# On SD-Prime Labelings of Some One Point Unions of Gears

^{1}

^{2}

^{3}

^{*}

## Abstract

**:**

## 1. Introduction

**Definition**

**1.**

**Definition**

**2.**

**Corollary**

**1.**

**Corollary**

**2.**

**Theorem**

**1**

**Theorem**

**2**

**Theorem**

**3**

## 2. Preliminary

**Theorem**

**4.**

**Proof.**

**Corollary**

**3.**

**Proof.**

**Corollary**

**4.**

**Proof.**

## 3. Merging of Type 1 Vertices

**Lemma**

**1.**

- (1)
- each pair of consecutive terms in S is of different parity and coprime,
- (2)
- each pair of consecutive terms in T is of different parity and coprime.

**Proof.**

- (1)
- Clearly, (1) is true when the two numbers come from the same subsequence ${s}_{l}$ for some l.Suppose $l\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. The last term of ${s}_{l-1}$ is $6(l-1)+2$ and the first term of ${s}_{l}$ is $6l+1$. Now, $(6l-4,6l+1)=(6l-4,5)=(l-4,5)=1$.Suppose $l\equiv 2,3,4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. The last term of ${s}_{l-1}$ is $6(l-1)+4$ and the first term of ${s}_{l}$ is $6l+1$. Clearly, $(6l-2,6l+1)=(6l-2,3)=(1,3)=1$.Suppose $l\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. The last term of ${s}_{l-1}$ is $6(l-1)+2$ and the first term of ${s}_{l}$ is $6l+5$. Now, $(6l-4,6l+5)=(6l-4,9)$. Since $(6l-4,3)=1$, $(6l-4,9)=1$.
- (2)
- Clearly, (2) is true when the two numbers come from the same subsequence ${t}_{l}$ for some l.Suppose $l\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. The last term of ${t}_{l-1}$ is $6(l-1)+6$ and first term of ${t}_{l}$ is $6l+5$. Clearly, $(6l,6l+5)=(6l,5)=(l,5)=1$.Suppose $l\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. The last term of ${t}_{l-1}$ is $6(l-1)+6$ and first term of ${t}_{l}$ is $6l+1$. Clearly, $(6l,6l+1)=(6l,1)=1$.

**Theorem**

**5.**

**Proof.**

- Case 1:
- Suppose ${n}_{1}=2k$ for some $k\ge 2$. We let $f\left({c}_{1}\right)=3$, $f\left({c}_{2}\right)=1$, $f\left({u}_{1}\right)=7=f\left({v}_{1}\right)$, $f\left({u}_{2}\right)=4$, $f\left({u}_{2{n}_{1}-1}\right)=9$, $f\left({u}_{2{n}_{1}}\right)=8$, $f\left({v}_{2}\right)=10$, $f\left({v}_{3}\right)=11$, $f\left({v}_{4}\right)=12$, $f\left({v}_{2{n}_{2}-2}\right)=2$, $f\left({v}_{2{n}_{2}-1}\right)=5$ and $f\left({v}_{2{n}_{2}}\right)=6$ (see Figure 2a).Now, there are $4k-4$ and $2{n}_{2}-7$ uncolored vertices in ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$, respectively. We label consecutively the vertices from ${u}_{3}$ to ${u}_{2{n}_{1}-2}$ by the first $4k-4$ terms of S and from ${v}_{5}$ to ${v}_{{n}_{1}+2}$ by the first $2k-2$ terms of T, respectively, where S and T are the sequences in Lemma 1.Note that $(f\left({u}_{2{n}_{1}-2}\right),f\left({u}_{2{n}_{1}-1}\right))=(6k+4,9)=1$ or $(f\left({u}_{2{n}_{1}-2}\right),f\left({u}_{2{n}_{1}-1}\right))=(6k+2,9)=1$, and the vertex labeled by $6l+3$ is not adjacent to the core ${c}_{1}$ for every l. So, f satisfies the requirement of SD-prime labeling for $G\left({n}_{1}\right)$. Up to now, we have used labels from 1 to $6k+6=3{n}_{1}+6$. Finally, we label the unlabeled vertices from ${v}_{{n}_{1}+3}$ to ${v}_{2{n}_{2}-3}$ by $3{n}_{1}+7$ to $2{n}_{1}+2{n}_{2}+1$ consecutively, if $2{n}_{2}\ge {n}_{1}+6$. Note that, $(f\left({v}_{2{n}_{2}-3}\right),f\left({v}_{2{n}_{2}-2}\right))=(2{n}_{1}+2{n}_{2}+1,2)=1$. By Theorem 3, we see that f is an SD-prime labeling of G.For the remaining case, i.e., $2{n}_{2}\le {n}_{1}+5$. Since ${n}_{2}\ge {n}_{1}$, ${n}_{1}\le 5$. Only ${n}_{2}={n}_{1}=4$ is a case. For this case, we have the labeling:
- Case 2:
- Suppose ${n}_{1}=2k+1$ for some $k\ge 1$. We let $f\left({c}_{1}\right)=3$, $f\left({c}_{2}\right)=1$, $f\left({u}_{1}\right)=7=f\left({v}_{1}\right)$, $f\left({u}_{2}\right)=8$, $f\left({u}_{3}\right)=5$, $f\left({u}_{4}\right)=4$, $f\left({u}_{2{n}_{1}-1}\right)=9$, $f\left({u}_{2{n}_{1}}\right)=10$, $f\left({v}_{2}\right)=6$, $f\left({v}_{3}\right)=11$, $f\left({v}_{4}\right)=12$ and $f\left({v}_{2{n}_{2}}\right)=2$ (see Figure 2b). When $k=1$, the graph $G\left({n}_{1}\right)$ is labeled completely. For this case, we simply label the vertices from ${v}_{5}$ to ${v}_{2{n}_{2}-1}$ by 13 to $2{n}_{2}+7$ consecutively.So, we assume that $k\ge 2$. Similar to Case 1, we may label the vertices from ${u}_{5}$ to ${u}_{2{n}_{1}-2}$ and from ${v}_{5}$ to ${v}_{2{n}_{2}-1}$ using the labels in $[13,2{n}_{1}+2{n}_{2}+1]$.

