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On Degree-Based Topological Indices of Petersen Subdivision Graph

Mukhtar Ahmad1, Saddam Hussain2, Ulfat Parveen3, Iqra Zahid3, Muhammad Sultan3,Ather Qayyum3,∗
1Department of Mathematics, Khawaja Fareed University of Engineering and Information Technology

Rahim Yar Khan, Pakistan
itxmemuktar@gmail.com

2Department of Statistics, University of Mian Wali, Pakistan
saddamhussain.stat885@gmail.com

3Department of Mathematics, Institute of Southern Punjab Multan, Pakistan
uulfat05@gmail.com, iqraimran57@gmail.com, sultan.sadeeq7866127@gmail.com, atherqayyum@isp.edu.pk

∗Correspondence: atherqayyum@isp.edu.pk

Abstract. In this paper, we adequately describe the generalised petersen graph, expanding to thecategories of graphs. We created a petersen graph, which is cyclic and has vertices that are arrangedin the centre and nine gons plus one vertex, leading to the factorization of regular graphs. Petersengraph is still shown in graph theory literature, nevertheless.

1. Introduction
Named after Julius Petersen, a Danish mathematician, the graph of Petersen is(from 1839 to1910). Petersen researched factorizations of normal factorizations during the 1890s. In 1891, asignificant paper of graphs was published which is commemorated in that volume. Petersen provedthat any graph of 3-regular with at a I-factor includes much of the two bridges. Tait had writtena few years ago that he had shown I-factorable for each 3-regular graph,but that this outcomeIt was not valid without restriction. But Tait’s comment in 1898 was interpreted by petersen toimply that each 3-regular bridge less graph is L-factorable. If this outcome were valid, then itwould have been stronger than Theorem for Petersen. The key characteristics of the petersengraph were examined in detail in 1985. The graph of petersen continously to express in the entiregraph-theory education. We update our previous analysis in the present article by denoting extrarecently findings concerning the petersen graph.Julius Petersen’s ’Die Theorie der regulken graphs’ is an exceptional paper that developed a new
Received: 15 Apr 2023.
Key words and phrases. atom-bond connectivity index; reduced zagreb; randic indices; general connectivity index;petersen graph. 1

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 2
theory in graph theory, based on the exchange property of trees spanning and the cyclomaticnumber of trees spanningresently resently Zaib Hassan Niazi et.al[15].
1.1. The graph of petersen. Every petersen graph is cyclic graph and the graph G′ in general formconsists V having set of vertex and E having set of edge, if the natural number, there exist n thegraph with vertices are V (G′)=4n, edges are E(G′)=6n, and the specific of this graph is thatabout degree of every each vertex is P(k,t) = [d(x1),d(x2)]= 3. Then this graphic which is said tobe petersen graphic. Then petersen graphic is denoted by P[V (G′),E(G′)] =(4n,6n) petersen mapexplored by 1985, updated by new analysis.Sylvester’s association to graphs of invariants and covariants requires interpretation of principleof invariants in 1880s.
1.2. Graphical idea of petersen graph. If the set of natural number is Tn = {1,2,3, ...}, if thereexists n then graph with vertices are V (G′) = 4n, and edges E(G′) = 6n, in general form ofpetersen expressed by P[V (G′),E(G′)] =(4n,6n). This graph having a specification, that degree ofevery each vertex is P(k,t)= [d(x1),d(x2)]= 3.Now we write;
Tn =1,2,3, ....

V (G′) = 4n → [1]
E(G′) = 6n → [2]
k = d(x1)=3

t = d(x2)=3

Then
P(k,t) = [d(x1),d(x2)]= 3

P[V (G′),E(G′)] =(4n,6n).
Next we discuss the topological indices

Zagreb indices of the group were recognized in the early 1980s and are now known as thefirst and second zagreb indices. They are important molecular descriptors and have been closelycorrelated with chemical properties.[Degree based topological indices]The first zagreb index M1(G) is equal to the sum of the squares of the degrees of the vertices for

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 3
the (molecular) graph G[1]. It can also be considered as the sum over the edges of G, and M1(G)is defined as:[The first and second zagreb indices of some graph operations]

M1(G,x)=
∑

[x1,x2∈E(G)]

[d(x1)+d(x2)] (1)
The second zagreb index M2(G) is equal to the sum of the products of the degrees of the adjacentvertices for the pair of vertices for the (molecular) graph G, and M2(G) is defined as:[the first andsecond zagreb indices of some graph operations]

