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Introduction

Throughout this paper, we consider only simple graphs. Let G be such a graph on n vertices and m edges. We denote the vertex set and edge set of G by V(G) and E(G), respectively. Thus, |V(G)| = n and |E(G)| = m. As usual, n is said to be the order and m the size of G. If u and ν are two adjacent vertices of G, then the edge connecting them will be denoted by . The degree of a vertex w ε V(G) is the number of vertices adjacent to w and is denoted by dG(w). The complement of G, denoted by G¯$\begin{array}{} \displaystyle \bar G \end{array}$, is a graph which has the same vertex set as G, in which two vertices are adjacent if and only if they are not adjacent in G. The line graph L(G) of a graph G is the graph with vertex set as the edge set of G and two vertices of L(G) are adjacent whenever the corresponding edges in G have a vertex in common. The subdivision graph S(G) of a graph G whose vertex set is V(G) ∪ E(G) where two vertices are adjacent if and only if one is a vertex of G and other is an edge of G incident with it. The partial complement of subdivision graph S¯(G)$\begin{array}{} \displaystyle \bar S(G) \end{array}$ of a graph G whose vertex set is V(G) ∪ E(G) where two vertices are adjacent if and only if one is a vertex of G and the other is an edge of G non incident with it. Please refer to [17, 25] for unexplained graph theoretic terminology and notation.

In theoretical chemistry, the physico-chemical properties of chemical compounds are often modeled by means of molecular-graph-based structure-descriptors which are also referred to topological indices [16, 30]. Topological indices are found to be very useful in chemistry, biochemistry and nanotechnology in isomer discrimination, structure-property relationship, structure-activity relationship and pharmaceutical drug design. The first and second Zagreb indices of a graph are among the most studied vertex degree based topological indices. The first and second Zagreb indices, respectively defined by

M1(G)=uV(G)dG(u)2=uνE(G)[dG(u)+dG(ν)] and M2(G)=uνE(G)dG(u)dG(ν)$$\begin{array}{} \displaystyle {M_1}(G) = \sum\limits_{u \in V(G)} {{d_G}} {(u)^2} = \sum\limits_{u\nu \in E(G)} {[{d_G}(u) + {d_G}(\nu )]} {\rm{ and }}{M_2}(G) = \sum\limits_{u\nu \in E(G)} {{d_G}} (u){d_G}(\nu ) \end{array}$$

are widely studied degree-based topological indices, that were introduced by Gutman and Trinajstić [15] in 1972.

The vertex-degree-based graph invariant

F(G)=νV(G)dG(ν)3=uνE(G)[dG(u)2+dG(ν)2]$$\begin{array}{} \displaystyle F(G) = \sum\limits_{\nu \in V(G)} {{d_G}} {(\nu )^3} = \sum\limits_{u\nu \in E(G)} {[{d_G}{{(u)}^2} + {d_G}{{(\nu )}^2}]} \end{array}$$

was encountered in [15]. Recently there has been some interest to F, called forgotten topological index or F-index [10].

Shirdel et al. [29] introduced a new Zagreb index of a graph G named hyper-Zagreb index and is defined as:

HM(G)=uνE(G)(dG(u)+dG(ν))2.$$\begin{array}{} \displaystyle HM(G) = \sum\limits_{u\nu \in E(G)} {({d_G}(} u) + {d_G}(\nu ){)^2}. \end{array}$$

Computation of these topological indices of graphs are reported in [24, 1113].

Li and Zhao [27] introduced the first general Zagreb index as follows

αλ(G)=uV(G)[dG(u)]λ.$$\begin{array}{} \displaystyle \alpha_{\lambda}(G)=\sum \limits_{u\in V(G)}[d_{G}(u)]^{\lambda}. \end{array}$$

It is easy to write that

αλ(G)=uνE(G)[(dG(u))λ1+(dG(ν))λ1].$$\begin{array}{} \displaystyle {\alpha _\lambda }(G) = \sum\limits_{u\nu \in E(G)} {[{{({d_G}(u))}^{\lambda - 1}} + {{({d_G}(\nu ))}^{\lambda - 1}}]} . \end{array}$$

The general sum connectivity index [31] was introduced by Zhou et al. and is defined as

M1α(G)=uνE(G)[dG(u)+dG(ν)]α.$$\begin{array}{} \displaystyle M_1^\alpha (G) = \sum\limits_{u\nu \in E(G)} {{{[{d_G}(u) + {d_G}(\nu )]}^\alpha }} . \end{array}$$

By Eq. (1), it is consistent to define M13(G)$\begin{array}{} \displaystyle M_{1}^3(G) \end{array}$ as

M13(G)=uνE(G)[dG(u)+dG(ν)]3.$$\begin{array}{} \displaystyle M_1^3(G) = \sum\limits_{u\nu \in E(G)} {{{[{d_G}(u) + {d_G}(\nu )]}^3}} . \end{array}$$

Here we note that, α2(G)=M11(G)=M1(G)$\begin{array}{} \displaystyle \alpha_{2}(G)=M_{1}^1(G)=M_{1}(G) \end{array}$, α3(G) = F(G) and M12(G)=HM(G)$\begin{array}{} \displaystyle M_1^2(G) = HM(G) \end{array}$.

Graph operations play a vital role in chemical graph theory. Different chemically important graphs can be obtained by applying graph operations on some general or particular graphs. One of the chemically interseting graph operation is Cartesian product of graphs. The Cartesian product G1 × G2 of graphs G1 and G2 has the vertex set V(G1 × G2) = V(G1) × V(G2) and (u1, ν1)(u2, ν2) is an edge of G1 × G2 if and only if [u1 = u2 and ν1ν2 ε E(G2)] or [ν1 = ν2 and u1u2 ε E(G1)].

Many of the chemically interesting graphs can be obtained by applying the Cartesian product of graphs. For example, the ladder graph Ln is the molecular graph related to the polynomial structure obtained by the Cartesian product of P2 and Pn+1. The C4 nanotube TUC4(m,n) is the Cartesian product of Pn and Pm and the C4 nanotorus TC4(m,n) is the Cartesian product of Cn and Cm.

Graovac and Pisanski [14] were the first to consider the problem of computing topological indices of graph operations. In their paper, they computed an exact formula for the Wiener index of the Cartesian product of graphs. In [24], Klavzar, Rajapakse and Gutman computed the Szeged index of the Cartesian product graphs. In a series of recent papers [1823], M. H. Khalifeh and his coworkers extended this program to other topological indices, such as the vertex and edge PI index, the first and second Zagreb index, the vertex and edge versions of Szeged index, and the hyper-Wiener and the edge-Wiener indices of several operations. The present work is the continuation of research along the same lines, and is concerned with additional types of graph operations.

New Cartesian products of graphs

Eliasi et al. in [9] generalized the concept of Cartesian products of graphs, and introduced four new sums of graphs called F-sums of graphs and studied the Wiener index of resulting graphs. Recently there has been some interest on computing topological indices of F-sums of graphs [1, 5, 8, 26, 28].

Motivated by applications of Cartesian product of graphs, here we are more generalize the concept of Cartesian products of graphs and introduce the new C-products of graphs. For this purpose we proceed to introduce some notions and definition of [7].

For a graph G = (V,E), let G0 be the graph with V(G0) = V(G) and with no edges, G1 the complete graph with V(G1) = V(G), G+ = G, and G=G¯$\begin{array}{} \displaystyle G^-=\overline{G} \end{array}$.

Definition 1.

[7] Given a graph G with vertex set V(G) and edge set E(G) and three variables x,y,z ∊ {0,1,+,−}, the xyz-transformation graph Txyz(G) of G is the graph with vertex set V (Txyz(G)) = V(G) ∪ E(G) and the edge set E(Txyz(G)) = E((G)x) ∪ E((L(G))y) ∪ E(W) where W = S(G) if z = +, W=S¯(G)$\begin{array}{} \displaystyle W=\overline{S}(G) \end{array}$ if z = −, W is the graph with V (W) = V(G) ∪ E(G) and with no edges if z = 0 and W is the complete bipartite graph with parts V(G) and E(G) if z = 1.

