j$ to meet this condition.

$\left(iv\right)$$\tau \left(e_j\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}{e}_j=2\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$

$\left(v\right)$$\delta \left(e_j\right)$ is the number of times an element is repeated in the $p-i$ elements $\{e_1,e_2,\dots ,e_{p-i}\}$. If some element occurs only once, at this point a certain $\delta \left(e_j\right)=1$. We need to consider all the elements. $eg:\{2,3,3,4,4,4\}$. At this point $p-i=6$, ${\prod }_{All\delta \left(e_j\right)}\delta \left(e_j\right)!=1!2!3!$.

Proof. 

We divide it into two cases and generalize to obtain $|A_m|$, $|A_k|$ and its formula individually.

Case 1. 

If $1\le m\le k$, then $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$.

• $k=1$, $m=1$,
  $T_1\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)$.
• $k=2$, $m=1,m=2$,
  $T_1\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)$;
  $T_2\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^2$.
• $k=3$, $m=1,m=2,m=3$,
  $T_1\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)$;
  $T_2\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^2$;
  $T_3\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{3}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^3$.
• $k=4$, $m=1,m=2,m=3,m=4$,
  $T_1\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)$;
  $T_2\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^2$;
  $T_3\left(k{Q}_3\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{3}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^3$;
  $T_4\left(k{Q}_3\right)={\sum }_{p=1}^4|A_p|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)·\left[\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)·2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2}{2}}\right)·{(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)}^2\right]+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{3}}\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{3}{2}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^2·(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{4}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^4$.

Thus, we can obtain $T_m\left(k{Q}_3\right)$ by simply giving the formula for $|A_p|$, where we compute $T_m$ in two subcases.

Subcase 1.1. 

$1\le m\le k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m-1\right]$.

By induction, we give the formula for $|A_p|$.

$$|A_p|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i\left(\sum _{All{e}_j}{\displaystyle \frac{(p-i)!}{{\prod }_{All\delta \left(e_j\right)}\delta \left(e_j\right)!}}\prod _{j=1}^{p-i}\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{e_j}}\right)+\tau \left(e_j\right)4\right)\right)\right]$$

Let $\xi ={\sum }_{All{e}_j}{\displaystyle \frac{(p-i)!}{{\prod }_{All\delta \left(e_j\right)}\delta \left(e_j\right)!}}{\prod }_{j=1}^{p-i}\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{e_j}}\right)+\tau \left(e_j\right)4\right)$. Note that we stipulate that $\xi =1$ when $p-i=0$ .

The case of the range of values of ℓ and i are given below. Since $\ell <e_1\le e_2\le \cdots \le e_{p-i}\le 4$, then

$$(p-i)\ell <\sum _{j=1}^{p-i}{e}_j=m-i\ell \le 4(p-i).$$

By simplifying, we get $\ell <{\displaystyle \frac{m}{p}}$, i.e. $\ell \le ⌊{\displaystyle \frac{m}{p}}⌋$. From $m-i\ell \le 4(p-i)$, we have $i\le {\displaystyle \frac{4p-m}{4-\ell }}$ and $\ell \ge {\displaystyle \frac{m-4p}{i}}+4$. We know that $\ell \le ⌊{\displaystyle \frac{m}{p}}⌋$ and $i,\ell$ are integers, thus $i\le ⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋$ and $\ell \ge ⌈{\displaystyle \frac{m-4p}{i}}⌉+4$. Similarly, let $\ell +1\le e_1\le e_2\le \cdots \le e_{p-i}\le 4$, then

$$(\ell +1)(p-i)\le m-i\ell ,$$

it can get

$$i\ge \ell p+p-m\ge 2p-m.$$

Since both p and i are positive integers, the selection of their range of values must this. So, we have $2p-m\ge 1$, then $p\ge ⌈{\displaystyle \frac{m+1}{2}}⌉$. Also, it must have ${\displaystyle \frac{m-4p}{i}}+4\ge 1$, which gives $p\le ⌊{\displaystyle \frac{m+3i}{4}}⌋$. In short, if $p\ge ⌈{\displaystyle \frac{m+1}{2}}⌉$, then $i\in \left[2p-m,⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋\right]$, otherwise, $i\in \left[1,⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋\right]$; when i is determined, for any i, if $p\le ⌊{\displaystyle \frac{m+3i}{4}}⌋$, then $\ell \in \left[⌈{\displaystyle \frac{m-4p}{i}}⌉+4,⌊{\displaystyle \frac{m}{p}}⌋\right]$, otherwise $\ell \in \left[1,⌊{\displaystyle \frac{m}{p}}⌋\right]$.