M2(G,x)=
∑

[x1,x2∈E(G)]

[d(x1)d(x2)] (2)
In 1972, the first zagreb index, a very old topological index, was launched and several variants ofthe zagreb index were subsequently proposed, e.g. Shirdel et al. described a novel index in 2013under the title of ’hyper-zagreb index’ and then it was identified as[2]: [A note on hyper-zagrebindex of graph operations]

HM1(G)=
∑

[x1,x2∈E(G)]

[d(x1)+d(x2)]
2 (3)

E. Deutshi and S. Klavzar,in 2015, defined a new polynomial, M-polynomial in the followingway, based on the degree of the vertex[3]:[COMPUTING HYPER ZAGREB INDEX AND M-POLYNOMIALS]
M1(G,y,z)=

∑
[x1,x2∈E(G)]

y [d(x1)]z[d(x2)] (4)
In Shuxian defined two polynomials related to the first zagreb index as in the form:

M∗1(G,x)=
∑

[xi∈V (G)]

[d(xi)][x
(xi)] (5)

M0(G,x)=
∑

[xi∈V (G)]

(x)[d(xi)] (6)
Two zagreb type polynomials are defined as follow:

Ma,b(G,x)=
∑

[xi,xj∈E(G)]

(x)[a{d(xi)}+b{d(xi)}] (7)
M′a,b(G,x)=

∑
[xi,xj∈E(G)]

(x)([a+{d(xi)}][b+{d(xi)}]) (8)
Todeshine et al. introduced two updated models of the zagreb index for moleculargraphs[4]:[MULTIPLICATIVE ZAGREB INDICES OF TREES] First multiplicative zagreb index formolecular graph G defined as follows:

PM1(G)=
∏

[x1,x2∈E(G)]

[d(x1)+d(x2)] (9)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 4
Second multiplicative zagreb index for molecular graph G defined as follows:

PM2(G)=
∏

[x1,x2∈E(G)]

[d(x1)×d(x2)] (10)
First multiplicative zagreb polynomial for molecular graph G defined as follows:

PM1(G,x)=
∏

[x1,x2∈E(G)]

X[d(x1)+d(x2)] (11)
Second multiplicative zagreb polynomial for molecular graph G defined as follows:

PM2(G,x)=
∏

[x1,x2∈E(G)]

X[d(x1)d(x2)] (12)
The first degree-based topological index was proposed by Milan Randic in 1975[5]:[Degree-BasedTopological Indices]

R1(α)(G)=
∑

[x1,x2∈E(G)]

[d(x1)+d(x2)]
α (13)

Atom-bond connectivity index (ABC) is a topological index used in chemistry, environmental sci-ences and pharmacology[6]: [Estrada, Torres, Rodriguez, and Gutman, 1998b]
ABC(G)=

∑
[x1,x2∈E(G)]

√
[d(x1)+d(x2)]−2
d(x1)×d(x2)

(14)
First, second and third reduced zagreb indices[7] are described as follow:

MR1(G)=
∑

[x1,x2∈E(G)]

|(d(x1)−1)+(d(x2)−1)| (15)
MR2(G)=

∑
[x1,x2∈E(G)]

[(d(x1)−1)(d(x2)−1)] (16)
MR3(G)=

∑
[x1,x2∈E(G)]

|(d(x1)−1)− (d(x2)−1)| (17)
RR(G)=

∑
[x1,x2∈E(G)]

√
d(x1)×d(x2) (18)

The reduced reciprocal randic index is defined as[8]:
RRR(G)=

∑
[x1,x2∈E(G)]

√
[d(x1)−1]× [d(x2)−1] (19)

Recently in 2015 Furtula and Gutman [8] introduced another topological index known as forgottenindex or F − index. For more detail on the F − index, we refer to the articles [9].The forgottenindex of a graph G is defined as[10, 11, 12].
F(G)=

∑
[x1,x2∈E(G)]

[(dx1)
2 +(dx2)

2] (20)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 5
The forgotten polynomial of a graph G is defined as:

F(G,x)=
∑

[x1,x2∈E(G)]

(x)[(dx1)
2+(dx2)

2] (21)
The symmetric division degree index of a connected graph G is defined as:

SDD(G)=
∑

[x1,x2∈E(G)]

mini(d(x1),d(x2))

max(d(x1),d(x2))
+
maxi(d(x1),d(x2))

mini(d(x1),d(x2))
(22)

There are two types of general connectivity index. The general randic index (or product-connectivityindex) was proposed by Bolloba and Erdos and is defined as follows:
M1(G)=

∑
[x2∈V (G)]

[dG(x2)]
2 (23)

where α is a real number. If α =−1
2
, then it becomes the randic index and if α =1 then it becomesthe second zagreb index. Zhou and Trinajstic developed the general sum-connectivity index: [Onthe general sum-connectivity index of trees]

M1(G)=
∑

[x1,x2∈E(G)]

[d(x1)+d(x2)]
α (24)

where α is a real number. If α = 1, then the general sum connectivity index becomes the firstzagreb index resently Asghar et.al[14].
2. Main Results

In this section, we established some results on degree based topological indices of Petersengraph.
Theorem 2.1 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, first zagrebpolynomials indices are,
M1(G,x) =[|6n|](x)(6)
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now first zagreb polynomials indices are i.e. ,
⇒ GA(R) = ∑y1,y2∈E(R) 2√dy1dy2dy1+dy2Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in first zagrebtopological index of the general form,

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 6
⇒ M1(G,x) =[|E(G)|](x)[(3)+(3)]

⇒ M1(G,x) =(6n)(x)(6)
⇒ M1(G,x) = (6n)(x)6.
M1(G,x) = (x)6× [General edges of Petersen graph]

Theorem 2.2 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, second zagrebpolynomials indices are,
M2(G,x) = (6n)(x)9
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now second zagreb polynomials indices are i.e. ,
M2(G,x) = ∑[x1,x2∈E(G)](x)[d(x1)×d(x2)] → [1]
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in second zagrebtopological index of the general form,
⇒ M2(G,x) =∑x1,x2∈E(G)(x)[(3)(3)]

⇒ M2(G,x) =[|E(G)|](x)9
⇒ M2(G,x) = (6n)(x)9.
M2(G,x) = (x)9× [General edges of petersen graph]

Theorem 2.3 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, randic indicesare,
R1(α)(G) = (6n)[6]α
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now randic indices are i.e. ,
R1(α)(G) = ∑[x1,x2∈E(G)][d(x1)+d(x2)]α
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in randic indices

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 7
topological index of the general form,
R1(α)(G) = ∑[x1,x2∈E(G)][d(x1)+d(x2)]α
In general form of topological index becomes;

⇒ R1(α)(G) = ∑[x1,x2∈E(G)][d(x1)+d(x2)]αNow putting values in above equation,
⇒ R1(α)(G) =[|E(G)|][(3)+(3)]α
⇒ R1(α)(G) =[|E(G)|][6]α
⇒ R1(α)(G) =[|6n|][6]α
⇒ R1(α)(G) = (6n)[6]α.
R1(α)(G) = [6α]×[ The general edges of petersen graph]

Theorem 2.4 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, reducedreciprocal randic are,
RRR(G) = 12n.
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now reduced reciprocal randic are i.e. ,
RR(G) = ∑[x1,x2∈E(G)] √d(x1)×d(x2)
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersengraph about every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in reducedreciprocal randic topological index of the general form,
⇒ RRR(G) = ∑[x1,x2∈E(G)] √[d(x1)−1]× [d(x2)−1]Now puttings the values then;

⇒ RRR(G) = ∑[x1,x2∈E(G)] √(3−1)× (3−1)
⇒ RRR(G) = [|E(G)|]√(4)
⇒ RRR(G) = [|6n|]√(4)
⇒ RRR(G) = (6n)√(4)
⇒ RRR(G) = 6n(2)
⇒ RRR(G) = 12n.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 8
Theorem 2.5 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, hyper zagrebindex are,
HM1(G) = 216n
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now hyper zagreb index are i.e. ,
HM1(G)= ∑[x1,x2∈E(G)][d(x1)+d(x2)]2 → [1]
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in hyper zagrebindex topological index of the general form,
⇒ HM1(G) =[|E(G)|][(3+3)]2

⇒ HM1(G) =[|6n|](6)2
⇒ HM1(G) = (6n)(6)2
⇒ HM1(G) = (6n)(36)
⇒ HM1(G) = 216n.
HM1(G)= Thirty six times to general edges of petersen graph.

