m-polynomials and degree-based topological indices of the

Research ArticleM-Polynomials and Degree-Based Topological Indices of theMolecule Copper(I) Oxide

Faryal Chaudhry1 Iqra Shoukat1 Deeba Afzal 1 Choonkil Park 2 Murat Cancan 3

and Mohammad Reza Farahani 4

1Department of Mathematics and Statistics University of Lahore Lahore 54000 Pakistan2Research Institute for Natural Sciences Hanyang University Seoul 04763 Republic of Korea3Faculty of Education Van Yuzuncu Yıl University Van 65080 Turkey4Department of Applied Mathematics Iran University of Science and Technology (IUST) Narmak Tehran 16844 Iran

Correspondence should be addressed to Choonkil Park baakhanyangackr and Mohammad Reza Farahani mrfarahani88gmailcom

Received 4 December 2020 Revised 20 December 2020 Accepted 28 January 2021 Published 24 February 2021

Academic Editor Teodorico C Ramalho

Copyright copy 2021 Faryal Chaudhry et al +is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

Topological indices are numerical parameters used to study the physical and chemical residences of compounds Degree-basedtopological indices have been studied extensively and can be correlated with many properties of the understudy compounds In thefactors of degree-based topological indices M-polynomial played an important role In this paper we derived closed formulas forsome well-known degree-based topological indices like first and second Zagreb indices the modified Zagreb index the symmetricdivision index the harmonic index the Randic index and inverse Randic index and the augmented Zagreb index using calculus

1 Introduction

11 Application Background A graph that represents theconstruction of a molecule and also their connectivity isknown as a molecular graph and such a representation isgenerally known as topological representations of moleculeMolecular graphs are normally characterized by means ofexclusive topological basis for parallel of chemicals shape ofa molecule with organic chemical or bodily homes Study ofgraph has some programs of various topological indices inquantitative structure-activity relationship (QSAR) andquantitative structure-property relationship (QSPR) digitalscreenings and computational drug designing citations asshown in [1 2] +us far several exclusive topological in-dices have been established and maximum of them are mosteffective graph descriptors in [3 4] apart some indices haveproven their parallel with organic chemical or physicalresidences of secure molecules in [5ndash17]

In the field of mathematics any graph has vertices andedges that are represented by the atoms and chemical bonds

Graph that represents the construction of molecules andtheir connectivity is known as a molecular graph and suchrepresentation is usually referred as topological represen-tation of molecules +ere are some significant topologicalindices like distance-based topological indices degree-basedtopological indices and primarily based topological indicesAmong these works distance primarily based topologicalindices unit works out a crucial task in a chemical graphstarted specifically in chemistry [1819] Many fields havemany features that can be solved with the help of graphs Inthe physiochemical compounds or network systems we havea tendency to abstractly outline exclusive ideas in modelingof mathematics We have a tendency to refer to as thedistinctive names such as Randic index and national capitalindex

A topological index is a numerical parameter of a graphand describes its topology It describes the molecular shapenumerically and is applied within the advancement ofqualitative structure-activity relationships (QSARs) +efollowing are the 3 types of topological indices

HindawiJournal of ChemistryVolume 2021 Article ID 6679819 12 pageshttpsdoiorg10115520216679819

(1) Degree-based(2) Distance-based(3) Spectral-based

Degree-based topological indices were studied exten-sively and may be correlated with many residences of theunderstudy molecular compounds +ere is a strong rela-tionship among distance-based and degree-based topolog-ical indices in [20] Most commonly known invariants ofsuch kinds are degree-based topological indices +ese areactually the numerical values that correlate the structurewith various physical properties chemical reactivities andbiological activities Topological indices are sincerely thenumerical values that relate the shape to one of a kind ofphysical residences artificial reactivity and natural bio-logical activities [2122]

Loads of research has been executed inside the course ofM-polynomial as in the case of Munir et al processedM-polynomial and related lists of triangular boron nanotubesin [6] polyhex nanotubes in [23] nanostar dendrimers in [4]and titania nanotubes in [5] M-Polynomials and topologicallists of V-phenylenic nanotubes and nanotori In this paper theobjective is to process theM-polynomial of the crystallographicrealistic structure of the atom copper(I) oxide (Cu2O) [824]

12 Crystallographic Structure of Cu2O(m n) Copper oxideis a p-type semiconductor and inorganic compound Copperoxide is a chemical element with formula Cu2O(m n)Cu2O(m n) is a certainly happening reddish coral that isparticularly used in chemical sensors and solar orientatedcells in [8 24] It has many advantages such as photo-chemical effects stability pigment a fungicide nontoxicityand low cost It has potential applications in new energysensing sterilization and other fields It has narrow bandgap and is easily excited by visible light

Cu2O(m n) is additionally responsible for the pinkshading in Benedictrsquos test and is the essential cause to selectCu2O (see Figures 1 and 2) +e promising projects ofCu2O(m n) are mainly on chemical sensors sunlight-basedcells photocatalysis lithium particle batteries and catalysisHere we have taken into consideration a monolayer ofCu2O(m n) for satisfaction To ultimate the basis forCu2O(m n) we pick out the setting of this graph as Cu2O(mn) be the chemical graph of copper(I) oxide with (m n) unitcells within the aircraft

2 Definitions and Literature Review

21 M-Polynomial M-Polynomial is defined by S Klavzaror E Deutsch in 2015 [3 8] Within the factors of degree-based topological indices we compete necessary role ofM-polynomial Readers can refer to [9ndash17 27ndash35] It is theforemost general progressive polynomial and an additionallyclosed formula alongside 10 distance-based topological in-dices is given by M-polynomial It is explained as

