**Lemma 1.** cr(*K*2,3 + *C*3) = 10*,* cr(*K*2,3 + *C*4) = 15*, and* cr(*K*2,3 + *C*5) = 24*.*

Recall that two vertices *ti* and *tj* of *K*2,3 + *Cn* are antipodal in a drawing of *K*2,3 + *Cn* if the subgraphs *T<sup>i</sup>* and *T<sup>j</sup>* do not cross, and a drawing is said to be antipodal-free if it does not have antipodal vertices. The result in the following Theorem 2 has already been claimed by Yuan [17]. The correctness of an article written in Chinese cannot be verified because compilers cannot handle it. Therefore, such results can only be considered as unconfirmed hypotheses.

$$\text{Theorem 2. } \operatorname{cr}(K\_{2,3} + \mathbb{C}\_n) = 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3 \text{ for } n \ge 3.$$

**Proof of Theorem 2.** In Figure 2 there is the drawing of *K*2,3 + *Cn* with 4 *<sup>n</sup>* 2 *n*−1 2 + *n* + 3 crossings. Thus, cr(*K*2,3 + *Cn*) ≤ 4 *n* 2 *n*−1 2 + *n* + 3. Theorem 2 is true for *n* = 3, *n* = 4, and *n* = 5 by Lemma 1. Assume *n* ≥ 6. We prove the reverse inequality by contradiction. Suppose now that there is a drawing *D* of *K*2,3 + *Cn* with

$$\text{scr}\_D(K\_{2,3} + \mathbb{C}\_n) < 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3 \tag{3}$$

and that

$$\text{crr}(K\_{2,3} + C\_m) \ge 4\left\lfloor \frac{m}{2} \right\rfloor \left\lfloor \frac{m-1}{2} \right\rfloor + m + 3 \quad \text{for each } 3 \le m < n. \tag{4}$$

**Figure 2.** The good drawing of *K*2,3 + *Cn* with 4 *<sup>n</sup>* 2 *n*−1 2 + *n* + 3 crossings.

Let us show that the considered drawing *D* must be antipodal-free. For a contradiction suppose, with the rest of paper, that cr*D*(*Tn*−1, *Tn*) = 0. It is not difficult to verify in Figure 1 that if at least one of *Tn*−<sup>1</sup> and *Tn*, say *Tn*, does not cross the edges of the graph *K*2,3, then the edges of *Tn*−<sup>1</sup> must cross the edges of *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>n</sup>* at least twice, that is, cr*D*(*K*2,3, *<sup>T</sup>n*−<sup>1</sup> <sup>∪</sup> *<sup>T</sup>n*) <sup>≥</sup> 2. By [5], we already know that cr(*K*5,3) = 4, which yields that each *<sup>T</sup>k*, *<sup>k</sup>* <sup>=</sup> *<sup>n</sup>* <sup>−</sup> 1, *<sup>n</sup>*, have to cross the edges of the subgraph *<sup>T</sup>n*−<sup>1</sup> <sup>∪</sup> *<sup>T</sup><sup>n</sup>* at least four times. On the basis of this, we have

$$\text{crr}\_{\mathcal{D}}(K\_{2,3} + \mathbb{C}\_{\mathbb{H}}) = \text{crr}\_{\mathcal{D}}\left(K\_{2,3} + \mathbb{C}\_{n-2}\right) + \text{cr}\_{\mathcal{D}}(K\_{5,n-2}, T^{n-1} \cup T^{n}) + \text{cr}\_{\mathcal{D}}(K\_{2,3}, T^{n-1} \cup T^{n}) + \text{cr}\_{\mathcal{D}}(T^{n-1} \cup T^{n})$$

$$\geq 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + n - 2 + 3 + 4(n-2) + 2 + 0 = 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

This contradicts the assumption (3) and consequently confirms that *D* must be antipodal-free. As the graph *K*2,3 + *Dn* is a subgraph of the graph *K*2,3 + *Cn*, by Theorem 1, the edges of *K*2,3 + *Cn* are crossed at least 4 *<sup>n</sup>* 2 *n*−1 2 + *n* times, and therefore, at most two edges of the cycle *C*∗ *<sup>n</sup>* can be crossed in *D*. This also enforces that the vertices *ti* of the cycle *C*<sup>∗</sup> *<sup>n</sup>* must be placed at most in two different regions in the considered good subdrawing of *K*2,3. If *r* = |*RD*| and *s* = |*SD*|, then the assumption (4) together with cr(*K*5,*n*) = 4 *n* 2 *n*−1 2 enforce that there is at least one subgraph *T<sup>i</sup>* which crosses the edges of *K*2,3 at most once in the drawing *D*. To be precise,