Examples of xyz−transformations of a 4-vertex path are given in Figure 1. We call vertex in xyz-transformation graphs corresponding to vertex of parent graph as point vertex whereas vertex in xyz-transformation graphs corresponding to edge of parent graph as line vertex.

Fig. 1

Some xyz−transformations of a 4-vertex path.

Now we give the definition of the C-product of graphs in the following.

Definition 2.

Let C ∊ {Txyz|x,y,z ∊ {0,1,+,−}}. The C-product of G1 and G2, denoted by G1 ×CG2, is a graph with the set of vertices V(G1 ×CG2) = (V(G1) ∪ E(G1)) × V(G2) and two vertices (u1,u2) and (ν1,ν2) of G1 ×CG2 are adjacent if and only if [u1 = ν1 ε V(G1) and u2ν2 ε E(G2)] or [u2 = ν2 ε V(G2) and u1ν1E(C(G1))].

Thus we obtain 64 new C− products of graphs in which G1 ×T00+G2, G1 ×T+0+G2, G1 ×T0++G2 and G1 ×T+++G2 are F− sums of graphs introduced by Eliasi and Taeri [9]. Examples of C-products of P4 and P2 are given in Figure 2. In this paper, we compute the expressions for first Zagreb index and F-index of the C−products of graphs.

Fig. 2

Some C-product of P4 and P2.

Main Results

We start by stating the following propositions, which are immediately from definitions and needed for the proving our main results.

Proposition 1.

Let G be a (n,m)-graph. Then the degree of point vertex u and line vertex e(= ab in G) in Txyz(G) when z = 0 are

dTxy0(G)(u)={0if x=0&y{0,1,+,}.n1if x=1&y{0,1,+,}.dG(u)if x=+&y{0,1,+,}.n1dG(u)if x=&y{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy0}}\left( G \right)}}\left( u \right) = \{ \begin{array}{*{20}{c}} 0 \hfill & {if{\rm{ }}x = 0\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n - 1} \hfill & {if{\rm{ }}x = 1\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {{d_G}\left( u \right)} \hfill & {if{\rm{ }}x = + \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n - 1 - {d_G}\left( u \right)} \hfill & {if{\rm{ }}x = - \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

dTxy0(G)(e)={0if y=0&x{0,1,+,}.m1if y=1&x{0,1,+,}.dG(a)+dG(b)2if y=+&x{0,1,+,}.m+1dG(a)dG(b)if y=&x{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy0}}\left( G \right)}}\left( e \right) = \{ \begin{array}{*{20}{c}} 0 \hfill & {if{\rm{ }}y = 0\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {m - 1} \hfill & {if{\rm{ }}y = 1\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {{d_G}\left( a \right) + {d_G}\left( b \right) - 2} \hfill & {if{\rm{ }}y = + \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {m + 1 - {d_G}\left( a \right) - {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = - \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

Proposition 2.

Let G be a (n,m)-graph. Then the degree of point vertex u and line vertex e(= ab in G) in Txyz(G) when z = 1 are

dTxy1(G)(u)={mif x=0&y{0,1,+,}.m+n1if x=1&y{0,1,+,}.dG(u)+mif x=+&y{0,1,+,}.n1dG(u)+mif x=&y{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy1}}\left( G \right)}}\left( u \right) = \{ \begin{array}{*{20}{c}} m \hfill & {if{\rm{ }}x = 0\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {m + n - 1} \hfill & {if{\rm{ }}x = 1\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {{d_G}\left( u \right) + m} \hfill & {if{\rm{ }}x = + \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n - 1 - {d_G}\left( u \right) + m} \hfill & {if{\rm{ }}x = - \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

dTxy1(G)(e)={nif y=0&x{0,1,+,}.n+m1if y=1&x{0,1,+,}.n2+dG(a)+dG(b)if y=+&x{0,1,+,}.n+m+1dG(a)dG(b)if y=&x{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy1}}\left( G \right)}}\left( e \right) = \{ \begin{array}{*{20}{c}} n \hfill & {if{\rm{ }}y = 0\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m - 1} \hfill & {if{\rm{ }}y = 1\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n - 2 + {d_G}\left( a \right) + {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = + \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m + 1 - {d_G}\left( a \right) - {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = - \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

Proposition 3.

Let G be a (n,m)-graph. Then the degree of point vertex u and line vertex e(= ab in G) in Txyz(G) when z = + are

dTxy(G)(u)={dG(u)if x=0&y{0,1,+,}.dG(u)+n1if x=1&y{0,1,+,}.2dG(u)if x=+&y{0,1,+,}.n1if x=&y{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy}}\left( G \right)}}\left( u \right) = \{ \begin{array}{*{20}{c}} {{d_G}\left( u \right)} \hfill & {if{\rm{ }}x = 0\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {{d_G}\left( u \right) + n - 1} \hfill & {if{\rm{ }}x = 1\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {2{d_G}\left( u \right)} \hfill & {if{\rm{ }}x = + \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n - 1} \hfill & {if{\rm{ }}x = - \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

dTxy(G)(e)={2if y=0&x{0,1,+,}.m+1if y=1&x{0,1,+,}.dG(a)+dG(b)if y=+&x{0,1,+,}.m+3dG(a)dG(b)if y=&x{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy}}\left( G \right)}}\left( e \right) = \{ \begin{array}{*{20}{c}} 2 \hfill & {if{\rm{ }}y = 0\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {m + 1} \hfill & {if{\rm{ }}y = 1\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {{d_G}\left( a \right) + {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = + \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {m + 3 - {d_G}\left( a \right) - {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = - \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

Proposition 4.

Let G be a (n,m)-graph. Then the degree of point vertex u and line vertex e(= ab in G) in Txyz(G) when z = − are

dTxy(G)(u)={mdG(u)if x=0&y{0,1,+,}.n+m1dG(u)if x=1&y{0,1,+,}.mif x=+&y{0,1,+,}.n+m12dG(u)if x=&y{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy - }}\left( G \right)}}\left( u \right) = \{ \begin{array}{*{20}{c}} {m - {d_G}\left( u \right)} \hfill & {if{\rm{ }}x = 0\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m - 1 - {d_G}\left( u \right)} \hfill & {if{\rm{ }}x = 1\& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ m \hfill & {if{\rm{ }}x = + \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m - 1 - 2{d_G}\left( u \right)} \hfill & {if{\rm{ }}x = - \& y\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

dTxy(G)(e)={n2if y=0&x{0,1,+,}.n+m3if y=1&x{0,1,+,}.n+dG(a)+dG(b)4if y=+&x{0,1,+,}.n+m1dG(a)dG(b)if y=&x{0,1,+,}.$\begin{array}{} \displaystyle {d_{{T^{xy - }}\left( G \right)}}\left( e \right) = \{ \begin{array}{*{20}{c}} {n - 2} \hfill & {if{\rm{ }}y = 0\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m - 3} \hfill & {if{\rm{ }}y = 1\& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + {d_G}\left( a \right) + {d_G}\left( b \right) - 4} \hfill & {if{\rm{ }}y = + \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ {n + m - 1 - {d_G}\left( a \right) - {d_G}\left( b \right)} \hfill & {if{\rm{ }}y = - \& x\varepsilon \left\{ {0,1, + , - } \right\}.} \hfill \\ \end{array} \end{array}$

Proposition 5.

Let G1and G2be the graphs. If (u,ν) is a vertex of G1 ×CG2, then

dG1×CG2(u,ν)={dC(G1)(u)+dG2(ν)ifuV(C(G1))V(G1),νV(G2)dC(G1)(u)ifuV(C(G1))E(G1),νV(G2).$$\begin{array}{} \displaystyle {d_{{G_1}{ \times _C}{G_2}}}(u,\nu ) = \left\{ {\begin{array}{*{20}{c}} {{d_{C({G_1})}}(u) + {d_{{G_2}}}(\nu )} & {if\;u \in V(C({G_1})) \cap V({G_1}),\;\nu \in V({G_2})} \\ {{d_{C({G_1})}}(u)} & {if\;u \in V(C({G_1})) \cap E({G_1}),\;\nu \in V({G_2}).} \\ \end{array}} \right. \end{array}$$

We are now prepared to state and prove our main results.