Since $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m-1\right]$, we can obtain the equation

$$\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_p|=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}$$

Note that if $4p-m=0$, i.e., $p={\displaystyle \frac{m}{4}}$, then $|A_p|=|A_{{\displaystyle \frac{m}{4}}}|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\frac{m}{4}}}\right){\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{4}}\right)\right)}^{{\displaystyle \frac{m}{4}}}$, in which case only $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉+1,m-1\right]$ is considered, otherwise $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m-1\right]$.

Subcase 1.2. 

$1\le m\le k$, $p=m$.

We can obtain the equation

$$|A_m|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^m$$

for $m=k$, we have

$$|A_k|={\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^k$$

Hence, when $1\le m\le k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$, we obtain the number of certain m-independent sets of $k{Q}_3$ from the equation

$$\begin{array}{cc}\hfill T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^m|A_p|=& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_p|+|A_m|\hfill \\ \hfill =& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}\hfill \\ & +\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^m\hfill \end{array}$$

Case 2. 

If $k<m\le 4k$, then $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$.

We explain why it is $m\le 4k$, since $⌈{\displaystyle \frac{m}{4}}⌉\le k$, i.e., $k\ge {\displaystyle \frac{m}{4}}$, so $m\le 4k$ . If $m=4k+1$,it means that at least one of the independent numbers in graph $Q_3$ is 5, but $\alpha \left(Q_3\right)=4$, so $m\le 4k$.

Similarly, we only need to compute $T_m$, which at this point is $T_m\left(k{Q}_3\right)={\sum }_{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|+|A_k|$, and we also compute them in two subcases.

Subcase 2.1. 

$k<m\le 4k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k-1\right]$.

$$\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}$$

Subcase 2.2. 

$k<m\le 4k$, $p=k$.

For this case of $p=k$, which is different from $p=m$ in Case 1, then for different values of m, we consider the results of $|A_k|$ by dividing them into three subcases.

Subcase 2.2.1. 

$k<m<2k$.

$$|A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{m-k}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\left[\sum _{All{\ell }_j}{\displaystyle \frac{i!}{{\prod }_{All\delta \left({\ell }_j\right)}\delta \left({\ell }_j\right)!}}\prod _{j=1}^i\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{{\ell }_j}}\right)+\tau \left({\ell }_j\right)4\right)\right]\right\}$$

We give a brief explanation of the notation in Eq, and give how the range of values of i is obtained.

$\left(i\right)$i is a positive integer which means that for any k graphs of which there are i in the set of points chosen, the number of points chosen for each graph is related to the fact that only one point is chosen for each of the remaining $k-i$ graphs. The number of points selected for each of these i graphs is at least 2, but not more than the maximum number of independents of the graphs. $\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)$ is the fact that for i graphs, there are $\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)$ ways of taking the graphs.

$\left(ii\right)$${\ell }_j$ is the number of points selected for each graph for these i graphs and all the points are combined in $(m-(k-i\left)\right)$ independent sets, where $j\in [1,i]$, and there are $1<{\ell }_1\le {\ell }_2\le \cdots \le {\ell }_i\le 4$, and ${\sum }_{j=1}^i{\ell }_j=m-(k-i)$. Because for $(k-i)$ graphs, only one point is selected for each graph, while for m independent set, $(m-(k-i\left)\right)$ independent sets are needed for these i graphs to select the point.

$\left(iii\right)$$\tau \left({\ell }_j\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}{\ell }_j=2\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}\right.$

$\left(iv\right)$ Since the definition of $\delta \left({\ell }_j\right)$ is similar to that of $\delta \left(e_j\right)$, we do not redefine it. Similarly the summation term on ${\ell }_j$ in Eq. means to pick all different ${\ell }_j$ to this condition.

