# On the Crossing Numbers of the Join Products of Six Graphs of Order Six with Paths and Cycles
## 2 Cyclic Permutations and Configurations
: \(\operatorname{cr}(G^{*}+P_{n})=6\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor \frac{n-1}{2}\right\rfloor+2\left\lfloor\frac{n}{2}\right\rfloor\) _for \(n\geq 2\)._  
The graph \(G^{*}\) is a subgraph of \(G_{2}\), and so \(\operatorname{cr}(G_{2}+P_{n})\geq\operatorname{cr}(G_{2}+D_{n})\geq \operatorname{cr}(G^{*}+D_{n})\). We can also obtain the drawings of \(G_{2}+P_{n}\) with exactly \(6\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n-1}{2}\right\rfloor+2 \left\lfloor\frac{n}{2}\right\rfloor\) crossings by adding the edge \(v_{4}v_{6}\) to the graph \(G^{*}\) without additional crossings into both Figures 6 and 7. **Corollary 3**. : \(\operatorname{cr}(G_{2}+D_{n})=6\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor \frac{n-1}{2}\right\rfloor+2\left\lfloor\frac{n}{2}\right\rfloor\) _for \(n\geq 1\)._  
**Corollary 4**. : \(\operatorname{cr}(G_{2}+P_{n})=6\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor \frac{n-1}{2}\right\rfloor+2\left\lfloor\frac{n}{2}\right\rfloor\) _for \(n\geq 2\)._  
**5. Three Graphs \(H^{*}\), \(H_{1}\), and \(H_{2}\)**  
In Figure 8, let \(H^{*}\) be the connected graph of order six consisting of one 4-cycle and two leaves adjacent to two different but not opposite vertices of the 4-cycle. Let \(v_{3}v_{4}v_{5}v_{6}v_{3}\) and \(v_{1},v_{2}\) be the vertex notation of the 4-cycle and two leaves of \(H^{*}\), respectively. The crossing number of \(H^{*}+D_{n}\) was established by [PERSON] [22]. **Theorem 2** (See [22] Theorem 1). : \(\operatorname{cr}(H^{*}+D_{n})=6\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor \frac{n-1}{2}\right\rfloor+2\left\lfloor\frac{n}{2}\right\rfloor\) _for \(n\geq 1\)._  
The same reasoning as for the graph \(G^{*}+P_{n}\) using the drawing of \(H^{*}+P_{n}\) in Figure 9 gives the following result. Figure 1.0 The planar drawing of the graph \(H^{*}\). **Corollary 5**. : \(\operatorname{cr}(H^{*}+P_{n})=6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{\lfloor} \frac{n-1}{2}\big{\rfloor}+2\big{\lfloor}\frac{n}{2}\big{\rfloor}\) _for \(n\geq 2\)._  
In Figure 10, let \(H_{1}\) be the graph obtained from the planar drawing of \(H^{*}\) in Figure 8 by adding the edge \(v_{1}v_{4}\), i.e., \(H_{1}=H^{*}\cup\{v_{1}v_{4}\}\). Similarly, let \(H_{2}=H^{*}\cup\{v_{1}v_{4},v_{2}v_{5}\}\). The join products of the graphs \(H_{1}\) and \(H_{2}\) with \(P_{n}\) were already investigated by [PERSON] [10] and [PERSON] [1], respectively. **Theorem 3** (See [10] Theorem 1). : \(\operatorname{cr}(H_{1}+P_{n})=6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{ \lfloor}\frac{n-1}{2}\big{\rfloor}+2\big{\lfloor}\frac{n}{2}\big{\rfloor}+1\) _for \(n\geq 2\)._  
**Theorem 4** (See [1] Theorem 3.1). : \(\operatorname{cr}(H_{2}+P_{n})=6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{ \lfloor}\frac{n-1}{2}\big{\rfloor}+2\big{\lfloor}\frac{n}{2}\big{\rfloor}+1\) _for \(n\geq 2\)._  
**Theorem 5**. : \(\operatorname{cr}(G_{1}+P_{n})=6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{ \lfloor}\frac{n-1}{2}\big{\rfloor}+2\big{\lfloor}\frac{n}{2}\big{\rfloor}+1\) _for \(n\geq 2\)._  
Proof. : The good drawing of the join product \(G_{1}+P_{n}\) with exactly \(6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{\lfloor}\frac{n-1}{2}\big{\rfloor} +2\big{\lfloor}\frac{n}{2}\big{\rfloor}+1\) crossings by adding \(n\) new edges \(v_{1}v_{6}\), \(t_{i}t_{i+1}\), \(i=1,\ldots,n-1\), into the drawing in Figure 4 enforces the required upper border, that is, \(\operatorname{cr}(G_{1}+P_{n})\leq 6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{ \lfloor}\frac{n-1}{2}\big{\rfloor}+2\big{\lfloor}\frac{n}{2}\big{\rfloor}+1\). To prove the lower border using Corollary 1, we assume a good drawing \(D\) of \(G_{1}+P_{n}\) with \(6\big{\lfloor}\frac{n}{2}\big{\rfloor}\big{\lfloor}\frac{n-1}{2}\big{\rfloor} +2\big{\lfloor}\frac{n}{2}\big{\rfloor}\) crossings.