Theorem 2.6 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, two polynomialrelated to the first zagreb index are,
M∗1(G,x) =(12n)x4n
M0(G,x) =4nx3
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersengraph Tn = {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. Thedegree of each vertex in P(k,t) is 3 and now two polynomial related to the first zagreb index are i.e. ,
M∗1(G,x) =∑[xi∈V (G)][d(xi)][x[xi]]
M0(G,x) =∑[xi∈V (G)](x)[d(xi)]
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in two polynomialrelated to the first zagreb index topological index of the general form,

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 9
⇒ M∗1(G,x) =∑[xi∈V (G)](3)[x[xi]]

⇒ M∗1(G,x) =∑[xi∈V (G)](3)x[4n]
⇒ M∗1(G,x) =[|V (G)|](3)x4n
⇒ M∗1(G,x) =4n(3)x4n
M∗1(G,x) =(12n)x4n.
M∗1(G,x) =[3x4n]× [General vertices of petersen graph]Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in two polynomialrelated to the first zagreb index topological index of the general form,
⇒ M0(G,x) =∑[xi∈V (G)](x)3

⇒ M0(G,x) =[|V (G)|](x)3
M0(G,x) =4nx3.
M0(G,x) =[x3]× [General vertices of petersen graph]

Theorem 2.7 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, zagreb typepolynomials are,
Ma,b(G,x) =6nx[3(a+b)]
M′a,b(G,x) = (6n)(x)[(a+3)(b+3)]
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now zagreb type polynomials are i.e. ,
Ma,b(G,x) = ∑[xi,xj∈E(G)](x)[a{d(xi)}+b{d(xi)}]
M′a,b(G,x) = ∑[xi,xj∈E(G)](x)([a+{d(xi)}][b+{d(xi)}])Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in zagreb typepolynomials topological index of the general form,
⇒ Ma,b(G,x) =∑[xi,xj∈E(G)](x)[a(3)+b(3)]

⇒ Ma,b(G,x) =∑[xi,yj∈E(G)](x)[3(a+b)]
⇒ Ma,b(G,x) =[|E(G)|](x)[3(a+b)]
Ma,b(G,x) =6nx[3(a+b)].Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in zagreb typepolynomials topological index of the general form,

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 10
⇒ M′a,b(G,x) =∑[xi,xj∈E(G)](x)([(a+(3))][(b+(3))])

⇒ M′a,b(G,x) =[|E(G)|](x)[(a+(3))(b+(3))]
⇒ M′a,b(G,x) =[|6n|](x)[(a+3)(b+3)]
M′a,b(G,x) =(6n)(x)[(a+3)(b+3)].
M′a,b(G,x) =[(x)[(a+3)(b+3)]]× [General edges of petersen graph]

Theorem 2.8 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, atomic-bond-connectivity (ABC) index are,
ABC(G) = (4n).
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now atomic-bond-connectivity (ABC) index are i.e. ,
ABC(G) = ∑[x1,x2∈E(G)] √[d(x1)+d(x2)]−2d(x1)×d(x2) → [1]
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersengraph about every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values inatomic-bond-connectivity (ABC) index topological index of the general form,
⇒ ABC(G) = ∑[x1,x2∈E(G)] √[d(x1)+d(x2)]−2d(x1)×d(x2)Putting values in above equation;

⇒ ABC(G) =[|E(G)|]√[(3)+(3)−2]
(3)×(3)

⇒ ABC(G) =[|6n|]√ (4)
(3)2

⇒ ABC(G) =(6n)√4
(3)

⇒ ABC(G) = (4n).
ABC(G) = General vertices of petersen graph

Theorem 2.9 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, geometricarithmetic(GA) index are,
GA(G) =6n
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now geometric arithmetic(GA) index are i.e. ,
GA(G) = ∑[x1,x2∈E(G)] 2√d(x1)×d(x2)d(x1)+d(x2)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 11
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in geometricarithmetic(GA) index topological index of the general form,
⇒ GA(G) =[|E(G)|]2√(3)×(3)

(3)+(3)

⇒ GA(G) =[|6n|]2√(3)2
(6)

⇒ GA(G) =(6n)2√(3)2
6

⇒ GA(G) =(6n)2(3)
(6)

⇒ GA(G) =6n.
GA(R)= General edges of petersen graph.