Theorem 6.

Let G1and G2be the graphs. Then

αλ(G1×CG2)=uV(C(G1))V(G1)νV(G2)[dC(G1)(u)+dG2(ν)]λ+νV(G2)eV(C(G1))E(G1)dC(G1)λ(e).$$\begin{array}{} \displaystyle {\alpha _\lambda }({G_1}{ \times _C}{G_2}) = \sum\limits_{u \in V(C({G_1})) \cap V({G_1})} {\sum\limits_{\nu \in V({G_2})} {{{[{d_{C({G_1})}}(u) + {d_{{G_2}}}(\nu )]}^\lambda }} } + \sum\limits_{\nu \in V({G_2})} {\sum\limits_{e \in V(C({G_1})) \cap E({G_1})} {d_{C({G_1})}^\lambda } } (e). \end{array}$$

Proof.

By the definition of first general Zagreb index, we have

αλ(G1×CG2)=(u,ν)V(G1×CG2)dG1×CG2λ(u,ν)$$\begin{array}{} \displaystyle {\alpha _\lambda }({G_1}{ \times _C}{G_2}) = \sum\limits_{(u,\nu ) \in V({G_1}{ \times _C}{G_2})} {d_{{G_1}{ \times _C}{G_2}}^\lambda } (u,\nu ) \end{array}$$

We partition V(G1 ×CG2) into V (C(G1)) ∩ V(G1) and V (C(G1)) ∩ E(G1) and from Proposition 5, we have

αλ(G1×CG2)=uV(C(G1))V(G1)νV(G2)[dC(G1)(u)+dG2(ν)]λ+νV(G2)eV(C(G1))E(G1)dC(G1)λ(e).$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{\alpha _\lambda }({G_1}{ \times _C}{G_2})} \hfill & = \hfill & {\sum\limits_{u \in V(C({G_1})) \cap V({G_1})} {\sum\limits_{\nu \in V({G_2})} {{{[{d_{C({G_1})}}(u) + {d_{{G_2}}}(\nu )]}^\lambda }} } } \hfill \\ {} \hfill & {} \hfill & { + \sum\limits_{\nu \in V({G_2})} {\sum\limits_{e \in V(C({G_1})) \cap E({G_1})} {d_{C({G_1})}^\lambda } } (e).} \hfill \\ \end{array} \end{array}$$

For λ = 2,3 in (2), we get the following equations.

M1(G1×CG2)=uV(C(G1))V(G1)νV(G2)[dC(G1)2(u)+dG22(ν)+2dC(G1)(u)dG2(ν)]+νV(G2)eV(C(G1))E(G1)dC(G1)2(e).$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{M_1}({G_1}{ \times _C}{G_2})} \hfill & = \hfill & {\sum\limits_{u \in V(C({G_1})) \cap V({G_1})} {\sum\limits_{\nu \in V({G_2})} {[d_{C({G_1})}^2(u) + d_{{G_2}}^2(\nu ) + 2{d_{C({G_1})}}(u){d_{{G_2}}}(\nu )]} } } \hfill \\ {} \hfill & {} \hfill & { + \sum\limits_{\nu \in V({G_2})} {\sum\limits_{e \in V(C({G_1})) \cap E({G_1})} {d_{C({G_1})}^2} } (e).} \hfill \\ \end{array} \end{array}$$

and

F(G1×CG2)=uV(C(G1))V(G1)νV(G2)[dC(G1)3(u)+dG23(ν)+3dC(G1)2(u)dG2(ν)+3dC(G1)(u)dG22(ν)] +νV(G2)eV(C(G1))E(G1)dC(G1)3(e).$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {F({G_1}{ \times _C}{G_2})} \hfill & = \hfill & {\sum\limits_{u \in V(C({G_1})) \cap V({G_1})} {\sum\limits_{\nu \in V({G_2})} {[d_{C({G_1})}^3(u) + d_{{G_2}}^3(\nu ) + 3d_{C({G_1})}^2(u){d_{{G_2}}}(\nu ) + 3{d_{C({G_1})}}(u)d_{{G_2}}^2(\nu )]} } {\rm{ }}} \hfill \\ {} \hfill & {} \hfill & { + \sum\limits_{\nu \in V({G_2})} {\sum\limits_{e \in V(C({G_1})) \cap E({G_1})} {d_{C({G_1})}^3} } (e).} \hfill \\ \end{array} \end{array}$$

For a given graph Gi, its vertex and edge sets will be denoted by V(Gi) and E(Gi), respectively, and their cardinalities by ni and mi, respectively, where i = 1,2.

By plugging the corresponding degrees of vertices of Txy0 from Proposition 1 in (3) and (4), bearing in mind that νV(G)dG(ν)=2m$\begin{array}{} \displaystyle \sum\limits_{\nu \in V(G)} {{d_G}} (\nu ) = 2m \end{array}$, νV(G)=n$\begin{array}{} \displaystyle \sum\limits_{\nu \in V(G)} = n \end{array}$ and eE(G)=m$\begin{array}{} \displaystyle \sum\limits_{e \in E(G)} = m \end{array}$, we get the following two theorems.

Theorem 7.

Let G1and G2be the graphs. Then

M1(G1 ×T000G2) = n1M1(G2)

M1(G1 ×T100G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1)

M1(G1 ×T+00G2) = n2M1(G1) + n1M1(G2) + 8m1m2

M1(G1 ×T−00G2) = n1n2(n1 − 1)2 + n2M1(G1) − 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) − 8m1m2

M1(G1 ×T010G2) = n1M1(G2) + n2m1(m1 − 1)2

M1(G1 ×T110G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2m1(m1 − 1)2

M1(G1 ×T+10G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2m1(m1 − 1)2

M1(G1 ×T−10G2) = n1n2(n1 − 1)2 + n2M1(G1) − 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) − 8m1m2 + n2m1(m1 − 1)2

M1(G1 ×T0+0G2) = n1M1(G2) + n2[HM(G1) + 4m1 − 4M1(G1)]

M1(G1 ×T1+0G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2[HM(G1) + 4m1 − 4M1(G1)]

M1(G1 ×T++0G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2[HM(G1) + 4m1 − 4M1(G1)]

M1(G1 ×T−+0G2) = n1n2(n1 − 1)2 + n2M1(G1) − 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) − 8m1m2 + n2[HM(G1) + 4m1 − 4M1(G1)]

M1(G1 ×T0−0G2) = n1M1(G2) + n2[m1(m1 + 1)2 + HM(G1) − 2(m1 + 1)M1(G1)]

M1(G1 ×T1−0G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2[m1(m1 + 1)2 + HM(G1) − 2(m1 + 1)M1(G1)]

M1(G1 ×T+−0G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2[m1(m1 + 1)2 + HM(G1) − 2(m1 + 1)M1(G1)]

M1(G1 ×T−−0G2) = n1n2(n1 − 1)2 + n2M1(G1) − 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) − 8m1m2 + n2[m1(m1 + 1)2 + HM(G1) − 2(m1 + 1)M1(G1)].

Theorem 8.