**Theorem**

**6.**

**Proof.**

**Theorem**

**7.**

**Proof.**

- Case 1:
- Suppose ${n}_{1}=2{k}_{1}$ and ${n}_{2}=2{k}_{2}$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 2$. The initial labeling for some vertices are listed as follows:When ${k}_{1}+{k}_{2}=3$, then ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ are already labeled. For this case, the set of unassigned labels is $[13,24]\cup [29,78]\cup [83,2{n}_{3}+13]$. Here, we may simply label the remaining vertices of ${G}_{{n}_{3}}$ consecutively.Suppose ${k}_{1}+{k}_{2}\ge 4$. We label consecutively the vertices from ${v}_{1,3}$ to ${v}_{1,2{n}_{1}-2}$ and then ${v}_{2,7}$ to ${v}_{2,2{n}_{2}-2}$ by the first $4({k}_{1}+{k}_{2}-3)$ terms of ${S}_{1}$ and from ${v}_{3,5}$ to ${v}_{3,{n}_{1}+{n}_{2}-4}$ by the first $2({k}_{1}+{k}_{2}-4)$ terms of T, respectively, where T is the sequence defined in Lemma 1. Note that, when ${k}_{1}+{k}_{2}=4$, no vertices of ${G}_{{n}_{3}}$ are labeled at this step.After this step, let R be the set of remaining labels arranged in natural order. Namely,$$R=\left\{\begin{array}{cc}[17,2({n}_{1}+{n}_{2}+{n}_{3})+1]\backslash {s}_{13}\hfill & \mathrm{if}\text{}{k}_{1}+{k}_{2}=4;\hfill \\ {t}_{{k}_{1}+{k}_{2}-2}\cup {t}_{{k}_{1}+{k}_{2}-1}\cup \hfill & \\ \phantom{\rule{2.em}{0ex}}\cup [6{k}_{1}+6{k}_{2}+1,2({n}_{1}+{n}_{2}+{n}_{3})+1]\backslash {s}_{13}\hfill & \mathrm{if}\text{}5\le {k}_{1}+{k}_{2}\le 12;\hfill \\ {t}_{{k}_{1}+{k}_{2}-2}\cup {t}_{{k}_{1}+{k}_{2}-1}\cup {t}_{{k}_{1}+{k}_{2}}\cup \hfill & \\ \phantom{\rule{2.em}{0ex}}\cup [6{k}_{1}+6{k}_{2}+7,2({n}_{1}+{n}_{2}+{n}_{3})+1]\hfill & \mathrm{if}\text{}{k}_{1}+{k}_{2}\ge 13.\hfill \end{array}\right.$$By part (2) of Lemma 1, we see that each pair of consecutive terms in R is of different parity and coprime.So, we may label the vertices from ${v}_{3,{n}_{1}+{n}_{2}-5}$ to ${v}_{3,2{n}_{3}-3}$.Hence, we obtain an SD-prime labeling for this case.
- Case 2:
- Suppose ${n}_{1}=2{k}_{1}$ and ${n}_{2}=2{k}_{2}+1$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 1$.If ${k}_{2}=1$, then ${n}_{1}=2$ and ${n}_{2}=3$. The labeling is defined as follows:Suppose ${k}_{2}\ge 2$. The initial labeling for some vertices are listed as follows:Similar to Case 1, we label consecutively the vertices from ${v}_{1,3}$ to ${v}_{1,2{n}_{1}-2}$ and then ${v}_{2,5}$ to ${v}_{2,2{n}_{2}-2}$ by the first $4({k}_{1}+{k}_{2}-2)$ terms of ${S}_{1}$ and from ${v}_{3,7}$ to ${v}_{3,{n}_{1}+{n}_{2}-1}$ by the first $2({k}_{1}+{k}_{2}-3)$ terms of T, respectively. The rest is similar to Case 1.