$$\text{crr}\_D(K\_{2,3}) + \text{crr}\_D(K\_{2,3}, K\_{5,n}) \le \text{crr}\_D(K\_{2,3}) + 0r + s + 2(n - r - s) < n + 3\epsilon$$

that is,

$$\text{crr}\_D(\mathbb{K}\_{2,3}) + s + 2(n - r - s) \le n + 2. \tag{5}$$

This implies that 2*r* + *s* ≥ *n* + cr*D*(*K*2,3) − 2. Further, if cr*D*(*K*2,3) = 0 and *r* = 0, then *s* ≥ *n* − 2. Now, we will deal with the possibilities of obtaining a subgraph *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *RD* <sup>∪</sup> *SD* in *<sup>D</sup>* and we will exhibit that in all mentioned cases a contradiction with the assumption (3) is achieved.

**Case 1:** cr*D*(*K*2,3) = 0. In this case, without lost of generality, we can assume the drawing with the vertex notation of *K*2,3 as shown in Figure 1a. The unique subdrawing of *K*2,3 induced by *D* contains three different regions. Hence, let us denote these three regions by *ω*1,4,3,2, *ω*1,4,5,2, and *ω*2,5,4,3 depending on which of vertices are located on the boundary of the corresponding region. Since the vertices of *C*∗ *<sup>n</sup>* do not have to be placed in the same region in the considered subdrawing of *K*2,3, two possible subcases may occur:

(a) All vertices of *C*∗ *<sup>n</sup>* are placed in two regions of subdrawing of *K*2,3 induced by *D*. In the rest of paper, based on their symmetry, we can suppose that all vertices *ti* of *C*<sup>∗</sup> *<sup>n</sup>* are placed in *<sup>ω</sup>*1,4,3,2 <sup>∪</sup> *<sup>ω</sup>*1,4,5,2. Of course, there is no possibility to obtain a subdrawing of *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* for a *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *RD*, that is, *r* = 0. Clearly, the edges of *C*∗ *<sup>n</sup>* must cross the edges of *K*2,3 exactly twice. This fact, with the property (5) in the form 0 + 1*s* + 2(*n* − *s*) < *n* + 1, confirms that *s* = *n*, which yields that each subgraph *T<sup>i</sup>* cross the edges of *K*2,3 just once. If some vertices *ti* of *C*<sup>∗</sup> *<sup>n</sup>* are placed in *ω*1,4,3,2, then we deal with the configurations A*k*, *k* ∈ {1, 2, 3, 4} (they have been already introduced in [14]). For *ti* ∈ *ω*1,4,5,2, there are four other ways for how to obtain the subdrawing of *F<sup>i</sup>* depending on which edge of *K*2,3 is crossed by the edge *tiv*<sup>3</sup> provided by there is only one subdrawing of *<sup>F</sup><sup>i</sup>* \ {*v*3} represented by the rotation (1452). These four possibilities can be denoted by A*k*, for *k* = 5, 6, 7, 8 and they are represent by the cyclic permutations (14532), (13452), (14523), and (14352), respectively. Consequently, we denote by M*<sup>D</sup>* the subset of all configurations that exist in the drawing *D* belonging to the set M = {A*<sup>k</sup>* : *k* = 1, ... , 8}. Using the same arguments like in [14], the resulting lower bounds for the number of crossings of two configurations from M can be established in Table 1 (here, A*<sup>k</sup>* and A*<sup>l</sup>* are configurations of the subgraphs *F<sup>i</sup>* and *F<sup>j</sup>* , where *k*, *l* ∈ {1, . . . , 8}).

**Table 1.** The necessary number of crossings between two different subgraphs *T<sup>i</sup>* and *T<sup>j</sup>* for the configurations A*<sup>k</sup>* and A*l*.


Let us first assume that {A*k*, A*k*+4}⊆M*<sup>D</sup>* for some *k* ∈ {1, 2, 3, 4}. In the rest of paper, let us assume two different subgraphs *<sup>T</sup>n*−1, *<sup>T</sup><sup>n</sup>* <sup>∈</sup> *SD* such that *<sup>F</sup>n*−<sup>1</sup> and *<sup>F</sup><sup>n</sup>* have different configurations <sup>A</sup><sup>1</sup> and <sup>A</sup>5, respectively. Then, cr*D*(*K*2,3 <sup>∪</sup> *<sup>T</sup>n*−<sup>1</sup> <sup>∪</sup> *<sup>T</sup>n*, *<sup>T</sup><sup>i</sup>* ) ≥ 1 + 5 = 6 holds for any *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* with *<sup>i</sup>* <sup>=</sup> *<sup>n</sup>* <sup>−</sup> 1, *<sup>n</sup>* by summing the values in all columns in two considered rows of Table 1. Hence, by fixing the subgraph *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup>n*−<sup>1</sup> <sup>∪</sup> *<sup>T</sup>n*, we have