Let G1and G2be the graphs. Then

F(G1 ×T000G2) = n1F(G2)

F(G1 ×T100G2) = n1n2(n1 − 1)3 + n1F(G2) + 6(n1 − 1)2n1m2 + 3(n1 − 1)n1M1(G2)

F(G1 ×T+00G2) = n2F(G1) + n1F(G2) + 6m2M1(G1) + 6m1M1(G2)

F(G1 ×T−00G2) = n1n2(n1 −1)3n2F(G1)−6n2m1(n1 −1)2 +3n2(n1 −1)M1(G1)+n1F(G2)+6n1m2(n1 −1)2 + 6m2M1(G1) − 24m1m2(n1 − 1) + 3n1(n1 − 1)M1(G2) − 6m1M1(G2)

F(G1 ×T010G2) = n1F(G2) + n2m1(m1 − 1)3

F(G1 ×T110G2) = n1n2(n1 − 1)3 + n1F(G2) + 6(n1 − 1)2n1m2 + 3(n1 − 1)n1M1(G2) + n2m1(m1 − 1)3

F(G1 ×T+10G2) = n2F(G1) + n1F(G2) + 6m2M1(G1) + 6m1M1(G2) + n2m1(m1 − 1)3

F(G1 ×T−10G2) = n1n2(n1 −1)3n2F(G1)−6n2m1(n1 −1)2 +3n2(n1 −1)M1(G1)+n1F(G2)+6n1m2(n1 −1)2 + 6m2M1(G1) − 24m1m2(n1 − 1) + 3n1(n1 − 1)M1(G2) − 6m1M1(G2) + n2m1(m1 − 1)3

F(G1×T0+0G2)=n1F(G2)+n2[M13(G1)8m16HM(G1)+12M1(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0+0}}G_{2})=n_{1}F(G_{2})+n_{2}[M_{1}^3(G_{1})-8m_{1}-6HM(G_{1})+12M_{1}(G_{1})] \end{array}$

F(G1×T1+0G2)=n1n2(n11)3+n1F(G2)+6(n11)2n1m2+3(n11)n1M1(G2)+n2[M13(G1)8m16HM(G1)+12M1(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1+0}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{1}F(G_{2})+6(n_{1}-1)^2n_{1}m_{2}+3(n_{1}-1)n_{1}M_{1}(G_{2})+n_{2}[M_{1}^3(G_{1})-8m_{1}-6HM(G_{1})+12M_{1}(G_{1})] \end{array}$

F(G1×T++0G2)=n2F(G1)+n1F(G2)+6m2M1(G1)+6m1M1(G2)+n2[M13(G1)8m16HM(G1)+12M1(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{++0}}G_{2})=n_{2}F(G_{1})+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})+6m_{1}M_{1}(G_{2})+n_{2}[M_{1}^3(G_{1})-8m_{1}-6HM(G_{1})+12M_{1}(G_{1})] \end{array}$

F(G1×T+0G2)=n1n2(n11)3n2F(G1)6n2m1(n11)2+3n2(n11)M1(G1)+n1F(G2)+6n1m2(n11)2+6m2M1(G1)24m1m2(n11)+3n1(n11)M1(G2)6m1M1(G2)+n2[M13(G1)8m16HM(G1)+12M1(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{-+0}}G_{2})=n_{1}n_{2}(n_{1}-1)^3-n_{2}F(G_{1})-6n_{2}m_{1}(n_{1}-1)^2+3n_{2}(n_{1}-1)M_{1}(G_{1})+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+6m_{2}M_{1}(G_{1})-24m_{1}m_{2}(n_{1}-1)+3n_{1}(n_{1}-1)M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}[M_{1}^3(G_{1})-8m_{1}-6HM(G_{1})+12M_{1}(G_{1})] \end{array}$

F(G1×T00G2)=n1F(G2)+n2[m1(m1+1)3M13(G1)3(m1+1)2M1(G1)+3(m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0-0}}G_{2})=n_{1}F(G_{2})+n_{2}[m_{1}(m_{1}+1)^3-M_{1}^3(G_{1})-3(m_{1}+1)^2M_{1}(G_{1})+3(m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T10G2)=n1n2(n11)3+n1F(G2)+6(n11)2n1m2+3(n11)n1M1(G2)+n2[m1(m1+1)3M13(G1)3(m1+1)2M1(G1)+3(m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1-0}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{1}F(G_{2})+6(n_{1}-1)^2n_{1}m_{2}+3(n_{1}-1)n_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+1)^3-M_{1}^3(G_{1})-3(m_{1}+1)^2M_{1}(G_{1})+3(m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T+0G2)=n2F(G1)+n1F(G2)+6m2M1(G1)+6m1M1(G2)+n2[m1(m1+1)3M13(G1)3(m1+1)2M1(G1)+3(m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+-0}}G_{2})=n_{2}F(G_{1})+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})+6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+1)^3-M_{1}^3(G_{1})-3(m_{1}+1)^2M_{1}(G_{1})+3(m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T0G2)=n1n2(n11)3n2F(G1)6n2m1(n11)2+3n2(n11)M1(G1)+n1F(G2)+6n1m2(n11)2+6m2M1(G1)24m1m2(n11)+3n1(n11)M1(G2)6m1M1(G2)+n2[m1(m1+1)3M13(G1)3(m1+1)2M1(G1)+3(m1+1)HM(G1)].$\begin{array}{} \displaystyle F(G_{1}\times_{T^{--0}}G_{2})=n_{1}n_{2}(n_{1}-1)^3-n_{2}F(G_{1})-6n_{2}m_{1}(n_{1}-1)^2+3n_{2}(n_{1}-1)M_{1}(G_{1})+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+6m_{2}M_{1}(G_{1})-24m_{1}m_{2}(n_{1}-1)+3n_{1}(n_{1}-1)M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+1)^3-M_{1}^3(G_{1})-3(m_{1}+1)^2M_{1}(G_{1})+3(m_{1}+1)HM(G_{1})]. \end{array}$

By plugging the corresponding degrees of vertices of Txy1 from Proposition 2 in (3) and (4), we get the following two theorems.

Theorem 9.

Let G1and G2be the graphs. Then

M1(G1×T001G2)=n1n2m12+n1M1(G2)+4n1m1m2+n2m1n12$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{001}}G_{2})=n_{1}n_{2}m_{1}^{2}+n_{1}M_{1}(G_{2})+4n_{1}m_{1}m_{2}+n_{2}m_{1}n_{1}^2 \end{array}$

M1(G1×T101G2)=n1n2(n1+m11)2+n1M1(G2)+4n1m2(n1+m11)+n2m1n12$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{101}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^2+n_{1}M_{1}(G_{2})+4n_{1}m_{2}(n_{1}+m_{1}-1)+n_{2}m_{1}n_{1}^2 \end{array}$

M1(G1×T+01G2)=n1n2m12+n2M1(G1)+4m12n2+n1M1(G2)+4m1m2n1+8m1m2+n2m1n12$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+01}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})+4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+8m_{1}m_{2}+n_{2}m_{1}n_{1}^2 \end{array}$

M1(G1×T01G2)=n1n2(n1+m11)2+n2M1(G1)4m1n2(n1+m11)+n1M1(G2)+4n1m2(n1+m11)8m1m2+n2m1n12$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{-01}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^2+n_{2}M_{1}(G_{1})-4m_{1}n_{2}(n_{1}+m_{1}-1)+n_{1}M_{1}(G_{2})+4n_{1}m_{2}(n_{1}+m_{1}-1)-8m_{1}m_{2}+n_{2}m_{1}n_{1}^2 \end{array}$

M1(G1×T011G2)=n1n2m12+n1M1(G2)+4n1m1m2+n2m1(n1+m11)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{011}}G_{2})=n_{1}n_{2}m_{1}^{2}+n_{1}M_{1}(G_{2})+4n_{1}m_{1}m_{2}+n_{2}m_{1}(n_{1}+m_{1}-1)^2 \end{array}$

M1(G1 ×T111G2) = n1n2(n1 + m1 − 1)2 + n1M1(G2) + 4n1m2(n1 + m1 − 1) + n2m1(n1 + m1 − 1)2

M1(G1×T+11G2)=n1n2m12+n2M1(G1)+4m12n2+n1M1(G2)+4m1m2n1+8m1m2+n2m1(n1+m11)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+11}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})+4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+8m_{1}m_{2}+n_{2}m_{1}(n_{1}+m_{1}-1)^2 \end{array}$

M1(G1 ×T−11G2) = n1n2(n1 + m1 − 1)2 + n2M1(G1) − 4m1n2(n1 + m1 − 1) + n1M1(G2) + 4n1m2(n1 + m1 −1) − 8m1m2 + n2m1(n1 + m1 − 1)2

M1(G1×T0+1G2)=n1n2m12+n1M1(G2)+4n1m1m2+n2[m1(n12)2+HM(G1)+2(n12)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{0+1}}G_{2})=n_{1}n_{2}m_{1}^{2}+n_{1}M_{1}(G_{2})+4n_{1}m_{1}m_{2}+n_{2}[m_{1}(n_{1}-2)^2+HM(G_{1})+2(n_{1}-2)M_{1}(G_{1})] \end{array}$