M(G a b) 1113944δleilejleΔ

mij(G)aib

J

(1)

and we have δ Min dr | r isin V(G)1113864 1113865 and Δ Max dr | r isin V1113864

(G) where mij(G) is the edge E(G) where ile j

22 Degree-Based Topological Indices Any purpose on agraph which does not build upon numbering of its vertices ismolecular descriptor +is is also called as topological indexTopological indices are most useful in the field of isomericdiscrimination chemical validation QSAR QSPR and apharmaceutical drug form Topological indices are accessedfrom the system of molecule

+ere are some important degree-based topologicalindices defined and the first Zagreb index was introduced byGutman and Trinajstic as follows

M1(G) 1113944rsεE(G)

dr + ds( 1113857 (2)

Gutman and Trinajstic proposed the second Zagrebindex in 1972 which is stated as

M2(G) 1113944rsεE(G)

dr times ds( 1113857 (3)

+e second modified Zagreb index is defined as

mM2(G) 1113944

rsεE(G)

1d(r)d(s)

(4)

General 1st and 2nd multiplicative Zagreb indices areintroduced by Kulli Stone Wang and Wei and are stated as

MZa1II(G) 1113945

rsisinE(G)

dr + ds( 1113857a

MZa2II(G) 1113945

rsisinE(G)

dr + ds( 1113857a

(5)

+e general 1st and 2nd Zagreb indices proposed by KulliStone Wang and Wei are stated as

Za1(G) 1113944

rsisinE(G)

dr + ds( 1113857a

Za2(G) 1113944

rsisinE(G)

drds( 1113857a

(6)

In 1987 Fajtlowicz in [36] proposed the harmonic indexand stated

H(G) 1113944rsisinE(G)

2dr + ds

(7)

+e inverse sum index is defined

I(G) 1113944rsisinE(G)

drds

dr + ds

(8)

Symmetric division index is described as

SS D(G) 1113944rsisinE(G)

min dr ds( 1113857

max dr ds( 1113857+max dr ds( 1113857

min dr ds( 1113857 (9)

SU and XU recognized general Randic index or generalmultiplicative Randic index stated as follows (Table 1)

2 Journal of Chemistry

Rα(G) 1113944rsisinE(G)

dr + dsα

RαII(G) 1113945rsisinE(G)

dr + dsα

(10)

Theorem 1 Crystallographic structure of the graph of cop-per(I) oxide GasympCu20[m n] where n mge 1 We have

M(G a b) f(a b) (4m + 4n minus 4)ab2

+(4mn minus 4n minus 4m + 4)a2b2

+ 4mna2b4

(11)

Proof suppose G be the crystallographic structure of Cu2O[lm n]+e edge set of Cu2O[lm n] has the following threepartitions by Figures 1 and 2

E1 E 12 e rs isin E(G)|dr 1 ds 21113864 1113865



(12)

such that

E1(G)1113868111386811138681113868

1113868111386811138681113868 4mm + 4n minus 4

E2(G)1113868111386811138681113868

1113868111386811138681113868 4mn minus 4m minus 4n + 4

E3(G)1113868111386811138681113868

1113868111386811138681113868 4mn

(13)

+us the M-polynomial of Cu2O[l m n] is

M(G a b) 1113944ile j

mij(G)aib

j

M(G a b) 11139441le 2

m12(G)ab2

+ 11139442le 2

m22(G)a2b2

+ 11139442le 4

m24(G)a2b4

M(G a b) 1113944rsisinE1

m12(G)ab2

+ 1113944uvisinE2

m22(G)a2b2

+ 1113944uvisinE3

m24(G)a2b4

M(G a b) E1(G)1113868111386811138681113868

1113868111386811138681113868ab2

+ E2(G)1113868111386811138681113868

1113868111386811138681113868a2b2

+ E3(G)1113868111386811138681113868

1113868111386811138681113868a2y4

M(G a b) (4m + 4n minus 4)ab2


+ 4mna2b4

(14)

Theorem 2 Crystallographic structure of the graph of cop-per(I) oxide GasympCu20[m n] where n mge 1 We haveM1(G) 40mn minus 4m minus 4n + 4

Proof suppose

M(G a b) f(a b) (4m + 4n minus 4) times ab2

+(4mn minus 4n minus 4m + 4) times a2b2

+ 4mn times a2b4

(15)

We have to find

Cu

O

(a) (b)

Figure 1 (a) Cu2O [1 1] [25] (b) Cu2O [2 2] [1]

Figure 2 Copper(I) oxide [4 4] [26]

Journal of Chemistry 3

Da zf

zaa

zf

za (4n + 4m minus 4)b

2+ 2(4mn minus 4n minus 4m + 4)ab

2+ 8mnab

4

(16)

Multiply a on both sides

Da azf

za (4m + 4n minus 4)ab

2

+ 2(4mn minus 4m minus 4n + 4)a2b2

+ 8mna2b4

(17)

Similarly

Dbf(a b) bzf

zb 2(4n + 4m minus 4)ab

2

+ 2(4mn minus 4n minus 4m + 4)a2b2

+ 16mnab4

M1(G) Da + Dy1113872 1113873f(a b)|ab1

(18)

Now the first Zagreb index is

M1(G) Da + Dy1113872 1113873f(a b)|ab1

M1(G) [(4m + 4n minus 4) + 2(4mn minus 4m minus 4n + 4)

+ 8mn] +[2(4n + 4m minus 4)]

+ 2(4mn minus 4m minus 4n + 4) + 16mn)

M1(G) [4m + 4n minus 4 + 8mn minus 8m minus 8n + 8 + 8mn

+ 8n + 8m minus 8 + 8mn minus 8m minus 8n + 8 + 16mn]

(19)

After solving the result is

M(G) 40mn minus 4m minus 4n + 4 (20)

+e 3D plot of first Zagreb index is given in Figure 3 (f oru 1 left v 1 middle and w 1 right) and we see thedependent variables of the first Zagreb index on the involvedparameters