$$\operatorname{crr}\_{\mathcal{D}}(\mathcal{K}\_{2,3} + \mathcal{C}\_{n}) \ge \operatorname{cr}\_{\mathcal{D}}(\mathcal{K}\_{5,n-2}) + \operatorname{cr}\_{\mathcal{D}}(\mathcal{K}\_{5,n-2}, \mathcal{K}\_{2,3} \cup T^{n-1} \cup T^n) + \operatorname{cr}\_{\mathcal{D}}(\mathcal{K}\_{2,3} \cup T^{n-1} \cup T^n)$$

$$\ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 6(n-2) + 2 + 1 \ge 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

Due to the symmetry of three remaining pairs of configurations, we also obtain a contradiction in *D* by applying the same process. Now, let us turn to the good drawing *D* of the graph *K*2,3 + *Cn* in which {A*k*, A*k*+4} ⊆ M*<sup>D</sup>* for any *k* = 1, 2, 3, 4. Further, let us also suppose that the number of subgraphs with the configuration A*<sup>k</sup>* ∈ M*<sup>D</sup>* is at least equal to the number of subgraphs with the configuration <sup>A</sup>*<sup>l</sup>* ∈ M*D*, for each possible *<sup>l</sup>* <sup>=</sup> *<sup>k</sup>*, and let *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* be such a subgraph with the configuration <sup>A</sup>*<sup>k</sup>* of *<sup>F</sup><sup>i</sup>* . Hence,

$$\text{crr}\_D(\mathbb{K}\_{5,n-1}, T^i) = \sum\_{j \neq i} \text{crr}\_D(T^j, T^i) \ge 3(n-2) + 2 - 2\left\lfloor \frac{n}{7} \right\rfloor \ge \frac{5}{2}n - \frac{5}{2}n$$

where an idea of the arithmetic mean of the values four, three and two of Table 1 could be exploited. Thus, by fixing the subgraph *T<sup>i</sup>* , we have

$$\text{cr}\_D(\mathcal{K}\_{2,3} + \mathcal{C}\_n) = \text{cr}\_D\left(\mathcal{K}\_{2,3} + \mathcal{C}\_{n-1}\right) + \text{cr}\_D(\mathcal{K}\_{5,n-1}, T^i) + \text{cr}\_D(\mathcal{K}\_{2,3}, T^i)$$

$$\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + n - 1 + 3 + \frac{5}{2}n - \frac{5}{2} + 1 \geq 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

(b) All vertices *ti* of *C*<sup>∗</sup> *<sup>n</sup>* are placed in the same region of subdrawing of *K*2,3 induced by *D*. In the rest of paper, based also on their symmetry, we suppose that *ti* ∈ *ω*1,4,3,2 for each *i* = 1, ... , *n*. Whereas the set *RD* is again empty, there are at least *<sup>n</sup>* <sup>−</sup> 2 subgraphs *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* provided by the property (5) in the form *<sup>s</sup>* <sup>+</sup> <sup>2</sup>(*<sup>n</sup>* <sup>−</sup> *<sup>s</sup>*) <sup>&</sup>lt; *<sup>n</sup>* <sup>+</sup> 3. For *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD*, we consider only one from the configurations <sup>A</sup>*k*, for *k* = 1, 2, 3, 4. Again, let us also assume that the number of subgraphs with the configuration A*<sup>k</sup>* is at least equal to the number of subgraphs with the configuration A*l*, for each possible *l* = *k*, and let *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* be such a subgraph with the configuration <sup>A</sup>*<sup>k</sup>* of *<sup>F</sup><sup>i</sup>* . Then, by fixing the subgraph *Ti* , we have

$$\text{cr}\_D(K\_{2,3} + \mathbb{C}\_n) = \text{cr}\_D\left(K\_{2,3} + \mathbb{C}\_{n-1}\right) + \text{cr}\_D(K\_{5,n-1}, T^i) + \text{cr}\_D(K\_{2,3}, T^i)$$

$$\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + n - 1 + 3 + 3(s - 2) + 2 + 1(n - s) + 1 = 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor$$

$$+ 2n + 2s - 1 \geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 2n + 2(n - 2) - 1 \geq 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

wherein a simplified form of the idea of the arithmetic mean of the values of Table 1 is applied.