M1(G1 ×T1+1G2) = n1n2(n1 + m1 − 1)2 + n1M1(G2) + 4n1m2(n1 + m1 − 1) + n2[m1(n1 − 2)2 + HM(G1) + 2(n1 − 2)M1(G1)]

M1(G1×T++1G2)=n1n2m12+n2M1(G1)+4m12n2+n1M1(G2)+4m1m2n1+8m1m2+n2[m1(n12)2+HM(G1)+2(n12)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{++1}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})+4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+8m_{1}m_{2}+n_{2}[m_{1}(n_{1}-2)^2+HM(G_{1})+2(n_{1}-2)M_{1}(G_{1})] \end{array}$

M1(G1 ×T−+1G2) = n1n2(n1 +m1 −1)2 +n2M1(G1)−4m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 8m1m2 + n2[m1(n1 − 2)2 + HM(G1) + 2(n1 − 2)M1(G1)]

M1(G1×T01G2)=n1n2m12+n1M1(G2)+4n1m1m2+n2[m1(n1+m1+1)2+HM(G1)2(n1+m1+1)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{0-1}}G_{2})=n_{1}n_{2}m_{1}^{2}+n_{1}M_{1}(G_{2})+4n_{1}m_{1}m_{2}+n_{2}[m_{1}(n_{1}+m_{1}+1)^2+HM(G_{1})-2(n_{1}+m_{1}+1)M_{1}(G_{1})] \end{array}$

M1(G1 ×T1−1G2) = n1n2(n1 +m1 +1)2 +n1M1(G2)+4n1m2(n1 +m1 −1)+n2[m1(n1 +m1 +1)2 +HM(G1)−2(n1 + m1 + 1)M1(G1)]

M1(G1×T+1G2)=n1n2m12+n2M1(G1)+4m12n2+n1M1(G2)+4m1m2n1+8m1m2+n2[m1(n1+m1+1)2+HM(G1)2(n1+m1+1)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+-1}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})+4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+8m_{1}m_{2}+n_{2}[m_{1}(n_{1}+m_{1}+1)^2+HM(G_{1})-2(n_{1}+m_{1}+1)M_{1}(G_{1})] \end{array}$

M1(G1 ×T−−1G2) = n1n2(n1 +m1 −1)2 +n2M1(G1)−4m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 8m1m2 + n2[m1(n1 + m1 + 1)2 + HM(G1) − 2(n1 + m1 + 1)M1(G1)].

Theorem 10.

Let G1and G2be the graphs. Then

F(G1×T001G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2m1n13$\begin{array}{} \displaystyle F(G_{1}\times_{T^{001}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}m_{1}n_{1}^3 \end{array}$

F(G1×T101G2)=n1n2(n1+m11)3+n1F(G2)+6n1m2(n1+m11)2+3n1(n1+m11)M1(G2)+n2m1n13$\begin{array}{} \displaystyle F(G_{1}\times_{T^{101}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}+m_{1}-1)^2+3n_{1}(n_{1}+m_{1}-1)M_{1}(G_{2})+n_{2}m_{1}n_{1}^3 \end{array}$

F(G1×T+01G2)=n1n2m13+n2F(G1)+6n2m13+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)+24m12m2+n2m1n13$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+01}}G_{2})=n_{1}n_{2}m_{1}^3+n_{2}F(G_{1})+6n_{2}m_{1}^3+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})+24m_{1}^2m_{2}+n_{2}m_{1}n_{1}^3 \end{array}$

F(G1×T01G2)=n1n2(n1+m11)3n2F(G1)+(6n1m26n2m1)(n1+m11)2+(3n2M1(G1)+3n1M1(G2)24m1m2)(n1+m11)+n1F(G2)+6m2M1(G1)6m1M1(G2)+n2m1n13$\begin{array}{} \displaystyle F(G_{1}\times_{T^{-01}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3-n_{2}F(G_{1})+(6n_{1}m_{2}-6n_{2}m_{1})(n_{1}+m_{1}-1)^2+(3n_{2}M_{1}(G_{1})+3n_{1}M_{1}(G_{2})-24m_{1}m_{2})(n_{1}+m_{1}-1)+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})-6m_{1}M_{1}(G_{2})+n_{2}m_{1}n_{1}^3 \end{array}$

F(G1×T011G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2m1(n1+m11)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{011}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}m_{1}(n_{1}+m_{1}-1)^3 \end{array}$

F(G1 ×T111G2) = n1n2(n1 +m1 −1)3 +n1F(G2)+6n1m2(n1 +m1 −1)2 +3n1(n1 +m1 −1)M1(G2)+n2m1(n1 + m1 − 1)3

F(G1×T+11G2)=n1n2m13+n2F(G1)+6n2m13+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)+24m12m2+n2m1(n1+m11)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+11}}G_{2})=n_{1}n_{2}m_{1}^3+n_{2}F(G_{1})+6n_{2}m_{1}^3+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})+24m_{1}^2m_{2}+n_{2}m_{1}(n_{1}+m_{1}-1)^3 \end{array}$

F(G1 ×T−11G2) = n1n2(n1 +m1 −1)3n2F(G1)+(6n1m2 −6n2m1)(n1 +m1 −1)2 +(3n2M1(G1)+3n1M1(G2)−24m1m2)(n1 + m1 − 1) + n1F(G2) + 6m2M1(G1) − 6m1M1(G2) + n2m1(n1 + m1 − 1)3

F(G1×T0+1G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2[m1(n12)3+M13(G1)+3(n12)2M1(G1)+3(n12)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0+1}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}-2)^3+M_{1}^3(G_{1})+3(n_{1}-2)^2M_{1}(G_{1})+3(n_{1}-2)HM(G_{1})] \end{array}$

F(G1×T1+1G2)=n1n2(n1+m11)3+n1F(G2)+6n1m2(n1+m11)2+3n1(n1+m11)M1(G2)+n2[m1(n12)3+M13(G1)+3(n12)2M1(G1)+3(n12)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1+1}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}+m_{1}-1)^2+3n_{1}(n_{1}+m_{1}-1)M_{1}(G_{2})+n_{2}[m_{1}(n_{1}-2)^3+M_{1}^3(G_{1})+3(n_{1}-2)^2M_{1}(G_{1})+3(n_{1}-2)HM(G_{1})] \end{array}$

F(G1×T++1G2)=n1n2m13+n2F(G1)+6n2m13+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)+24m12m2+n2[m1(n12)3+M13(G1)+3(n12)2M1(G1)+3(n12)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{++1}}G_{2})=n_{1}n_{2}m_{1}^3+n_{2}F(G_{1})+6n_{2}m_{1}^3+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})+24m_{1}^2m_{2}+n_{2}[m_{1}(n_{1}-2)^3+M_{1}^3(G_{1})+3(n_{1}-2)^2M_{1}(G_{1})+3(n_{1}-2)HM(G_{1})] \end{array}$

F(G1×T+1G2)=n1n2(n1+m11)3n2F(G1)+(6n1m26n2m1)(n1+m11)2+(3n2M1(G1)+3n1M1(G2)24m1m2)(n1+m11)+n1F(G2)+6m2M1(G1)6m1M1(G2)+n2[m1(n12)3+M13(G1)+3(n12)2M1(G1)+3(n12)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{-+1}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3-n_{2}F(G_{1})+(6n_{1}m_{2}-6n_{2}m_{1})(n_{1}+m_{1}-1)^2+(3n_{2}M_{1}(G_{1})+3n_{1}M_{1}(G_{2})-24m_{1}m_{2})(n_{1}+m_{1}-1)+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})-6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}-2)^3+M_{1}^3(G_{1})+3(n_{1}-2)^2M_{1}(G_{1})+3(n_{1}-2)HM(G_{1})] \end{array}$