- Case 3:
- Suppose ${n}_{1}=2{k}_{1}+1$ and ${n}_{2}=2{k}_{2}$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 2$. The initial labeling for some vertices are listed as follows:This case is similar to Case 2.
- Case 4:
- Suppose ${n}_{1}=2{k}_{1}+1$ and ${n}_{2}=2{k}_{2}+1$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 1$. The initial labeling for some vertices are listed as follows:Similar to Case 1, we label consecutively the vertices from ${v}_{1,5}$ to ${v}_{1,2{n}_{1}-2}$ and then ${v}_{2,5}$ to ${v}_{2,2{n}_{2}-2}$ by the first $4({k}_{1}+{k}_{2}-2)$ terms of ${S}_{1}$ and from ${v}_{3,5}$ to ${v}_{3,{n}_{1}+{n}_{2}-4}$ by the first $2({k}_{1}+{k}_{2}-3)$ terms of T, respectively. The rest is similar to Case 1.

**Example**

**1.**

**Example**

**2.**

**Corollary**

**5.**

**Proof.**

**Corollary**

**6.**

**Theorem**

**8.**

**Proof.**

- Case 1:
- ${n}_{1}+{n}_{2}=6$. We have $3\le {n}_{3}\le 33$.When ${n}_{3}=3$. This implies that ${n}_{1}={n}_{2}=3$. For this case, following is an SD-prime labeling for $G(3,3,3)$:Now, we assume that ${n}_{3}\ge 4$.When $({n}_{1},{n}_{2})=(3,3)$. We label ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ as follows:When $({n}_{1},{n}_{2})=(2,4)$. We label ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ as follows:The initial labeling for some vertices of ${G}_{{n}_{3}}$ are as follows:The unassigned labels in $[22,2{n}_{3}+13]$, if any, can be labeled from ${v}_{3,8}$ to ${v}_{3,2{n}_{3}-1}$ consecutively.
- Case 2:
- ${n}_{1}+{n}_{2}=7$.When $({n}_{1},{n}_{2})=(3,4)$.The labeling for ${G}_{{n}_{2}}$ is as in Case 1.When $({n}_{1},{n}_{2})=(2,5)$. We label ${G}_{{n}_{1}}$ as in Case 1 and label ${G}_{{n}_{2}}$ asThe initial labeling for ${G}_{{n}_{3}}$ is as in Case 1. The unassigned labels in $[24,2{n}_{3}+15]$ can be labeled from ${v}_{3,8}$ to ${v}_{3,2{n}_{3}-1}$ consecutively.
- Case 3:
- ${n}_{1}+{n}_{2}=8$.Suppose $({n}_{1},{n}_{2})=(4,4)$. If ${n}_{3}=4$, we label ${G}_{{n}_{2}}$ as in Case 1.The labeling for ${G}_{{n}_{1}}$ and ${G}_{{n}_{3}}$ are as follows:If ${n}_{3}\ge 5$, the labeling for ${G}_{{n}_{1}}$ is as follows:The labeling for ${G}_{{n}_{2}}$ is as in Case 1.When $({n}_{1},{n}_{2})=(3,5)$. We label ${G}_{{n}_{1}}$ asThe labeling for ${G}_{{n}_{2}}$ is as in Case 2.When $({n}_{1},{n}_{2})=(2,6)$. We label ${G}_{{n}_{1}}$ as in Case 1 and label ${G}_{{n}_{2}}$ asThe initial labeling for ${G}_{{n}_{3}}$ is as follows:The unassigned labels in $[28,2{n}_{3}+17]$, if any, can be labeled from ${v}_{3,10}$ to ${v}_{3,2{n}_{3}-1}$ consecutively.