Proof suppose

M(G a b) (4m + 4n minus 4) times ab2

+(4mn minus 4n minus 4m + 4)

times a2b2

+ 4mn times a2b4

(21)

We have to find DbDa first we take Da

Da (4m + 4n minus 4) times ab2

+(4mn minus 4m minus 4n + 4)2a

times a times b2

+ 4mn2a times a times b

4


+ 2(4mn minus 4m minus 4n + 4)

times a2b2

+ 8mn times a2b4

(22)

Now take Db

DbDaf(a b) 2(4m + 4n minus 4)ab

+ 2(4mn minus 4m minus 4n + 4)a2

times 2b b

+ 8mna2

times 4b3

times b

DbDaf(a b) 2(4m + 4n minus 4) times ab2

+ 4(4mn minus 4m minus 4n + 4) times a2b2

+ 32mn times a2b4

(23)

+e second Zagreb index is

M2(G) DbDa(f(a b))|ab1

M2(G) 2(4m + 4n minus 4) + 4(4mn minus 4m minus 4n + 4) + 32mn

M2(G) 8nn + 8m minus 8 + 16mn minus 16m minus 16n + 16 + 32mn

(24)


M2(G) 48mn minus 8m minus 8n + 8 (25)

+e 3D plot of second Zagreb index is given in Figure 4 (for u 1 left v 1 middle and w 1 right) and we see thedependent variables of the second Zagreb index on theinvolved parameters

Theorem 4 Crystallographic structure of the graph of cop-per(I) oxide GasympCu20[m n] where n mge 1 and we have

mM

2(G)

32

mn + m + n minus 13 (26)

Table 1 Formulas of degree-based topological indices from M-polynomial

Topological Indices f(t s) M(G t s)First Zagreb index t + s M1(G t s) (Dt + Ds)M(G t s)|ts1Second Zagreb index Ts M2(G t s) (DtDs)M(G t s)|ts1Second modified Zagreb index 1ts mM2(G t s) (δtδs)M(G t s)|ts1General Randic index αne 0 (ts)α Rα(G) (Dα

t Dαs )M(G t s)|ts1

Inverse general Randic index αne 0 1(ts)α RRα(G)(δαt δαs )M(G t s)|ts1

Symmetric division index (t2 + s2)ts SS D(G) |Dtδs δsDt|ts1Harmonic index 2(t + s) H(G) 2δtJM(G t s)|t1Inverse sum index ts(t + s) I(G) δtJDtDsM(G t s)|t1

Ds s(zzs)M(G t s)|t s 1 Dt t(zzt)M(G t s)|t s 1 δt 1113938t

0(M(G y s))ydt δs 1113938t

0 M(G t y)yds J M(G t t) Qα xαM(G t s) αne 0


Proof suppose


+(4mn minus 4m minus 4n + 4)a2b2

+ 4mna2b4

(27)

Now we have to find SaSb first we find Sa

Sa 1113946a

0

f(x b)

xdx

f(x b) (4n + 4m minus 4)xb2

+(4mn minus 4m minus 4n + 4)x2b2

+ 4mnx2b4

f(x b)

x (4m + 4n minus 4)b

2+(4mn minus 4n minus 4m + 4)xb

2

+ 4mnxb4

(28)

Taking integration on both sides

1113946a

0

f(x b)

xdx 1113946

a

0(4m + 4n minus 4)b

2dx

+ 1113946a

0(4mn minus 4m minus 4n + 4)xb

2dx

+ 4mn 1113946a

0xdxb

4

Sa (4m + 4n minus 4)ab2

+12

(4mn minus 4n minus 4m + 4)a2b2

+ 2mna2b4

(29)

Now take Sb and then

50

0

ndash50 ndash50

0

50

15

1

05

0

ndash05

ndash1

times105

(a)

50

0

ndash50 ndash500

50

15

1

05

0

ndash05

ndash1

times105

(b)

Figure 3 First Zagreb index plotted in 3D

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(a)

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(b)

Figure 4 Second Zagreb index plotted in 3D


SaSbf(a b) (4m + 4n minus 4)ax2

+12

(4mn minus 4m minus 4n + 4)a2x2

+ 2mna2x4

SaSbf(a b) 12

(4m + 4n minus 4)ab2

+14

(4mn minus 4m minus 4n + 4)a2b2

+12

mna2b4

(30)

Now the second modified Zagreb index is

mM2(G) SaSbf(a b)|ab1

12

(4m + 4n minus 4) +14

(4mn minus 4m minus 4n + 4) +12

mn

(2m + 2n minus 2) +(mn minus m minus n + 1) +12

mn

2m minus m + 2n minus n minus 2 + 1 + mn 1 +12

1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)

+e 3D plot of modified second Zagreb index is given inFigure 5 (f or u 1 left v 1 middle and w 1 right) and wesee the dependent variables of the modified second Zagrebindex on the involved parameters


Rα(G) 2α+2minus 22α+2

1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose


+(4mn minus 4m minus 4n + 4)

times a2b2

+(4mn) times a2b4

(34)

We have to find DaDb first and we find Da



times a2b2

+ 8mn times a2b4

(35)

Now take Db

DaDb (4m + 4n minus 4)a times 2b times b + 2(4mn minus 4n minus 4m + 4)a2

times 2b times b + 2(4mn)a2

times 4b3

times b

(36)

Take α on the above equation

DαaD

αb 2α(4m + 4n minus 4)ab

2+ 4α(4mn minus 4m minus 4n + 4)a

2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2

+ 22α+2(mn minus m minus n + 1)a

2b2

+ 23α+2mna

2b4

(37)

Now the general Randic index is

Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2

(mn minus m minus n + 1) + 23α+2mn

Rα(G) 2α+2m + 2α+2

n minus 2α+2+ 22α+2

mn minus 22α+2m

minus 22α+2n + 22α+2

+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)