F(G1×T01G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2[m1(n1+m1+1)3M13(G1)3(n1+m1+1)2M1(G1)+3(n1+m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0-1}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}+m_{1}+1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}+1)^2M_{1}(G_{1})+3(n_{1}+m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T11G2)=n1n2(n1+m11)3+n1F(G2)+6n1m2(n1+m11)2+3n1(n1+m11)M1(G2)+n2[m1(n1+m1+1)3M13(G1)3(n1+m1+1)2M1(G1)+3(n1+m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1-1}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}+m_{1}-1)^2+3n_{1}(n_{1}+m_{1}-1)M_{1}(G_{2})+n_{2}[m_{1}(n_{1}+m_{1}+1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}+1)^2M_{1}(G_{1})+3(n_{1}+m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T+1G2)=n1n2m13+n2F(G1)+6n2m13+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)+24m12m2+n2[m1(n1+m1+1)3M13(G1)3(n1+m1+1)2M1(G1)+3(n1+m1+1)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+-1}}G_{2})=n_{1}n_{2}m_{1}^3+n_{2}F(G_{1})+6n_{2}m_{1}^3+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})+24m_{1}^2m_{2}+n_{2}[m_{1}(n_{1}+m_{1}+1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}+1)^2M_{1}(G_{1})+3(n_{1}+m_{1}+1)HM(G_{1})] \end{array}$

F(G1×T1G2)=n1n2(n1+m11)3n2F(G1)+(6n1m26n2m1)(n1+m11)2+(3n2M1(G1)+3n1M1(G2)24m1m2)(n1+m11)+n1F(G2)+6m2M1(G1)6m1M1(G2)+n2[m1(n1+m1+1)3M13(G1)3(n1+m1+1)2M1(G1)+3(n1+m1+1)HM(G1)].$\begin{array}{} \displaystyle F(G_{1}\times_{T^{--1}}G_{2})=n_{1}n_{2}(n_{1}+m_{1}-1)^3-n_{2}F(G_{1})+(6n_{1}m_{2}-6n_{2}m_{1})(n_{1}+m_{1}-1)^2+(3n_{2}M_{1}(G_{1})+3n_{1}M_{1}(G_{2})-24m_{1}m_{2})(n_{1}+m_{1}-1)+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})-6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}+m_{1}+1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}+1)^2M_{1}(G_{1})+3(n_{1}+m_{1}+1)HM(G_{1})]. \end{array}$

By plugging the corresponding degrees of vertices of Txy+ from Proposition 3 in (3) and (4), we get the following two theorems.

Theorem 11.

Let G1and G2be the graphs. Then

M1(G1 ×T00+G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + 4n2m1

M1(G1 ×T10+G2) = n1n2(n1 − 1)2 + n2M1(G1) + 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) + 8m1m2 + 4n2m1

M1(G1 ×T+0+G2) = 4n2M1(G1) + n1M1(G2) + 16m1m2 + 4n2m1

M1(G1 ×T−0+G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + 4n2m1

M1(G1 ×T01+G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2m1(m1 + 1)2

M1(G1 ×T11+G2) = n1n2(n1 − 1)2 + n2M1(G1) + 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) + 8m1m2 + n2m1(m1 + 1)2

M1(G1 ×T+1+G2) = 4n2M1(G1) + n1M1(G2) + 16m1m2 + n2m1(m1 + 1)2

M1(G1 ×T−1+G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2m1(m1 + 1)2

M1(G1 ×T0++G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2HM(G1)

M1(G1 ×T1++G2) = n1n2(n1 − 1)2 + n2M1(G1) + 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) + 8m1m2 + n2HM(G1)

M1(G1 ×T+++G2) = 4n2M1(G1) + n1M1(G2) + 16m1m2 + n2HM(G1)

M1(G1 ×T−++G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2HM(G1)

M1(G1 ×T0−+G2) = n2M1(G1) + n1M1(G2) + 8m1m2 + n2[m1(m1 + 3)2 + HM(G1) − 2(m1 + 3)M1(G1)]

M1(G1 ×T1−+G2) = n1n2(n1 − 1)2 + n2M1(G1) + 4n2m1(n1 − 1) + n1M1(G2) + 4n1m2(n1 − 1) + 8m1m2 + n2[m1(m1 + 3)2 + HM(G1) − 2(m1 + 3)M1(G1)]

M1(G1 ×T+−+G2) = 4n2M1(G1) + n1M1(G2) + 16m1m2 + n2[m1(m1 + 3)2 + HM(G1) − 2(m1 + 3)M1(G1)]

M1(G1 ×T−−+G2) = n1n2(n1 − 1)2 + n1M1(G2) + 4n1m2(n1 − 1) + n2[m1(m1 + 3)2 + HM(G1) − 2(m1 + 3)M1(G1)].

Theorem 12.

Let G1and G2be the graphs. Then

F(G1 ×T00+G2) = n2F(G1) + n1F(G2) + 6m2M1(G1) + 6m1M1(G2) + 8n2m1

F(G1 ×T10+G2) = n1n2(n1 −1)3 +n2F(G1)+6(n1 −1)2n2m1 +3(n1 −1)n2M1(G1)+n1F(G2)+6n1m2(n1 −1)2 + 6m2M1(G1) + 24m1m2(n1 − 1) + 3n1(n1 − 1)M1(G2) + 6m1M1(G2) + 8n2m1

F(G1 ×T+0+G2) = 8n2F(G1) + n1F(G2) + 24m2M1(G1) + 12m1M1(G2) + 8n2m1

F(G1 ×T−0+G2) = n1n2(n1 − 1)3 + n1F(G2) + 6n1m2(n1 − 1)2 + 3(n1 − 1)n1M1(G2) + 8n2m1

F(G1 ×T01+G2) = n2F(G1) + n1F(G2) + 6m2M1(G1) + 6m1M1(G2) + n2m1(m1 + 1)3

F(G1 ×T11+G2) = n1n2(n1 −1)3 +n2F(G1)+6(n1 −1)2n2m1 +3(n1 −1)n2M1(G1)+n1F(G2)+6n1m2(n1 −1)2 + 6m2M1(G1) + 24m1m2(n1 − 1) + 3n1(n1 − 1)M1(G2) + 6m1M1(G2) + n2m1(m1 + 1)3

F(G1 ×T+1+G2) = 8n2F(G1) + n1F(G2) + 24m2M1(G1) + 12m1M1(G2) + n2m1(m1 + 1)3

F(G1 ×T−1+G2) = n1n2(n1 − 1)3 + n1F(G2) + 6n1m2(n1 − 1)2 + 3(n1 − 1)n1M1(G2) + n2m1(m1 + 1)3

F(G1×T0++G2)=n2F(G1)+n1F(G2)+6m2M1(G1)+6m1M1(G2)+n2M13(G1)$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0++}}G_{2})=n_{2}F(G_{1})+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})+6m_{1}M_{1}(G_{2})+n_{2}M_{1}^{3}(G_{1}) \end{array}$

F(G1×T1++G2)=n1n2(n11)3+n2F(G1)+6(n11)2n2m1+3(n11)n2M1(G1)+n1F(G2)+6n1m2(n11)2+6m2M1(G1)+24m1m2(n11)+3n1(n11)M1(G2)+6m1M1(G2)+n2M13(G1)$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1++}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{2}F(G_{1})+6(n_{1}-1)^2n_{2}m_{1}+3(n_{1}-1)n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+6m_{2}M_{1}(G_{1})+24m_{1}m_{2}(n_{1}-1)+3n_{1}(n_{1}-1)M_{1}(G_{2})+6m_{1}M_{1}(G_{2})+n_{2}M_{1}^{3}(G_{1}) \end{array}$

F(G1×T+++G2)=8n2F(G1)+n1F(G2)+24m2M1(G1)+12m1M1(G2)+n2M13(G1)$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+++}}G_{2})=8n_{2}F(G_{1})+n_{1}F(G_{2})+24m_{2}M_{1}(G_{1})+12m_{1}M_{1}(G_{2})+n_{2}M_{1}^{3}(G_{1}) \end{array}$