**Theorem**

**9.**

**Proof.**

- Case 1:
- Suppose ${n}_{1}=2{k}_{1}$ and ${n}_{2}=2{k}_{2}$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 3$. In this case, $5\le {k}_{1}+{k}_{2}\le 13$. The initial labeling for some vertices in ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ are listed as follows:We label consecutively the vertices from ${v}_{2,11}$ to ${v}_{2,2{n}_{2}-2}$ and then ${v}_{1,3}$ to ${v}_{2,2{n}_{1}-2}$, if any, by the first $4({k}_{1}+{k}_{2}-4)$ terms of ${S}_{2}$. It is easy to see that the last term of ${s}_{l}$ is not a multiple of 17 for $3\le l\le 12$. The maximum used label is at most $6({k}_{1}+{k}_{2}-1)+5\le 2({n}_{1}+{n}_{2}+{n}_{3})-1$. So, we may label ${G}_{{n}_{3}}$ initially as follows:We label consecutively the vertices from ${v}_{3,5}$ to ${v}_{3,{n}_{1}+{n}_{2}-2}$ by the first $2({k}_{1}+{k}_{2}-3)$ terms of ${T}_{2}$. The rest is similar to Case 1 of the proof of Theorem 7. Since ${n}_{1}+{n}_{2}-2\le 2{n}_{3}-2$, $f\left({v}_{3,2{n}_{3}-2}\right)=2({n}_{1}+{n}_{2}+{n}_{3})$. The above labeling is SD-prime if ${n}_{1}+{n}_{2}+{n}_{3}\not\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$.If ${n}_{1}+{n}_{2}+{n}_{3}\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$, then ${n}_{1}+{n}_{2}+{n}_{3}\in \{16,19,22,25,28,31,34,37\}$. We swap the labels of ${v}_{3,2{n}_{3}-1}$ and ${v}_{2,9}$. One may check that the resulting labeling is SD-prime.
- Case 2:
- Suppose ${n}_{1}=2{k}_{1}$ and ${n}_{2}=2{k}_{2}+1$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 2$. In this case, $4\le {k}_{1}+{k}_{2}\le 12$. The initial labeling for some vertices in ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ are listed as follows:We label consecutively the vertices from ${v}_{2,9}$ to ${v}_{2,2{n}_{2}-2}$ and then ${v}_{1,3}$ to ${v}_{1,2{n}_{1}-2}$, if any, by the first $4({k}_{1}+{k}_{2}-3)$ terms of ${S}_{2}$. Note that ${n}_{1}+1\le {n}_{2}\le {n}_{3}$. The maximum used label is at most $6({k}_{1}+{k}_{2})+4=3({n}_{1}+{n}_{2})+1\le 2({n}_{1}+{n}_{2}+{n}_{3})$ when ${k}_{1}+{k}_{2}\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. So, we may label ${G}_{{n}_{3}}$ initially as follows:We label consecutively the vertices from ${v}_{3,7}$ to ${v}_{3,{n}_{1}+{n}_{2}-1}$ by the first $2({k}_{1}+{k}_{2}-3)$ terms of ${T}_{2}$. The rest is similar to Case 1. Since ${n}_{1}+{n}_{2}-1\le 2{n}_{3}-2$, $f\left({v}_{3,2{n}_{3}-2}\right)=2({n}_{1}+{n}_{2}+{n}_{3})$, the above labeling is SD-prime if ${n}_{1}+{n}_{2}+{n}_{3}\not\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$.When ${n}_{1}+{n}_{2}+{n}_{3}\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. We swap the labels of ${v}_{3,2{n}_{3}-1}$ and ${v}_{2,7}$. Similar to Case 1, one may check that the resulting labeling is SD-prime.When ${k}_{1}+{k}_{2}\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$, there are only two cases: ${k}_{1}+{k}_{2}=5,10$.Suppose ${k}_{1}+{k}_{2}=5$. After labeling ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$, the maximum used labels is 35. Since $2{k}_{2}+1\ge 2{k}_{1}$, ${k}_{2}\ge 3$, we have ${n}_{1}+3\le {n}_{2}\le {n}_{3}$ so that $2({n}_{1}+{n}_{2}+{n}_{3})+1\ge 37$. Hence, the labeling method is as the case when ${k}_{1}+{k}_{2}\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$.Suppose ${k}_{1}+{k}_{2}=10$. After labeling ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$, the maximum used labels is 65. In this case, we have $2({n}_{1}+{n}_{2}+{n}_{3})+1\ge 65$. When $2({n}_{1}+{n}_{2}+{n}_{3})+1>65$, the labeling method is as the case when ${k}_{1}+{k}_{2}\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}5)$. We now deal with $2({n}_{1}+{n}_{2}+{n}_{3})+1=65$. It implies that $({n}_{1},{n}_{2},{n}_{3})=(10,11,11)$. Actually, it is not a case according to the previous construction so we list a required labeling as follows:
- Case 3:
- Suppose ${n}_{1}=2{k}_{1}+1$ and ${n}_{2}=2{k}_{2}$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 3$. In this case, $4\le {k}_{1}+{k}_{2}\le 12$. The initial labeling for some vertices in ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ are listed as follows:We label consecutively the vertices from ${v}_{2,7}$ to ${v}_{2,2{n}_{2}-2}$ and then ${v}_{1,5}$ to ${v}_{1,2{n}_{1}-2}$, if any, by the first $4({k}_{1}+{k}_{2}-3)$ terms of ${S}_{2}$. Note that ${n}_{1}+1\le {n}_{2}\le {n}_{3}$. The rest is as Case 2.
- Case 4:
- Suppose ${n}_{1}=2{k}_{1}+1$ and ${n}_{2}=2{k}_{2}+1$ for some ${k}_{1}\ge 1$ and ${k}_{2}\ge 2$. In this case, $4\le {k}_{1}+{k}_{2}\le 12$. The initial labeling for some vertices in ${G}_{{n}_{1}}$ and ${G}_{{n}_{2}}$ are listed as follows:We label consecutively the vertices from ${v}_{2,9}$ to ${v}_{2,2{n}_{2}-2}$ and then ${v}_{1,5}$ to ${v}_{1,2{n}_{1}-2}$, if any, by the first $4({k}_{1}+{k}_{2}-3)$ terms of ${S}_{2}$. The maximum used label is at most $6({k}_{1}+{k}_{2})+4=3({n}_{1}+{n}_{2})-2\le 2({n}_{1}+{n}_{2}+{n}_{3})-2$. So, we may label ${G}_{{n}_{3}}$ initially as follows:The rest is similar to Case 1.