Theorem 2.10 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, first multiplezagreb index are,
PM1(G) = (6)6n
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now first multiple zagreb index are i.e. ,
PM1(G) = ∏[x1,x2∈E(G)][d(x1)+d(x2)]Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in first multiplezagreb index topological index of the general form,
⇒ PM1(G) = ∏[x1,x2∈E(G)][(3+3)]

⇒ PM1(G) = (6)[|E(G)|]
⇒ PM1(G) = (6)[|6n|]
⇒ PM1(G) = (6)6n.
PM1(G) = General edges of petersen graph to the power of six.

Theorem 2.11 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, secondmultiple zagreb index are,
PM2(G) = (9)[6n]
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now second multiple zagreb index are i.e. ,
PM2(G) = ∏[x1,x2∈E(G)][d(x1)×d(x2)]

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Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in second multiplezagreb index topological index of the general form,
⇒ PM2(G) = ∏[x1,x2∈E(R)][(3)× (3)]

⇒ PM2(G) = (9)[|E(R)|]
⇒ PM2(G) = (9)[6n].
PM2(G) = (9)6n PM2(G) = General edges of petersen graph to the power of nine.

Theorem 2.12 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, forgottenpolynomial are,
F(R) = (6n)x18
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now forgotten polynomial are i.e. ,
F(G,x) =∑[x1,x2∈E(G)](x)[(dx1)2+(dx2)2]Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in forgottenpolynomial topological index of the general form,
⇒ F(G,x) =∑[x1,x2∈E(G)](x)[(3)2+(3)2]

⇒ F(G,x) =[|E(G)|](x)[9+9]
⇒ F(G) =[6n](x)18
⇒ F(R) =(6n)x18.
F(R) =[x18]× [General edges of petersen graph]

Theorem 2.13 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, symmetricdivision deg. index are,
SDD(G) = 12n
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now symmetric division deg. index are i.e. ,
SDD(G) = ∑[x1,x2∈E(G)][d(x1)2+d(x2)2d(x1)d(x2) ]Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersen graphabout every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in symmetric

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division deg. index topological index of the general form,
⇒ SDD(G) = ∑[x1,x2∈E(G)] mini(3,3)max(3,3) + maxi(3,3)mini(3,3)

⇒ SDD(G) =[|E(G)|][3
3
+ 3
3
]

⇒ SDD(G) =[|E(G)|][(3)+(3)
3
]

⇒ SDD(G) =[|6n|][(6)
3
]

⇒ SDD(G) =(6n)[2]
⇒ SDD(G) =12n.
SDD(G) = Two times of general edges of petersen graph.

Theorem 2.14 Let P(k,t) be petersen subdivision graph. Then, for Tn = {1,2,3, ...}, generalconnectivity index are,
SDD(G) = 12n
Proof: The petersen graph Tn = {1,2,3, ...} appears in figure(graph). The petersen graph Tn= {1,2,3, ...} contains V (G′) = 4n no of vertices and E(G′) = 6n no of edges. The degree ofeach vertex in P(k,t) is 3 and now general connectivity index are i.e. ,
M1(G) = ∑[x2∈V (G)][dG(x2)]2
M2(G) = ∑[x1,x2∈E(G)][(dG(x1))× (dG(x2))] → [1]
Now we suppose vertices are V (G) = 4n, edges are E(G) = 6n and degree of petersengraph about every each vertices is P(k,t) =[d(x1),d(x2)] = 3. Now putting the values in generalconnectivity index topological index of the general form,
⇒ M1(G) = ∑[x2∈V (G)] [(3)2]

⇒ M1(G) = [|V (G)|](3)2
⇒ M1(G) = (4n)(9)
⇒ M1(G) = 36n.
M1(G) = Nine times to vertices of petersen graph.

In general form of real number index is, in equation [1] becomes;
⇒ M2(G) = ∑[x1,x2∈E(G)][dG(x1)×dG(x2)]Now putting values in above equation.
⇒ M2(G) = ∑[x1,x2∈E(G)](3)(3)
⇒ M2(G) = [|E(G)|](3)(3)
⇒ M2(G) = (6n)(9)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 14
⇒ M2(G) = 54n.
M2(G) = Nine times to edges of petersen graph.