F(G1×T++G2)=n1n2(n11)3+n1F(G2)+6n1m2(n11)2+3(n11)n1M1(G2)+n2M13(G1)$\begin{array}{} \displaystyle F(G_{1}\times_{T^{-++}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+3(n_{1}-1)n_{1}M_{1}(G_{2})+n_{2}M_{1}^{3}(G_{1}) \end{array}$

F(G1×T0+G2)=n2F(G1)+n1F(G2)+6m2M1(G1)+6m1M1(G2)+n2[m1(m1+3)3M13(G1)3(m1+3)2M1(G1)+3(m1+3)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0-+}}G_{2})=n_{2}F(G_{1})+n_{1}F(G_{2})+6m_{2}M_{1}(G_{1})+6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+3)^3-M_{1}^{3}(G_{1})-3(m_{1}+3)^2M_{1}(G_{1})+3(m_{1}+3)HM(G_{1})] \end{array}$

F(G1×T1+G2)=n1n2(n11)3+n2F(G1)+6(n11)2n2m1+3(n11)n2M1(G1)+n1F(G2)+6n1m2(n11)2+6m2M1(G1)+24m1m2(n11)+3n1(n11)M1(G2)+6m1M1(G2)+n2[m1(m1+3)3M13(G1)3(m1+3)2M1(G1)+3(m1+3)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{1-+}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{2}F(G_{1})+6(n_{1}-1)^2n_{2}m_{1}+3(n_{1}-1)n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+6m_{2}M_{1}(G_{1})+24m_{1}m_{2}(n_{1}-1)+3n_{1}(n_{1}-1)M_{1}(G_{2})+6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+3)^3-M_{1}^{3}(G_{1})-3(m_{1}+3)^2M_{1}(G_{1})+3(m_{1}+3)HM(G_{1})] \end{array}$

F(G1×T++G2)=8n2F(G1)+n1F(G2)+24m2M1(G1)+12m1M1(G2)+n2[m1(m1+3)3M13(G1)3(m1+3)2M1(G1)+3(m1+3)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+-+}}G_{2})=8n_{2}F(G_{1})+n_{1}F(G_{2})+24m_{2}M_{1}(G_{1})+12m_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+3)^3-M_{1}^{3}(G_{1})-3(m_{1}+3)^2M_{1}(G_{1})+3(m_{1}+3)HM(G_{1})] \end{array}$

F(G1×T+G2)=n1n2(n11)3+n1F(G2)+6n1m2(n11)2+3(n11)n1M1(G2)+n2[m1(m1+3)3M13(G1)3(m1+3)2M1(G1)+3(m1+3)HM(G1)].$\begin{array}{} \displaystyle F(G_{1}\times_{T^{--+}}G_{2})=n_{1}n_{2}(n_{1}-1)^3+n_{1}F(G_{2})+6n_{1}m_{2}(n_{1}-1)^2+3(n_{1}-1)n_{1}M_{1}(G_{2})+n_{2}[m_{1}(m_{1}+3)^3-M_{1}^{3}(G_{1})-3(m_{1}+3)^2M_{1}(G_{1})+3(m_{1}+3)HM(G_{1})]. \end{array}$

By plugging the corresponding degrees of vertices of Txy from Proposition 4 in (3) and (4), we reach the following two theorems.

Theorem 13.

Let G1and G2be the graphs. Then

M1(G1×T00G2)=n1n2m12+n2M1(G1)4m12n2+n1M1(G2)+4m1m2n18m1m2+n2m1(n12)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{00-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})-4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}-8m_{1}m_{2}+n_{2}m_{1}(n_{1}-2)^2 \end{array}$

M1(G1 ×T10−G2) = n1n2(n1 + m1 − 1)2 + n2M1(G1) − 4m1n2(n1 + m1 − 1) + n1M1(G2) + 4n1m2(n1 + m1 −1) − 8m1m2 + n2m1(n1 − 2)2

M1(G1×T+0G2)=n1n2m12+n1M1(G2)+4m1m2n1+n2m1(n12)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+0-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+n_{2}m_{1}(n_{1}-2)^2 \end{array}$

M1(G1 ×T−0−G2) = n1n2(n1 +m1 −1)2 +4n2M1(G1)−8m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 16m1m2 + n2m1(n1 − 2)2

M1(G1×T01G2)=n1n2m12+n2M1(G1)4m12n2+n1M1(G2)+4m1m2n18m1m2+n2m1(n1+m13)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{01-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})-4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}-8m_{1}m_{2}+n_{2}m_{1}(n_{1}+m_{1}-3)^2 \end{array}$

M1(G1 ×T11−G2) = n1n2(n1 + m1 − 1)2 + n2M1(G1) − 4m1n2(n1 + m1 − 1) + n1M1(G2) + 4n1m2(n1 + m1 −1) − 8m1m2 + n2m1(n1 + m1 − 3)2

M1(G1×T+1G2)=n1n2m12+n1M1(G2)+4m1m2n1+n2m1(n1+m13)2$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+1-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+n_{2}m_{1}(n_{1}+m_{1}-3)^2 \end{array}$

M1(G1 ×T−1−G2) = n1n2(n1 +m1 −1)2 +4n2M1(G1)−8m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 16m1m2 + n2m1(n1 + m1 − 3)2

M1(G1×T0+G2)=n1n2m12+n2M1(G1)4m12n2+n1M1(G2)+4m1m2n18m1m2+n2[m1(n14)2+HM(G1)+2(n14)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{0+-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})-4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}-8m_{1}m_{2}+n_{2}[m_{1}(n_{1}-4)^2+HM(G_{1})+2(n_{1}-4)M_{1}(G_{1})] \end{array}$

M1(G1 ×T1+−G2) = n1n2(n1 +m1 −1)2 +n2M1(G1)−4m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 8m1m2 + n2[m1(n1 − 4)2 + HM(G1) + 2(n1 − 4)M1(G1)]

M1(G1×T++G2)=n1n2m12+n1M1(G2)+4m1m2n1+n2[m1(n14)2+HM(G1)+2(n14)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{++-}}G_{2})=n_{1}n_{2}m_{1}^2+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+n_{2}[m_{1}(n_{1}-4)^2+HM(G_{1})+2(n_{1}-4)M_{1}(G_{1})] \end{array}$

M1(G1 ×T−+−G2) = n1n2(n1 + m1 − 1)2 + 4n2M1(G1) − 8m1n2(n1 + m1 − 1) + n1M1(G2) + 4n1m2(n1 + m1 − 1) − 16m1m2 + n2[m1(n1 − 4)2 + HM(G1) + 2(n1 − 4)M1(G1)]

M1(G1×T0G2)=n1n2m12+n2M1(G1)4m12n2+n1M1(G2)+4m1m2n18m1m2+n2[m1(n1+m11)2+HM(G1)2(n1+m11)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{0--}}G_{2})=n_{1}n_{2}m_{1}^2+n_{2}M_{1}(G_{1})-4m_{1}^2n_{2}+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}-8m_{1}m_{2}+n_{2}[m_{1}(n_{1}+m_{1}-1)^2+HM(G_{1})-2(n_{1}+m_{1}-1)M_{1}(G_{1})] \end{array}$

M1(G1 ×T1−−G2) = n1n2(n1 +m1 −1)2 +n2M1(G1)−4m1n2(n1 +m1 −1)+n1M1(G2)+4n1m2(n1 +m1 −1) − 8m1m2 + n2[m1(n1 + m1 − 1)2 + HM(G1) − 2(n1 + m1 − 1)M1(G1)]

M1(G1×T+G2)=n1n2m12+n1M1(G2)+4m1m2n1+n2[m1(n1+m11)2+HM(G1)2(n1+m11)M1(G1)]$\begin{array}{} \displaystyle M_{1}(G_{1}\times_{T^{+--}}G_{2})=n_{1}n_{2}m_{1}^2+n_{1}M_{1}(G_{2})+4m_{1}m_{2}n_{1}+n_{2}[m_{1}(n_{1}+m_{1}-1)^2+HM(G_{1})-2(n_{1}+m_{1}-1)M_{1}(G_{1})] \end{array}$

M1(G1 ×T−−−G2) = n1n2(n1 + m1 − 1)2 + 4n2M1(G1) − 8m1n2(n1 + m1 − 1) + n1M1(G2) + 4n1m2(n1 + m1 − 1) − 16m1m2 + n2[m1(n1 + m1 − 1)2 + HM(G1) − 2(n1 + m1 − 1)M1(G1)].