+e 3D plot of Randic index is given in Figure 6 (f oru 1 left v 1 middle and w 1 right) and we see thedependent variables of the Randic index on the involvedparameters


RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)

Now we have to find SaSb and first we find Sa

Sa (4m + 4n minus 4) 1113946a

0dxb

2+(4mn minus 4m minus 4n + 4)

middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4


+ 2(mn minus m minus n + 1)a2b2

+ 4mna2b4

(42)

Similarly take Sb

SaSb 4(m + n minus 1)a 1113946b

0xdx + 2(mn minus n minus m + 1)a

2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

SaSb 2(m + n minus 1)ab2

+(mn minus m minus n + 1)a2b2

+ mna2b4

(43)Take α on the above equation

SαaS

αb

12αminus 2 (m + n minus 1)ab

2+

122αminus 2 (mn minus m minus n + 1)a

2b2

+1

23αminus 2 mna2b4

(44)

+e inverse Randic is

50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)

Figure 5 Modified the second Zagreb index plotted in 3D

6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)

Figure 6 Randic index plotted in 3D


RRα(G) (f(a b))|ab1 1

2αminus 2 (m + n minus 1)

+1

22αminus 2 (mn minus m minus n + 1)

+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)

+e 3D plot of inverse Randic index is represented inFigure 7 (f or u 1 left v 1 middle and w 1 right) and wesee the dependent variables of the inverse Randic index onthe involved parameters

Theorem 7 Crystallographic structure of the graph of cop-per(I) oxide GasympCu20[m n] where n mge 1 and we haveSS D(G) 18mn + 2m + 2n minus 2

Proof suppose



times a2b2

+(4mn) times a2b4

(46)

First we have to find Sb

Sb (4n + 4m minus 4)a 1113946b

0xdx +(4mn minus 4m minus 4n + 4)a

2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12

(4n + 4m minus 4)ab2

+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db

SaDb(f(a b)) 2(4m + 4n minus 4)ab2


+ 8mna2b4

(50)

Now the symmetric division index is

SS D(G) SbDa + SaDb( 1113857(f(a b))|ab1 (51)

Put the values

SS D(G) 12

(4m + 4n minus 4) +(4mn minus 4m minus 4n + 4) + 2mn1113876 1113877 +[2(4m + 4n minus 4) +(4mn minus 4m minus 4n + 4) + 8mn]

SS D(G) (2m + 2n minus 2) +(4mn minus 4m minus 4n + 4) +(2mn + 8m + 8n minus 8) +(4mn minus 4m minus 4n + 4) + 8mn

SS D(G) (2m minus 4m + 8m minus 4m) +(2n minus 4n + 8n minus 4n) minus (2 minus 4 + 8 minus 8) +(4mn + 2mn + 4mn + 8mn)

(52)

After the calculation the result is

SS D(G) 18mn + 2m + 2n minus 2 (53)

+e 3D plot of symmetric division index is given inFigure 8 (f or u 1 left v 1 middle and w 1 right) and wesee the dependent variables of the symmetric division indexon the involved parameters


H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)

First we have to find Jf(ab)

Jf(a b) Jf(a a) 4(m + n minus 1)a3

+ 4(mn minus m minus n + 1)a4

+ 8mna6

(56)

Take Sa


SaJf(x b) 4(m + n minus 1) 1113946a

0x2dx + 4(mn minus m minus n + 1)

middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12

(mn minus m minus n + 1)a4

+23

mna6

(57)

+e harmonic index is

H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12

(mn minus m minus n + 1) +23

mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)

+e 3D plot of harmonic index is given in Figure 9 (f oru 1 left v 1 middle and w 1 right) and we see thedependent variables of the harmonic index on the involvedparameters


SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)

First we have to find Db

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)

Figure 7 Inverse Randic index plotted in 3D

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)

Figure 8 Symmetric division index plotted in 3D


Dbf(a b) 8(m + n minus 1)ab2

+ 8(mn + m + n minus 1)a2b2

+ 32mna2b4

(62)Take Da

DaDbf(a b) 8(m + n minus 1)ab2

+ 16(mn minus n minus m + 1)a2b2

+ 64mna2b4

(63)Take Jf(a b)

JDaDbf(a b) 8(m + n minus 1)x3

+ 16(mn minus m minus n + 1)x4

+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3

+ 4(mn minus n minus m + 1)a4

+323

mna6

(65)+e inverse sum index is

SaJDaDb(f(a b))|a1 83

(m + n minus 1)

+ 4(mn minus m minus n + 1) +323

mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000

0ndash2000ndash4000ndash6000

(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)

Figure 9 Harmonic index plotted in 3D

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)

Figure 10 Inverse sum index plotted in 3D



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)

+e 3D plot of inverse sum index is given in Figure 10 (for u 1 left v 1 middle and w 1 right) and we see thedependent variables of the inverse sum index on the in-volved parameters

Data Availability

No data were used in this study

Disclosure

All authors have not any fund grant and sponsor forsupporting publication charges

Conflicts of Interest

+e authors declare that there are no conflicts of interestregarding the publication of this paper

References

[1] A Q Baig M Naeem M Naeem W Gao and J-B LiuldquoGeneral fifth M-Zagreb indices and fifth M-Zagreb poly-nomials of carbon graphiterdquo Eurasian Chemical Communi-cations vol 2 no 3 pp 634ndash640 2020

[2] F A Cotton G Wilkinson C A Murillo and M BochmannAdvanced Inorganic Chemistry John Wiley amp Sons HobokenNJ USA 6th edition 1999

[3] H Deng J Yang and F Xia ldquoA general modeling of somevertex-degree based topological indices in benzenoid systemsand phenylenesrdquo Computers amp Mathematics with Applica-tions vol 61 no 10 pp 3017ndash3023 2011

[4] W Gao Y Wang B Basavanagoud and M K JamilldquoCharacteristics studies of molecular structures in drugsrdquoSaudi Pharmaceutical Journal vol 25 no 4 pp 580ndash5862017