Since $1<{\ell }_1\le {\ell }_2\le \cdots \le {\ell }_i\le 4$, we have

$$i<\sum _{j=1}^i{\ell }_j=m-k+i\le 4i,$$

which by the inequality that follows gives $i\ge {\displaystyle \frac{m-k}{3}}$, and i must be a positive integer, hence $i\ge ⌈{\displaystyle \frac{m-k}{3}}⌉$. We let $2\le {\ell }_1\le {\ell }_2\le \cdots \le {\ell }_i\le 4$, in which case we have

$$2i\le m-k+i,$$

so $i\le m-k$. Since $i<k$, hence $m-k<k$, i.e. $m<2k$.

So this formula holds under the condition that $k<m<2k$ and the range of values of i is $\left[⌈{\displaystyle \frac{m-k}{3}}⌉,m-k\right]$.

Subcase 2.2.2. 

$m=2k$.

$$|A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\left[\sum _{All{\ell }_j}{\displaystyle \frac{i!}{{\prod }_{All\delta \left({\ell }_j\right)}\delta \left({\ell }_j\right)!}}\prod _{j=1}^i\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{{\ell }_j}}\right)+\tau \left({\ell }_j\right)4\right)\right]\right\}+{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4\right)}^k$$

Subcase 2.2.3. 

$2k<m\le 4k$.

The range of m at this point is very large and complex to compute, so we consider a simple computation. Since i cannot exceed k, the range of i is $\left[⌈{\displaystyle \frac{m-k}{3}}⌉,k\right]$. Hence

$$|A_k|=\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^k\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\left[\sum _{All{\ell }_j}{\displaystyle \frac{i!}{{\prod }_{All\delta \left({\ell }_j\right)}\delta \left({\ell }_j\right)!}}\prod _{j=1}^i\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{{\ell }_j}}\right)+\tau \left({\ell }_j\right)4\right)\right]\right\}$$

Set $\zeta ={\sum }_{All{\ell }_j}{\displaystyle \frac{i!}{{\prod }_{All\delta \left({\ell }_j\right)}\delta \left({\ell }_j\right)!}}{\prod }_{j=1}^i\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{{\ell }_j}}\right)+\tau \left({\ell }_j\right)4\right)$, the number of points selected for these i graphs goes to the $(m-(k-i\left)\right)$ independent sets is more difficult, so just simplify this summation formula.

We know that $1<{\ell }_1\le {\ell }_2\le \cdots \le {\ell }_i\le 4$, i.e., $2\le {\ell }_1\le {\ell }_2\le \cdots \le {\ell }_i\le 4$ holds, so the number obtained by ${\ell }_j$ is chosen in [2,4], in order to meet

$$\sum _{j=1}^i{\ell }_j=m-k+i,$$

we first consider certain ${\ell }_j$, where $j\in [1,i]$, we set the first n elements are 2, i.e., ${\sum }_{j=1}^n{\ell }_j=2n$, at this point, the remaining $i-n$ elements need to be

$$\sum _{j=n+1}^i{\ell }_j=m-k+i-2n.$$

At this point there are $i-n$ elements that are not 2, and for the selection of n, which has a range, we consider the range of values it takes. Because there are $i-n$ elements that are not 2, then the minimum of these $i-n$ elements is 3, and the maximum cannot be more than 4, so the condition

$$3(i-n)\le m-k+i-2n\le 4(i-n),$$

is needed. And the inequality $3(i-n)\le m-k+i-2n$ on the left side gives $n\ge 2i-m+k$; and the inequality $m-k+i-2n\le 4(i-n)$ on the right side gives $n\le {\displaystyle \frac{3i-m+k}{2}}$, and because n is a positive integer, so $n\le ⌊{\displaystyle \frac{3i-m+k}{2}}⌋$.

Since n is a positive integer, it stands to reason that $2i-m+k\ge 0$, but if it is less than 0, then n is in the range $\left[0,⌊{\displaystyle \frac{3i-m+k}{2}}⌋\right]$, otherwise $n\in \left[2i-m+k,⌊{\displaystyle \frac{3i-m+k}{2}}⌋\right]$.