**Case 2:** cr*D*(*K*2,3) = 1. We can choose the drawing with the vertex notation of *K*2,3 like in Figure 1b. Similarly as in the previous case, we will discuss two subcases:

(a) The cycle *C*∗ *<sup>n</sup>* is crossed by some edge of the graph *K*2,3. As the edges of *C*<sup>∗</sup> *<sup>n</sup>* cross the edges of *K*2,3 exactly twice, there is a subgraph *T<sup>i</sup>* which does not cross the edges of *K*2,3 provided by the property (5) in the form 1 <sup>+</sup> *<sup>s</sup>* <sup>+</sup> <sup>2</sup>(*<sup>n</sup>* <sup>−</sup> *<sup>r</sup>* <sup>−</sup> *<sup>s</sup>*) <sup>≤</sup> *<sup>n</sup>*. For a *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *RD*, the reader can easily verify that the subgraph *<sup>F</sup><sup>i</sup>* <sup>=</sup> *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* is uniquely represented by rot*D*(*ti*)=(12345), and cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) <sup>≥</sup> 4 holds for any *<sup>T</sup><sup>j</sup>* <sup>∈</sup> *RD* with *<sup>j</sup>* <sup>=</sup> *<sup>i</sup>* provided that rot*D*(*ti*) = rot*D*(*tj*), for more see [18]. Moreover, it is not difficult to verify in possible regions of *<sup>D</sup>*(*K*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* ) that cr*D*(*K*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* , *T<sup>j</sup>* ) <sup>≥</sup> 4 is true for each *<sup>T</sup><sup>j</sup>* <sup>∈</sup> *SD*. Thus, by fixing the subgraph *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* , we have

$$\text{crr}\_D(K\_{2,3} + \mathcal{C}\_n) \ge \text{crr}\_D(K\_{5,n-1}) + \text{crr}\_D(K\_{5,n-1}, K\_{2,3} \cup T^i) + \text{crr}\_D(K\_{2,3} \cup T^i)$$

$$\ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4(r-1) + 4s + 3(n-r-s) + 1 = 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor$$

$$+ 3n + (r+s) - 3 \ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 3n + 4 - 3 \ge 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3/2$$

where *r* + *s* ≥ 4 holds also due to the property (5).

(b) None edge of *C*∗ *<sup>n</sup>* is crossed by the edges of *K*2,3. Since all vertices *ti* of the cycle *C*<sup>∗</sup> *<sup>n</sup>* are placed in the same region of subdrawing of *K*2,3 induced by *D*, they must be placed in the outer region of *<sup>D</sup>*(*K*2,3). If there is a *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *RD*, then the edges of *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* are crossed at least four times by any subgraph *T<sup>j</sup>* with the placement of the vertex *tj* in the outer region of *D*(*K*2,3), which yields that the similar idea as in the previous subcase can be used by fixing the subgraph *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>*

$$\text{crr}\_D(\mathcal{K}\_{2,3} + \mathcal{C}\_\mathbb{II}) \ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4(r-1) + 4(n-r) + 1$$

$$= 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4n - 3 \ge 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

To finish the proof of this case, let us suppose that the set *RD* is empty. Whereas the set *RD* is empty, there are at least *<sup>n</sup>* <sup>−</sup> 1 subgraphs *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* according to the property (5). Since the edges *v*2*v*3, *v*3*v*4, *v*2*v*5, and *v*1*v*<sup>4</sup> of *K*2,3 can be crossed by the edges *tiv*4, *tiv*2, *tiv*4, and *tiv*2, respectively, these four ways can be denoted by B*k*, for *k* = 1, 2, 3, 4. So, the configurations B1, B2, B3, and B<sup>4</sup> are uniquely described by the cyclic permutations (12435), (13245), (12354), and (13452), respectively., and the aforementioned properties of the cyclic rotations imply all lower-bounds of number of crossings in Table 2.

**Table 2.** The necessary number of crossings between two different subgraphs *T<sup>i</sup>* and *T<sup>j</sup>* for the configurations B*<sup>k</sup>* and B*l*.