[5] G Gayathri U Priyanka and U Priyanka ldquoDegree basedtopological indices of zig zag chainrdquo Journal of Mathematicsand Informatics vol 11 pp 83ndash93 2017

[6] I Gutman ldquoSome properties of the Winer polynomialrdquoGraph Ieory Notes N Yvol 125 pp 13ndash18 1993

[7] L B Kier and L H Hall Molecular Connectivity in Struc-tureActivity Analysis Wiley New York NY USA 1986

[8] X Li and Y Shi ldquoA survey on the Randic indexrdquo MATCHCommunications in Mathematical and in Computer Chem-istry vol 59 pp 127ndash156 2008

[9] M Alaeiyan C Natarajan C Natarajan G Sathiamoorthyand M R Farahani ldquo+e eccentric connectivity index ofpolycyclic aromatic hydrocarbons (PAHs)rdquo EurasianChemical Communications vol 2 no 6 pp 646ndash651 2020

[10] M Cancan S Ediz H Mutee-Ur-Rehman and D Afzal ldquoM-polynomial and topological indices Poly (E+yleneAmido-Amine) dendrimersrdquo Journal of Information and Optimiza-tion Sciences vol 41 no 4 pp 1117ndash1131 2020

[11] M Cancan S Ediz S Ediz and M R Farahani ldquoOn ve-degree atom-bond connectivity sum-connectivity geomet-ric-arithmetic and harmonic indices of copper oxiderdquo Eur-asian Chemical Communications vol 2 no 5 pp 641ndash6452020

[12] M Imran S A Bokhary S A U h Bokhary S Manzoor andM K Siddiqui ldquoOn molecular topological descriptors ofcertain families of nanostar dendrimersrdquo Eurasian ChemicalCommunications vol 2 no 6 pp 680ndash687 2020

[13] M R Farahani ldquoSome connectivity indices and Zagreb indexof polyhex nanotubesrdquo Acta Chimica Slovenica vol 59pp 779ndash783 2012

[14] M Randic ldquoCharacterization of molecular branchingrdquoJournal of the American Chemical Society vol 97 no 23pp 6609ndash6615 1975

[15] B Zhou and I Gutman ldquoRelations between wiener hyper-wiener and Zagreb indicesrdquo Chemical Physics Letters vol 394no 1ndash3 pp 93ndash95 2004

[16] Z Ahmad M Naseem M Naseem M K JamilM F Nadeem and S Wang ldquoEccentric connectivity indicesof titania nanotubes TiO2[m n]rdquo Eurasian Chemical Com-munications vol 2 no 6 pp 712ndash721 2020

[17] Z Ahmad M Naseem M Naseem M Kamran JamilM Kamran Siddiqui and M Faisal Nadeem ldquoNew results oneccentric connectivity indices of V-Phenylenic nanotuberdquoEurasian Chemical Communications vol 2 no 6 pp 663ndash671 2020

[18] J B Babujee and S Ramakrishnan ldquoTopological indices andnew graph structuresrdquo Applied Mathematical Sciences vol 6no 108 pp 5383ndash5401 2012

[19] E Deutsch and S Klavzar ldquoM-polynomial and degree-basedtopological indicesrdquo Iranian Journal of MathematicalChemistry vol 6 p 93102 2015

[20] A A Dobrynin R Entringer and I Gutman ldquoWiener indexof trees theory and applicationsrdquo Acta Applicandae Mathe-maticae vol 66 no 3 pp 211ndash249 2001

[21] D Dimitrov ldquoOn structural properties of trees with minimalatom-bond connectivity index IV solving a conjecture aboutthe pendent paths of length threerdquo Applied Mathematics andComputation vol 313 pp 418ndash430 2017

[22] F Afzal M A Razaq M Abdul Razaq D Afzal andS Hameed ldquoWeighted entropy of penta chains graphrdquoEurasian Chemical Communications vol 2 no 6 pp 652ndash662 2020

[23] I Gutman and B Furtula Recent Results in the Ieory ofRandic Index University of Kragujevac Kragujevac Serbia2008

[24] L B Kier and L H HallMolecular Connectivity in Chemistryand Drug Research Academic Press New York NY USA1976

[25] M S Ahmad W Nazeer S M Kang M Imran and W GaoldquoCalculating degree-based topological indices of dominatingDavid derived networksrdquo Open Physics vol 15 no 1pp 1015ndash1021 2017

[26] A Q Baig M Imran W Khalid and M Naeem ldquoMoleculardescription of carbon graphite and crystal cubic carbonstructuresrdquo Canadian Journal of Chemistry vol 95 no 6p 674 2017

[27] F M Bruckler T Doslic A Graovac and I Gutman ldquoOn aclass of distance-based molecular structure descriptorsrdquoChemical Physics Letters vol 503 no 4ndash6 pp 336ndash338 2011

[28] E A Ekimov V A Sidorov E D Bauer N N MelnikN J Curro and J D +ompson ldquoSuperconductivity in di-amondrdquo Nature vol 428 p 5425 2004

[29] F Afzal S Hussain D Afzal and S Razaq ldquoSome new degreebased topological indices via M-polynomialrdquo Journal of In-formation and Optimization Sciences vol 41 no 4pp 1061ndash1076 2020


[30] W Gao Z Iqbal M Ishaq R Sarfraz M Aamir andA Aslam ldquoOn eccentricity-based topological indices study ofa class of porphyrin-cored dendrimersrdquo Biomolecules vol 8no 3 p 71 2018

[31] W Gao H Wu M K Siddiqui and A Q Baig ldquoStudy ofbiological networks using graph theoryrdquo Saudi Journal ofBiological Sciences vol 25 no 6 pp 1212ndash1219 2018