Again, n is not more than k, so it can obtain

$${\displaystyle \frac{3i-m+k}{2}}\le {\displaystyle \frac{3k-m+k}{2}}<k$$

from $i\le k$, and the result is $m>2k$. Thus the range of m is $(2k,4k]$, and as m increases, the maximum value of n is not more than k.

Hence, when $k<m<2k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$, we obtain the number of certain m-independent sets of $k{Q}_3$ from the equation

$$\begin{array}{cc}\hfill T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^k|A_p|=& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|+|A_k|\hfill \\ \hfill =& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}\hfill \\ & +\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{m-k}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}.\hfill \end{array}$$

when $m=2k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$,

$$\begin{array}{cc}\hfill T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^k|A_p|=& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|+|A_k|\hfill \\ \hfill =& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}\hfill \\ & +\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}+{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4\right)}^k.\hfill \end{array}$$

when $2k<m\le 4k$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$,

$$\begin{array}{cc}\hfill T_m\left(k{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^k|A_p|=& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}|A_p|+|A_k|\hfill \\ \hfill =& \sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\xi \right]\right\}\hfill \\ & +\sum _{i=⌈{\displaystyle \frac{m-k}{3}}⌉}^k\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^{k-i}·\zeta \right\}.\hfill \end{array}$$

While a portion of the graphs select only one point, or only two points as part of their independent sets in the process of selecting points, the remaining graphs can only select 3 or 4 points for the whole m independent sets, which is still complicated and therefore still to be solved.

In fact, for the case of $n=0$, let both ends of the equation of n are 0, that is,

$$2i-m+k=0,{\displaystyle \frac{3i-m+k}{2}}=0.$$

the extremes of both sides of n is actually for i graphs are all selected 3 points (the left side of the equation), or all selected 4 points (the right side of the equation) to m-independence. Therefore, in the process of calculation, if i satisfies the condition of the equation, the selection of ${\ell }_j$ becomes more straightforward and favorable to get the result.

Next, we consider the number of certain m-independent sets in $K{Q}_3$, where K can be obtained from $[1,k-1]$, and we explore its connection with the number of corresponding m-independent sets of $k{Q}_3$, for what deformation of the previously discussed formulas yields the new needed formulae.

Before the discussion of the situation, we need to compare how p changes for $K{Q}_3$ when m is taken from a different range than when it is taken from the same m for $k{Q}_3$. Since our previous formulas have been based on the final quantity based on the value of p taken, we consider the quantity of the m-independent set of $K{Q}_3$, i.e., $S_m\left(K{Q}_3\right)$, in the same way. Let’s make a distinction here by setting the previous p to $p_k$ and the p here to $p_K$.

Now we have

$$\left\{\begin{array}{cc}1\le m\le k,\hfill & p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]\hfill \\ k<m\le 4k,\hfill & p_k\in \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]\hfill \end{array}\right.$$

(1)

similarly,

$$\left\{\begin{array}{cc}1\le m\le K,\hfill & p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]\hfill \\ K<m\le 4K,\hfill & p_K\in \left[⌈{\displaystyle \frac{m}{4}}⌉,K\right]\hfill \end{array}\right.$$

(2)

where K can be obtained from $[1,k-1]$. Since K is smaller than k, this shows that $[1,K]\subseteq [1,k]$, then for $K{Q}_3$, when picking m, the same m is picked as in $k{Q}_3$, and their p belong to the same range, i.e., $\left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$. Therefore when calculating the number of m-independent sets in $K{Q}_3$, where $1\le m\le K$, we only need to compare it with the same m in $k{Q}_3$, which has the number of m-independent sets, so as to be able to obtain the relationship between them.