Theorem 14.

Let G1and G2be the graphs. Then

F(G1×T00G2)=n1n2m13n2F(G1)6m13n2+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)24m12m2+3m1n1M1(G2)6m1M1(G2)+n2m1(n12)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{00-}}G_{2})=n_{1}n_{2}m_{1}^3-n_{2}F(G_{1})-6m_{1}^3n_{2}+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})-24m_{1}^2m_{2}+3m_{1}n_{1}M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}m_{1}(n_{1}-2)^3 \end{array}$

F(G1 ×T10−G2) = n1n2(n1 +m1 −1)3n2F(G1)+(6n1m2 −6m1n2)(n1 +m1 −1)2 +(3n2M1(G1)+3n1M1(G2)−24m1m2)(n1 + m1 − 1) + n1F(G2) − 6m1M1(G2) + 6m2M1(G1) + n2m1(n1 − 2)3

F(G1×T+0G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2m1(n12)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+0-}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}m_{1}(n_{1}-2)^3 \end{array}$

F(G1 ×T−0−G2) = n1n2(n1 + m1 − 1)3 + (6n1m2 − 12n2m1)(n1 + m1 − 1)2 + (12n2M1(G1) + 3n1M1(G2) −48m1m2)(n1 + m1 − 1) − 8n2F(G1) + n1F(G2) + 24m2M1(G1) − 12m1M1(G2) + n2m1(n1 − 2)3

F(G1×T01G2)=n1n2m13n2F(G1)6m13n2+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)24m12m2+3m1n1M1(G2)6m1M1(G2)+n2m1(n1+m13)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{01-}}G_{2})=n_{1}n_{2}m_{1}^3-n_{2}F(G_{1})-6m_{1}^3n_{2}+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})-24m_{1}^2m_{2}+3m_{1}n_{1}M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}m_{1}(n_{1}+m_{1}-3)^3 \end{array}$

F(G1 ×T11−G2) = n1n2(n1 +m1 −1)3n2F(G1)+(6n1m2 −6m1n2)(n1 +m1 −1)2 +(3n2M1(G1)+3n1M1(G2)−24m1m2)(n1 + m1 − 1) + n1F(G2) − 6m1M1(G2) + 6m2M1(G1) + n2m1(n1 + m1 − 3)3

F(G1×T+1G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2m1(n1+m13)3$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+1-}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}m_{1}(n_{1}+m_{1}-3)^3 \end{array}$

F(G1 ×T−1−G2) = n1n2(n1 + m1 − 1)3 + (6n1m2 − 12n2m1)(n1 + m1 − 1)2 + (12n2M1(G1) + 3n1M1(G2) −48m1m2)(n1 + m1 − 1) − 8n2F(G1) + n1F(G2) + 24m2M1(G1) − 12m1M1(G2) + n2m1(n1 + m1 − 3)3

F(G1×T0+G2)=n1n2m13n2F(G1)6m13n2+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)24m12m2+3m1n1M1(G2)6m1M1(G2)+n2[m1(n14)3+M13(G1)+3(n14)2M1(G1)+3(n14)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0+-}}G_{2})=n_{1}n_{2}m_{1}^3-n_{2}F(G_{1})-6m_{1}^3n_{2}+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})-24m_{1}^2m_{2}+3m_{1}n_{1}M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}-4)^3+M_{1}^3(G_{1})+3(n_{1}-4)^2M_{1}(G_{1})+3(n_{1}-4)HM(G_{1})] \end{array}$

F(G1 ×T1+−G2) = n1n2(n1 +m1 −1)3n2F(G1)+(6n1m2 −6m1n2)(n1 +m1 −1)2 +(3n2M1(G1)+3n1M1(G2)−24m1m2)(n1 +m1 −1)+n1F(G2)−6m1M1(G2)+6m2M1(G1)+n2[m1(n1 −4)3 +M31(G1)+3(n1 −4)2M1(G1)+ 3(n1 − 4)HM(G1)]

F(G1×T++G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2[m1(n14)3+M13(G1)+3(n14)2M1(G1)+3(n14)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{++-}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}-4)^3+M_{1}^3(G_{1})+3(n_{1}-4)^2M_{1}(G_{1})+3(n_{1}-4)HM(G_{1})] \end{array}$

F(G1 ×T−+−G2) = n1n2(n1 + m1 − 1)3 + (6n1m2 − 12n2m1)(n1 + m1 − 1)2 + (12n2M1(G1) + 3n1M1(G2) −48m1m2)(n1 +m1 −1)−8n2F(G1)+n1F(G2)+24m2M1(G1)−12m1M1(G2)+n2[m1(n1 −4)3 +M31(G1)+ 3(n1 − 4)2M1(G1) + 3(n1 − 4)HM(G1)]

F(G1×T0G2)=n1n2m13n2F(G1)6m13n2+3m1n2M1(G1)+n1F(G2)+6m12n1m2+6m2M1(G1)24m12m2+3m1n1M1(G2)6m1M1(G2)+n2[m1(n1+m11)3M13(G1)3(n1+m11)2M1(G1)+3(n1+m11)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{0--}}G_{2})=n_{1}n_{2}m_{1}^3-n_{2}F(G_{1})-6m_{1}^3n_{2}+3m_{1}n_{2}M_{1}(G_{1})+n_{1}F(G_{2})+6m_{1}^2n_{1}m_{2}+6m_{2}M_{1}(G_{1})-24m_{1}^2m_{2}+3m_{1}n_{1}M_{1}(G_{2})-6m_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}+m_{1}-1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}-1)^2M_{1}(G_{1})+3(n_{1}+m_{1}-1)HM(G_{1})] \end{array}$

F(G1 ×T1−−G2) = n1n2(n1 +m1 −1)3n2F(G1)+(6n1m2 −6m1n2)(n1 +m1 −1)2 +(3n2M1(G1)+3n1M1(G2)−24m1m2)(n1 + m1 − 1) + n1F(G2) − 6m1M1(G2) + 6m2M1(G1) + n2[m1(n1 + m1 − 1)3M31(G1) − 3(n1 + m1 − 1)2M1(G1) + 3(n1 + m1 − 1)HM(G1)]

F(G1×T+G2)=n1n2m13+n1F(G2)+6n1m2m12+3m1n1M1(G2)+n2[m1(n1+m11)3M13(G1)3(n1+m11)2M1(G1)+3(n1+m11)HM(G1)]$\begin{array}{} \displaystyle F(G_{1}\times_{T^{+--}}G_{2})=n_{1}n_{2}m_{1}^3+n_{1}F(G_{2})+6n_{1}m_{2}m_{1}^2+3m_{1}n_{1}M_{1}(G_{2})+n_{2}[m_{1}(n_{1}+m_{1}-1)^3-M_{1}^3(G_{1})-3(n_{1}+m_{1}-1)^2M_{1}(G_{1})+3(n_{1}+m_{1}-1)HM(G_{1})] \end{array}$

F(G1 ×T−−−G2) = n1n2(n1 + m1 − 1)3 + (6n1m2 − 12n2m1)(n1 + m1 − 1)2 + (12n2M1(G1) + 3n1M1(G2) −48m1m2)(n1 + m1 − 1) − 8n2F(G1) + n1F(G2) + 24m2M1(G1) − 12m1M1(G2) + n2[m1(n1 + m1 − 1)3M31(G1) − 3(n1 + m1 − 1)2M1(G1) + 3(n1 + m1 − 1)HM(G1)].

The expression for first Zagreb index of eighteen graph operations G1 ×TxyzG2 for x,y,z ∊ {+,−}, x,z ∊ {+,−} with y = 0, y,z ∊ {+,−} with x = 0 and z ∊ {+,−} with x = y = 0 are obtained by Basavanagoud and Patil in [6]. We include these results for the sake of completeness.

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