[32] Y Hu X Li Y Shi T Xu and I Gutman ldquoOn moleculargraphs with smallest and greatest zeroth-order general Randicindexrdquo MATCH Communications in Mathematical and inComputer Chemistry vol 54 p 425 2005

[33] H Yang and X Zhang ldquo+e independent dominationnumbers of strong product of two cyclesrdquo Journal of DiscreteMathematical Sciences and Cryptography vol 21 no 7-8pp 1495ndash1507 2018

[34] O Ivanciuc ldquoChemical graphs molecular matrices and to-pological indices in chemoinformatics and quantitativestructure-activity relationshipssectrdquo Current Computer Aided-Drug Design vol 9 no 2 pp 153ndash163 2013

[35] S M Kang M A Zahid A u R Virk W Nazeer andW Gao ldquoCalculating the degree-based topological indices ofdendrimersrdquo Open Chemistry vol 16 no 1 pp 681ndash6882018

[36] G Caporossi I Gutman P Hansen and L Pavlovic ldquoGraphswith maximum connectivity indexrdquo Computational Biologyand Chemistry vol 27 no 1 pp 85ndash90 2003


(1) Degree-based(2) Distance-based(3) Spectral-based

Degree-based topological indices were studied exten-sively and may be correlated with many residences of theunderstudy molecular compounds +ere is a strong rela-tionship among distance-based and degree-based topolog-ical indices in [20] Most commonly known invariants ofsuch kinds are degree-based topological indices +ese areactually the numerical values that correlate the structurewith various physical properties chemical reactivities andbiological activities Topological indices are sincerely thenumerical values that relate the shape to one of a kind ofphysical residences artificial reactivity and natural bio-logical activities [2122]

Loads of research has been executed inside the course ofM-polynomial as in the case of Munir et al processedM-polynomial and related lists of triangular boron nanotubesin [6] polyhex nanotubes in [23] nanostar dendrimers in [4]and titania nanotubes in [5] M-Polynomials and topologicallists of V-phenylenic nanotubes and nanotori In this paper theobjective is to process theM-polynomial of the crystallographicrealistic structure of the atom copper(I) oxide (Cu2O) [824]

12 Crystallographic Structure of Cu2O(m n) Copper oxideis a p-type semiconductor and inorganic compound Copperoxide is a chemical element with formula Cu2O(m n)Cu2O(m n) is a certainly happening reddish coral that isparticularly used in chemical sensors and solar orientatedcells in [8 24] It has many advantages such as photo-chemical effects stability pigment a fungicide nontoxicityand low cost It has potential applications in new energysensing sterilization and other fields It has narrow bandgap and is easily excited by visible light

Cu2O(m n) is additionally responsible for the pinkshading in Benedictrsquos test and is the essential cause to selectCu2O (see Figures 1 and 2) +e promising projects ofCu2O(m n) are mainly on chemical sensors sunlight-basedcells photocatalysis lithium particle batteries and catalysisHere we have taken into consideration a monolayer ofCu2O(m n) for satisfaction To ultimate the basis forCu2O(m n) we pick out the setting of this graph as Cu2O(mn) be the chemical graph of copper(I) oxide with (m n) unitcells within the aircraft

2 Definitions and Literature Review

21 M-Polynomial M-Polynomial is defined by S Klavzaror E Deutsch in 2015 [3 8] Within the factors of degree-based topological indices we compete necessary role ofM-polynomial Readers can refer to [9ndash17 27ndash35] It is theforemost general progressive polynomial and an additionallyclosed formula alongside 10 distance-based topological in-dices is given by M-polynomial It is explained as

M(G a b) 1113944δleilejleΔ

mij(G)aib

J

(1)

and we have δ Min dr | r isin V(G)1113864 1113865 and Δ Max dr | r isin V1113864

(G) where mij(G) is the edge E(G) where ile j

22 Degree-Based Topological Indices Any purpose on agraph which does not build upon numbering of its vertices ismolecular descriptor +is is also called as topological indexTopological indices are most useful in the field of isomericdiscrimination chemical validation QSAR QSPR and apharmaceutical drug form Topological indices are accessedfrom the system of molecule

+ere are some important degree-based topologicalindices defined and the first Zagreb index was introduced byGutman and Trinajstic as follows

M1(G) 1113944rsεE(G)

dr + ds( 1113857 (2)

Gutman and Trinajstic proposed the second Zagrebindex in 1972 which is stated as

M2(G) 1113944rsεE(G)

dr times ds( 1113857 (3)

+e second modified Zagreb index is defined as

mM2(G) 1113944

rsεE(G)

1d(r)d(s)

(4)

General 1st and 2nd multiplicative Zagreb indices areintroduced by Kulli Stone Wang and Wei and are stated as

MZa1II(G) 1113945

rsisinE(G)

dr + ds( 1113857a

MZa2II(G) 1113945

rsisinE(G)

dr + ds( 1113857a

(5)

+e general 1st and 2nd Zagreb indices proposed by KulliStone Wang and Wei are stated as

Za1(G) 1113944

rsisinE(G)

dr + ds( 1113857a

Za2(G) 1113944

rsisinE(G)

drds( 1113857a

(6)

In 1987 Fajtlowicz in [36] proposed the harmonic indexand stated

H(G) 1113944rsisinE(G)

2dr + ds

(7)

+e inverse sum index is defined

I(G) 1113944rsisinE(G)

drds

dr + ds

(8)

Symmetric division index is described as

SS D(G) 1113944rsisinE(G)

min dr ds( 1113857

max dr ds( 1113857+max dr ds( 1113857

min dr ds( 1113857 (9)

SU and XU recognized general Randic index or generalmultiplicative Randic index stated as follows (Table 1)



dr + dsα


dr + dsα

(10)




+ 4mna2b4

(11)