For $K<m\le 4K$, we can divide it into

$$\left\{\begin{array}{cc}K<m_1\le k,\hfill & p_K\in \left[⌈{\displaystyle \frac{m_1}{4}}⌉,K\right]\hfill \\ k<m_2\le 4K,\hfill & p_K\in \left[⌈{\displaystyle \frac{m_2}{4}}⌉,K\right]\hfill \end{array}\right.$$

(3)

we can see that the range of values of $m_2$ at this point has an inclusive relationship with the range of values of m in the second case of $k{Q}_3$, including the range of values of p, i.e. $(k,4K]\subseteq (k,4k]$ and $\left[⌈{\displaystyle \frac{m_2}{4}}⌉,K\right]\subseteq \left[⌈{\displaystyle \frac{m}{4}}⌉,k\right]$, where $m=m_2$ . Thus in calculating the number of $m_2$-independent sets in $K{Q}_3$, where $k<m_2\le 4K$. Again we only need to compare the same m in $k{Q}_3$ with the formula for solving for the number of m-independent sets in it.

Theorem 3.2.

Let $m,k,K,p$ be positive integers with $1\le m\le 4K$, where $1\le K\le k-1$. $S_m\left(k{Q}_3\right)$ is the number of a certain m-independent set for graph $K{Q}_3$.

(i) If $1\le m\le K$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$, then

$$\begin{array}{cc}\hfill S_m\left(K{Q}_3\right)& =\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_{p_K}|+|A_{m_K}|\hfill \\ \hfill & =\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|+{\displaystyle \frac{(k-m)!K!}{(K-m)!k!}}|A_m|.\hfill \end{array}$$

(ii) If $K<m\le 4K$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,K\right]$, then

$$S_m\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m}{4}}⌉}^K|A_{p_K}|=\sum _{p_K=⌈{\displaystyle \frac{m}{4}}⌉}^K{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

where $|A_p|$ and $|A_m|$ are the corresponding formulas of Theorem 3.1.

Proof. 

We still split into two cases according to (3.2) and consider the following case. □

Case 1. 

If $1\le m\le K$, where $K\in [1,k-1]$, then $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m\right]$.

At this point there is the formula

$$\begin{array}{c}\hfill S_m\left(K{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^m|A_p|=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_p|+|A_m|\end{array}$$

To distinguish this from the previous formula, we let ${\sum }_{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_p|={\sum }_{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_{p_K}|$ and $|A_m|=|A_{m_K}|$, thus

$$\begin{array}{c}\hfill S_m\left(K{Q}_3\right)=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^m|A_p|=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_{p_K}|+|A_{m_K}|\end{array}$$

Subcase 1.1. 

$1\le m\le K$, $p\in \left[⌈{\displaystyle \frac{m}{4}}⌉,m-1\right]$.

Since

$$|A_{p_K}|=\left({\displaystyle \genfrac{}{}{0pt}{}{K}{p}}\right)\left[\sum _{\ell =1or⌈{\displaystyle \frac{m-4p}{i}}⌉+4}^{⌊{\displaystyle \frac{m}{p}}⌋}\sum _{i=1or2p-m}^{⌊{\displaystyle \frac{4p-m}{4-⌊\frac{m}{p}⌋}}⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{\ell }}\right)+\tau (\ell )4\right)}^i·\zeta \right].$$

Set $\zeta ={\sum }_{All{e}_j}{\displaystyle \frac{(p-i)!}{{\prod }_{All\delta \left(e_j\right)}\delta \left(e_j\right)!}}{\prod }_{j=1}^{p-i}\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{e_j}}\right)+\tau \left(e_j\right)4\right)$. For finding the number of m-independent sets of $k{Q}_3$, the $|A_p|$ we are seeking is compared to it, then we will obtain

$${\displaystyle \frac{|A_p|}{|A_{p_K}|}}={\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{k}{p}\right)}{\left(\genfrac{}{}{0pt}{}{K}{p}\right)}}={\displaystyle \frac{(K-p)!k!}{(k-p)!K!}},$$

thus

$$|A_{p_K}|={\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

Then we can get

$$\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_{p_K}|=\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

Subcase 1.2. 

$1\le m\le K$, $p=m$.