**Example**

**3.**

**Conjecture**

**1.**

## 4. Merging of Type 1 and Exactly One Type 2 Vertices

**Theorem**

**10.**

**Proof.**

**Theorem**

**11.**

**Proof.**

- Case 1:
- Suppose $b\not\equiv 2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. Let$$\begin{array}{cc}\hfill f\left({u}_{i}\right)& =2b+i+1\text{}\mathrm{for}\text{}2\le i\le 2a,\hfill \\ \hfill f\left({v}_{1}\right)& =3,f\left({v}_{j}\right)=j+2\text{}\mathrm{for}\text{}3\le j\le 2b.\hfill \end{array}$$Here, $(f\left({v}_{1}\right),f\left({v}_{2b}\right))=(3,2b+2)=1$ since $b\not\equiv 2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$.
- Case 2:
- Suppose $a=2=b$. Let$$\begin{array}{cc}\hfill f\left({u}_{2}\right)& =9,f\left({u}_{3}\right)=8,f\left({u}_{4}\right)=3,\hfill \\ \hfill f\left({v}_{1}\right)& =5,f\left({v}_{3}\right)=7,f\left({v}_{4}\right)=6.\hfill \end{array}$$
- Case 3:
- Suppose $a=2$ and $b\equiv 2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$, $b\ge 5$. Let$$\begin{array}{cc}\hfill f\left({u}_{2}\right)& =3,f\left({u}_{3}\right)=8,f\left({u}_{4}\right)=9,\hfill \\ \hfill f\left({v}_{3}\right)& =5,f\left({v}_{4}\right)=6,f\left({v}_{5}\right)=7,\hfill \\ \hfill f\left({v}_{1}\right)& =2b+5,f\left({v}_{j}\right)=j+4\text{}\mathrm{for}\text{}6\le j\le 2b.\hfill \end{array}$$
- Case 4:
- Suppose $a=3$ and $b\equiv 2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. Let$$\begin{array}{cc}\hfill f\left({u}_{2}\right)& =3,f\left({u}_{3}\right)=8,f\left({u}_{4}\right)=5,f\left({u}_{5}\right)=6,f\left({u}_{6}\right)=7,\hfill \\ \hfill f\left({v}_{1}\right)& =2b+7,f\left({v}_{j}\right)=j+6\text{}\mathrm{for}\text{}3\le j\le 2b.\hfill \end{array}$$
- Case 5:
- Suppose $a\ge 4$ and $b\equiv 2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. Let$$\begin{array}{cc}\hfill f\left({u}_{2}\right)& =3,f\left({u}_{2a}\right)=9,f\left({u}_{2a-1}\right)=10,f\left({u}_{2a-2}\right)=7,\hfill \\ \hfill f\left({u}_{2a-3}\right)& =6,f\left({u}_{2a-4}\right)=5,f\left({u}_{2a-5}\right)=8,\hfill \\ \hfill f\left({u}_{i}\right)& =2b+i+7\text{}\mathrm{for}\text{}3\le i\le 2a-6(\mathrm{this}\text{}\mathrm{line}\text{}\mathrm{can}\text{}\mathrm{be}\text{}\mathrm{ignored}\text{}\mathrm{when}a=4),\hfill \\ \hfill f\left({v}_{1}\right)& =9+2b,f\left({v}_{j}\right)=j+8\text{}\mathrm{for}\text{}3\le j\le 2b.\hfill \end{array}$$Now, $(f\left({u}_{2}\right),f\left({u}_{3}\right))=(3,2b+10)=(3,2)=1$.

**Theorem**

**12.**

**Proof.**

**Theorem**

**13.**

**Proof.**

- Case 1:
- Suppose ${n}_{1}\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. We define the labeling f by the following table.Since ${n}_{1}\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$, we have $2{n}_{1}+12\equiv 1,2\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. So, $(3,2{n}_{1}+12)=1$ and $(f\left({v}_{2,6}\right),f\left({v}_{2,7}\right))=1$ when ${n}_{2}\ge 4$.
- Case 2:
- Suppose ${n}_{1}\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$. We list the labeling as in the following table.Clearly, $(f\left({v}_{2,4}\right),f\left({v}_{2,5}\right))=(3,2{n}_{1}+10)=(3,10)=1$.

**Example**

**4.**

**Theorem**

**14.**

**Proof.**

**Conjecture**

**2.**

## 5. Merging of Type 3 Vertices

**Theorem**

**15.**

**Proof.**

**Theorem**

**16.**

**Proof.**

**Remark**

**1.**

**Conjecture**

**3.**

**Theorem**

**17.**

- (1)
- $\sum _{i=1}^{k+1}}{n}_{i}\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}3)$,
- (2)
- there is an SD-prime labeling h of H such that $h\left(v\right)={2}^{a}$ for some $a\ge 1$.

**Proof.**

**Theorem**

**18.**

**Proof.**

**Example**

**5.**

## 6. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## References

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Shiu, W.-C.; Lau, G.-C. On SD-Prime Labelings of Some One Point Unions of Gears. *Symmetry* **2021**, *13*, 849.
https://doi.org/10.3390/sym13050849

**AMA Style**

Shiu W-C, Lau G-C. On SD-Prime Labelings of Some One Point Unions of Gears. *Symmetry*. 2021; 13(5):849.
https://doi.org/10.3390/sym13050849

**Chicago/Turabian Style**

Shiu, Wai-Chee, and Gee-Choon Lau. 2021. "On SD-Prime Labelings of Some One Point Unions of Gears" *Symmetry* 13, no. 5: 849.
https://doi.org/10.3390/sym13050849