(12)

such that

E1(G)1113868111386811138681113868

1113868111386811138681113868 4mm + 4n minus 4

E2(G)1113868111386811138681113868

1113868111386811138681113868 4mn minus 4m minus 4n + 4

E3(G)1113868111386811138681113868

1113868111386811138681113868 4mn

(13)


M(G a b) 1113944ile j

mij(G)aib

j

M(G a b) 11139441le 2

m12(G)ab2

+ 11139442le 2

m22(G)a2b2

+ 11139442le 4

m24(G)a2b4


m12(G)ab2

+ 1113944uvisinE2

m22(G)a2b2

+ 1113944uvisinE3

m24(G)a2b4

M(G a b) E1(G)1113868111386811138681113868

1113868111386811138681113868ab2

+ E2(G)1113868111386811138681113868

1113868111386811138681113868a2b2

+ E3(G)1113868111386811138681113868

1113868111386811138681113868a2y4



+ 4mna2b4

(14)


Proof suppose



+ 4mn times a2b4

(15)

We have to find

Cu

O

(a) (b)

Figure 1 (a) Cu2O [1 1] [25] (b) Cu2O [2 2] [1]



Da zf

zaa

zf



2+ 8mnab

4

(16)


Da azf


2


+ 8mna2b4

(17)

Similarly

Dbf(a b) bzf


2


+ 16mnab4

M1(G) Da + Dy1113872 1113873f(a b)|ab1

(18)


M1(G) Da + Dy1113872 1113873f(a b)|ab1


+ 8mn] +[2(4n + 4m minus 4)]




(19)





Proof suppose



times a2b2

+ 4mn times a2b4

(21)




times a times b2


4



times a2b2

+ 8mn times a2b4

(22)

Now take Db



times 2b b

+ 8mna2

times 4b3

times b



+ 32mn times a2b4

(23)





(24)





mM

2(G)

32

mn + m + n minus 13 (26)







0(M(G y s))ydt δs 1113938t



Proof suppose



+ 4mna2b4

(27)


Sa 1113946a

0

f(x b)

xdx



+ 4mnx2b4

f(x b)



2

+ 4mnxb4

(28)


1113946a

0

f(x b)

xdx 1113946

a

0(4m + 4n minus 4)b

2dx

+ 1113946a


2dx

+ 4mn 1113946a

0xdxb

4


+12


+ 2mna2b4

(29)


50

0

ndash50 ndash50

0

50

15

1

05

0

ndash05

ndash1

times105

(a)

50

0

ndash50 ndash500

50

15

1

05

0

ndash05

ndash1

times105

(b)


50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(a)

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(b)




+12


+ 2mna2x4

SaSbf(a b) 12


+14


+12

mna2b4

(30)



12

(4m + 4n minus 4) +14


mn


mn


1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)




1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose



times a2b2

+(4mn) times a2b4

(34)




times a2b2

+ 8mn times a2b4

(35)

Now take Db



times 4b3

times b

(36)


DαaD



2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2


2b2

+ 23α+2mna

2b4

(37)


Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2


Rα(G) 2α+2m + 2α+2


mn minus 22α+2m


+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)



RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References








































dr + dsα


dr + dsα

(10)




+ 4mna2b4

(11)





(12)

such that

E1(G)1113868111386811138681113868

1113868111386811138681113868 4mm + 4n minus 4

E2(G)1113868111386811138681113868

1113868111386811138681113868 4mn minus 4m minus 4n + 4

E3(G)1113868111386811138681113868

1113868111386811138681113868 4mn

(13)


M(G a b) 1113944ile j

mij(G)aib

j

M(G a b) 11139441le 2

m12(G)ab2

+ 11139442le 2

m22(G)a2b2

+ 11139442le 4

m24(G)a2b4


m12(G)ab2

+ 1113944uvisinE2

m22(G)a2b2

+ 1113944uvisinE3

m24(G)a2b4

M(G a b) E1(G)1113868111386811138681113868

1113868111386811138681113868ab2

+ E2(G)1113868111386811138681113868

1113868111386811138681113868a2b2

+ E3(G)1113868111386811138681113868

1113868111386811138681113868a2y4



+ 4mna2b4

(14)


Proof suppose



+ 4mn times a2b4

(15)

We have to find

Cu

O

(a) (b)

Figure 1 (a) Cu2O [1 1] [25] (b) Cu2O [2 2] [1]



Da zf

zaa

zf



2+ 8mnab

4

(16)


Da azf


2


+ 8mna2b4

(17)

Similarly

Dbf(a b) bzf


2


+ 16mnab4

M1(G) Da + Dy1113872 1113873f(a b)|ab1

(18)


M1(G) Da + Dy1113872 1113873f(a b)|ab1


+ 8mn] +[2(4n + 4m minus 4)]




(19)





Proof suppose



times a2b2

+ 4mn times a2b4

(21)




times a times b2


4



times a2b2

+ 8mn times a2b4

(22)

Now take Db



times 2b b

+ 8mna2

times 4b3

times b



+ 32mn times a2b4

(23)





(24)





mM

2(G)

32

mn + m + n minus 13 (26)







0(M(G y s))ydt δs 1113938t



Proof suppose



+ 4mna2b4

(27)


Sa 1113946a

0

f(x b)

xdx



+ 4mnx2b4

f(x b)



2

+ 4mnxb4

(28)


1113946a

0

f(x b)

xdx 1113946

a

0(4m + 4n minus 4)b

2dx

+ 1113946a


2dx

+ 4mn 1113946a

0xdxb

4


+12


+ 2mna2b4

(29)


50

0

ndash50 ndash50

0

50

15

1

05

0

ndash05

ndash1

times105

(a)

50

0

ndash50 ndash500

50

15

1

05

0

ndash05

ndash1

times105

(b)


50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(a)

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(b)




+12


+ 2mna2x4

SaSbf(a b) 12


+14


+12

mna2b4

(30)



12

(4m + 4n minus 4) +14


mn


mn


1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)