At this point, we consider $|A_{m_K}|$, it has

$$|A_{m_K}|=\left({\displaystyle \genfrac{}{}{0pt}{}{K}{m}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^m,$$

similarly, we know

$$|A_m|=\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)·{\left(2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right)}^m,$$

they are compared to get

$${\displaystyle \frac{|A_m|}{|A_{m_K}|}}={\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{k}{m}\right)}{\left(\genfrac{}{}{0pt}{}{K}{m}\right)}}={\displaystyle \frac{(K-m)!k!}{(k-m)!K!}},$$

then we can get

$$|A_{m_K}|={\displaystyle \frac{(k-m)!K!}{(K-m)!k!}}|A_m|,$$

hence,

$$\begin{array}{cc}\hfill S_m\left(K{Q}_3\right)& =\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}|A_{p_K}|+|A_{m_K}|\hfill \\ \hfill & =\sum _{p=⌈{\displaystyle \frac{m}{4}}⌉}^{m-1}{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|+{\displaystyle \frac{(k-m)!K!}{(K-m)!k!}}|A_m|.\hfill \end{array}$$

Case 2. 

If $K<m\le 4K$, where $K\in [1,k-1]$, then $p_K\in \left[⌈{\displaystyle \frac{m}{4}}⌉,K\right]$.

We divide the range of values of m.

Subcase 2.1. 

$K<m_1\le k$, $p_K\in \left[⌈{\displaystyle \frac{m_1}{4}}⌉,K\right]$.

We know that the range of values of $m_1$ at this point is actually a subset of the range of values of m of Case 1. in $k{Q}_3$, at this point we have $(K,k]\subseteq (1,k]$, and we find that when we take the same number, the ranges of values of their corresponding p are also a containment relation, which means that when we calculate the number of a certain $m_1$-independent set, we only need to calculate a certain part of p about the same m of $k{Q}_3$. In fact, p is taken up to K.

Since

$$S_{m_1}\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m_1}{4}}⌉}^K|A_{p_K}|,$$

we know

$$|A_{p_K}|={\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|,$$

hence,

$$S_{m_1}\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m_1}{4}}⌉}^K|A_{p_K}|=\sum _{p_K=⌈{\displaystyle \frac{m_1}{4}}⌉}^K{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

Subcase 2.2. 

$k<m_2\le 4K$, $p_K\in \left[⌈{\displaystyle \frac{m_2}{4}}⌉,K\right]$.

For the range of values of $m_2$ , we can easily interrelate with the range of values of m of case 2. in the previous $k{Q}_3$, since at this point we have $(k,4K]\subseteq (k,4k]$. And, for them to take the same m, the range of $p_K$ in $K{Q}_3$ happens to be a subset of the range of p in $k{Q}_3$, and the range of $p_K$ happens to be in the range of p up to the element K. Therefore, we do not need to consider the case where $p=k$, but only subcase2.1.

Similarly,

$$S_{m_2}\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m_2}{4}}⌉}^K|A_{p_K}|.$$

we know

$$|A_{p_K}|={\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

hence,

$$S_{m_2}\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m_2}{4}}⌉}^K|A_{p_K}|=\sum _{p_K=⌈{\displaystyle \frac{m_2}{4}}⌉}^K{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

This shows that for Case 2., regardless of the value of m, we have the formula

$$S_m\left(K{Q}_3\right)=\sum _{p_K=⌈{\displaystyle \frac{m}{4}}⌉}^K|A_{p_K}|=\sum _{p_K=⌈{\displaystyle \frac{m}{4}}⌉}^K{\displaystyle \frac{(k-p)!K!}{(K-p)!k!}}|A_p|.$$

holds as long as $K<m\le 4K$, where $K\in [1,k-1]$.

4. Conclusions

In this paper, we obtain for K disjoint regular graphs $Q_3$ with 8 vertices, defined as $K{Q}_3$, the formula for the number of certain m-independent sets about it, where $1\le K\le k$ and $1\le m\le 4K$. We first explore the number of certain m-independent sets about $k{Q}_3$, where $1\le m\le 4k$. By taking different values of m, we make a case-by-case discussion and obtain the formula for its correlation. Under this condition, we again considered the case of less than k disjoint graphs $Q_3$, i.e., $K{Q}_3$, where $1\le K\le k-1$, and obtained formulas for the number of certain m-independent sets of them. We obtained the simple formula for $K{Q}_3$ by finding the relation between $k{Q}_3$ and the number of the same m-independent set of $K{Q}_3$.