Now, let us also suppose that the number of subgraphs with the configuration B*<sup>k</sup>* is at least equal to the number of subgraphs with the configuration <sup>B</sup>*l*, for each possible *<sup>l</sup>* <sup>=</sup> *<sup>k</sup>*, and let *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* be such a subgraph with the configuration <sup>B</sup>*<sup>k</sup>* of *<sup>F</sup><sup>i</sup>* . Hence,

$$\sum\_{\substack{T^j \in S\_{D,\,\,j\neq i}}} \text{cr}\_D(T^i, T^j) \ge 3(s-2) + 2 - 2\left\lfloor \frac{s}{4} \right\rfloor \ge \frac{5}{2}s - 4\sqrt{s}$$

where again an idea of the arithmetic mean of the values four and two of Table 2 could be exploited. Thus, by fixing the subgraph *T<sup>i</sup>* , we have

$$\text{cr}\_D(\mathcal{K}\_{2,3} + \mathcal{C}\_n) = \text{cr}\_D\left(\mathcal{K}\_{2,3} + \mathcal{C}\_{n-1}\right) + \text{cr}\_D(\mathcal{K}\_{5,n-1}, T^i) + \text{cr}\_D(\mathcal{K}\_{2,3}, T^i)$$

$$\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + n - 1 + 3 + \frac{5}{2}s - 4 + 1(n-s) + 1 = 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor$$

$$+ 2n + \frac{3}{2}s - 1 \geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 2n + \frac{3}{2}(n-1) - 1 \geq 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

**Case 3:** cr*D*(*K*2,3) = 2. In the rest of paper, we choose the drawing with the vertex notation of *K*2,3 like in Figure 1c. Obviously the set *RD* must be empty. As *s* = *n* by the property (5), all vertices *ti* of the subgraphs *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* must be placed in the region of *<sup>D</sup>*(*K*2,3) with four vertices *<sup>v</sup>*1, *<sup>v</sup>*2, *<sup>v</sup>*4, and *<sup>v</sup>*<sup>5</sup> of the graph *<sup>K</sup>*2,3 on its boundary. For *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD*, there is only one possible subdrawing of *<sup>F</sup><sup>i</sup>* \ {*v*3} described by the rotation (1245), which yields that there are exactly three ways of obtaining the subdrawing of *<sup>K</sup>*2,3 <sup>∪</sup> *<sup>T</sup><sup>i</sup>* depending on which edge of *<sup>K</sup>*2,3 may be crossed by *tiv*3. In all cases of *<sup>T</sup><sup>i</sup>* <sup>∈</sup> *SD* represented by either (12345) or (12453) or (12435), it is not difficult to verify using cyclic permutations that cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) <sup>≥</sup> 2 is fulfilling for each *<sup>T</sup><sup>j</sup>* <sup>∈</sup> *SD*, *<sup>j</sup>* <sup>=</sup> *<sup>i</sup>*. Thus, by fixing the subgraph *<sup>T</sup><sup>i</sup>* , we have

$$\text{crr}\_D(\mathcal{K}\_{2,3} + \mathcal{C}\_n) = \text{crr}\_D\left(\mathcal{K}\_{2,3} + \mathcal{C}\_{n-1}\right) + \text{cr}\_D(\mathcal{K}\_{5,n-1}, T^i) + \text{cr}\_D(\mathcal{K}\_{2,3}, T^i)$$

$$\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + n - 1 + 3 + 2(n-1) + 1 \geq 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3.$$

**Case 4:** cr*D*(*K*2,3) = 3. We assume the drawing with the vertex notation of *K*2,3 like in Figure 1d. As the property (5) enforces *r* ≥ 1 and *r* + *s* ≥ 4, the proof can proceed in the similar way as in the Subcase 2a).

We have shown, in all cases, that there is no good drawing *D* of the graph *K*2,3 + *Cn* with fewer than 4 *<sup>n</sup>* 2 *n*−1 2 + *n* + 3 crossings. This completes the proof.

Finally, in Figure 2, we are able to add the edge *v*1*v*<sup>5</sup> to the graph *K*2,3 without additional crossings, and we obtain one new graph *H* in Figure 3. So, the drawing of the graph *H* + *Cn* with 4 *n* 2 *n*−1 2 + *n* + 3 crossings is obtained. On the other hand, *K*2,3 + *Cn* is a subgraph of *H* + *Cn*, and therefore, cr(*H* + *Cn*) ≥ cr(*K*2,3 + *Cn*). Thus, the next result is an immediate consequence of Theorem 2.

**Figure 3.** The graph *H* by adding one edge to the graph *K*2,3.

$$\text{Corollary 1. } \text{cr}(H + \mathbb{C}\_n) = 4\left\lfloor \frac{n}{2} \right\rfloor \left\lfloor \frac{n-1}{2} \right\rfloor + n + 3 \text{ for any } n \ge 3.$$

#### **4. Conclusions**

We suppose that the application of various forms of arithmetic means can be used to estimate the unknown values of the crossing numbers for join products of some graphs on five vertices with the paths, and also with the cycles. The same we expect for larger graphs, namely for a lot of symmetric graphs of order six.

**Funding:** This research received no external funding.

**Conflicts of Interest:** The author declares no conflict of interest.

#### **References**


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