3. Numerical Examples
Every petersen graph is cyclic graph and the graph G′ in general form consists V having set ofvertex and E having set of edge, if the natural number, there exist n the graph with vertices are

V (G′)=4n, edges are E(G′)=6n, and the specific of this graph is that about degree of everyeach vertex is P(k,t) = [d(x1),d(x2)] = 3. Then this graphic which is said to be petersen graphic.Then petersen graphic is denoted by P[V (G′),E(G′)] =(4n,6n)The core features of petersen map explored by length in 1985. However, the petersen line contin-uously arise in literature of the theoretical graphing. By this article, we update previous analysisto introduce additionally fresh findings on the petersen mapping.Sylvester’s association to graphs of invariants and covariants requires interpretation of principleof invariants in 1880s.
Example 3.1. If there exists n is positive natural number then Tn = {1,2,3, ...}, so graphwith vertices are V (G′) = 4n, and edges E(G′) = 6n, in general form of petersen expressed by

P[V (G′),E(G′)] =(4n,6n). This graph having a specification, that degree of every each vertex is
P(k,t)= [d(x1),d(x2)]= 3.Now we write;
Tn =1,2,3, ....

V (G′)=4n .........(1)
E(G′)=6n .........(2)
k = d(x1)=3

t = d(x2)=3Put n =3 in equations (1) and (2) and these equations become;
T3 =3

V (G′)=12

E(G′)=18

P(k,t) = [d(x1),d(x2)]= 3

P[V (G′),E(G′)] =(12,18)Now figure is;
Example 3.2. If there exists n is positive natural number then Tn = {1,2,3, ...}, so graphwith vertices are V (G′) = 4n, and edges E(G′) = 6n, in general form of petersen expressed by

P[V (G′),E(G′)] =(4n,6n). This graph having a specification, that degree of every each vertex is
P(k,t)= [d(x1),d(x2)]= 3.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.20 15

Figure 1. Petersen graph
Now we write;
Tn =1,2,3, ....

V (G′)=4n .........(1)
E(G′)=6n .........(2)
k = d(x1)=3

t = d(x2)=3Put n =4 in equations (1) and (2) and these equations become;
T4 =4

V (G′)=16

E(G′)=24

P(k,t) = [d(x1),d(x2)]= 3

P[V (G′),E(G′)] =(16,24)Now figure is;

Figure 2. Petersen graph
Example 3.3. If there exists n is positive natural number then Tn = {1,2,3, ...}, so graphwith vertices are V (G′) = 4n, and edges E(G′) = 6n, in general form of petersen expressed by

P[V (G′),E(G′)] =(4n,6n). This graph having a specification, that degree of every each vertex is
P(k,t)= [d(x1),d(x2)]= 3.Now we write;

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Tn =1,2,3, ....

V (G′)=4n .........(1)
E(G′)=6n .........(2)
k = d(x1)=3

t = d(x2)=3Put n =5 in equations (1) and (2) and these equations become;
T5 =5

V (G′)=20

E(G′)=30

P(k,t) = [d(x1),d(x2)]= 3

P[V (G′),E(G′)] =(20,30)Now figure is;

Figure 3. Petersen graph
Example 3.4. If there exists n is positive natural number then Tn = {1,2,3, ...}, so graphwith vertices are V (G′) = 4n, and edges E(G′) = 6n, in general form of petersen expressed by

P[V (G′),E(G′)] =(4n,6n). This graph having a specification, that degree of every each vertex is
P(k,t)= [d(x1),d(x2)]= 3.Now we write;
Tn =1,2,3, ....

V (G′)=4n .........(1)
E(G′)=6n .........(2)
k = d(x1)=3

t = d(x2)=3Put n =10 in equations (1) and (2) and these equations become;
T10 =10

V (G′)=40

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E(G′)=60

P(k,t) = [d(x1),d(x2)]= 3

P[V (G′),E(G′)] =(40,60)Now figures are;

Figure 4. Petersen graph

4. Conclusion and Future Studies
Frequently, graph theory is refuted using the petersen graph. In this paper, the general petersengraph was constructed, and the exact expressions of the first and second zagreb indices, the forgottentopological index, the hyper zagreb index, the reduced second zagreb index and the petersen graphin terms of cyclic graph were then examined. The future work will concentrate on topologicalindeces, then generalised petersen via graph operations.

References
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	1. Introduction
	1.1. The graph of petersen
	1.2. Graphical idea of petersen graph

	2. Main Results
	3. Numerical Examples
	4. Conclusion and Future Studies
	References