1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose



times a2b2

+(4mn) times a2b4

(34)




times a2b2

+ 8mn times a2b4

(35)

Now take Db



times 4b3

times b

(36)


DαaD



2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2


2b2

+ 23α+2mna

2b4

(37)


Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2


Rα(G) 2α+2m + 2α+2


mn minus 22α+2m


+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)



RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References







































Da zf

zaa

zf



2+ 8mnab

4

(16)


Da azf


2


+ 8mna2b4

(17)

Similarly

Dbf(a b) bzf


2


+ 16mnab4

M1(G) Da + Dy1113872 1113873f(a b)|ab1

(18)


M1(G) Da + Dy1113872 1113873f(a b)|ab1


+ 8mn] +[2(4n + 4m minus 4)]




(19)





Proof suppose



times a2b2

+ 4mn times a2b4

(21)




times a times b2


4



times a2b2

+ 8mn times a2b4

(22)

Now take Db



times 2b b

+ 8mna2

times 4b3

times b



+ 32mn times a2b4

(23)





(24)





mM

2(G)

32

mn + m + n minus 13 (26)







0(M(G y s))ydt δs 1113938t



Proof suppose



+ 4mna2b4

(27)


Sa 1113946a

0

f(x b)

xdx



+ 4mnx2b4

f(x b)



2

+ 4mnxb4

(28)


1113946a

0

f(x b)

xdx 1113946

a

0(4m + 4n minus 4)b

2dx

+ 1113946a


2dx

+ 4mn 1113946a

0xdxb

4


+12


+ 2mna2b4

(29)


50

0

ndash50 ndash50

0

50

15

1

05

0

ndash05

ndash1

times105

(a)

50

0

ndash50 ndash500

50

15

1

05

0

ndash05

ndash1

times105

(b)


50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(a)

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(b)




+12


+ 2mna2x4

SaSbf(a b) 12


+14


+12

mna2b4

(30)



12

(4m + 4n minus 4) +14


mn


mn


1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)




1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose



times a2b2

+(4mn) times a2b4

(34)




times a2b2

+ 8mn times a2b4

(35)

Now take Db



times 4b3

times b

(36)


DαaD



2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2


2b2

+ 23α+2mna

2b4

(37)


Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2


Rα(G) 2α+2m + 2α+2


mn minus 22α+2m


+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)



RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References







































Proof suppose



+ 4mna2b4

(27)


Sa 1113946a

0

f(x b)

xdx



+ 4mnx2b4

f(x b)



2

+ 4mnxb4

(28)


1113946a

0

f(x b)

xdx 1113946

a

0(4m + 4n minus 4)b

2dx

+ 1113946a


2dx

+ 4mn 1113946a

0xdxb

4


+12


+ 2mna2b4

(29)


50

0

ndash50 ndash50

0

50

15

1

05

0

ndash05

ndash1

times105

(a)

50

0

ndash50 ndash500

50

15

1

05

0

ndash05

ndash1

times105

(b)


50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(a)

50

0

ndash50 ndash50

0

50

151

050

ndash05

ndash15ndash1

times105

(b)




+12


+ 2mna2x4

SaSbf(a b) 12


+14


+12

mna2b4

(30)



12

(4m + 4n minus 4) +14


mn


mn


1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)




1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose



times a2b2

+(4mn) times a2b4

(34)




times a2b2

+ 8mn times a2b4

(35)

Now take Db



times 4b3

times b

(36)


DαaD



2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2


2b2

+ 23α+2mna

2b4

(37)


Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2


Rα(G) 2α+2m + 2α+2


mn minus 22α+2m


+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)



RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References








































+12


+ 2mna2x4

SaSbf(a b) 12


+14


+12

mna2b4

(30)



12

(4m + 4n minus 4) +14


mn


mn


1113874 1113875

(31)


mM2(G)

32

mn + m + n minus 1 (32)




1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(33)

Proof suppose



times a2b2

+(4mn) times a2b4

(34)




times a2b2

+ 8mn times a2b4

(35)

Now take Db



times 4b3

times b

(36)


DαaD



2b2

+ 8α(4mn)a2b4

DαaD

αb 2α+2

m + n minus 1ab2


2b2

+ 23α+2mna

2b4

(37)


Rα(G) DαaD

αb(f(a b))|ab1

Rα(G) 2α+2(m + n minus 1) + 22α+2


Rα(G) 2α+2m + 2α+2


mn minus 22α+2m


+ 23α+2mn

(38)

+e result is


1113872 1113873(m + n minus 1) + 22α+2+ 23α+2

1113872 1113873mn

(39)



RRα(G) 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n) +1

22αminus 2 +2

23αminus 21113890 1113891(mn)

+1

2αminus 2 +1

22αminus 21113890 1113891

(40)


Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References







































Proof suppose



times a2b2

+(4mn) times a2b4

(41)


Sa (4m + 4n minus 4) 1113946a

0dxb


middot 1113946a

0xdxb

2+ 8mn 1113946

a

0xdxb

4



+ 4mna2b4

(42)

Similarly take Sb



2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx



+ mna2b4


SαaS

αb


2+


2b2

+1

23αminus 2 mna2b4

(44)


50

0

ndash50 ndash500

50

4000

2000

0

ndash2000

ndash4000

(a)

50

0

ndash50 ndash50

0

50

4000

2000

0

ndash2000

ndash4000

(b)


6040

200

1500

1000

500

0

010

2030

4050

(a)

6040

200

1500

1000

500

0

010

2030

4050

(b)





+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References









































+1


+2

23αminus 2 mn 1

2αminus 2 minus1

22αminus 21113890 1113891(m + n)

+1

22αminus 2 +2

23αminus 21113890 1113891(mn) +1

2αminus 2 +1

22αminus 21113890 1113891

(45)



Proof suppose



times a2b2

+(4mn) times a2b4

(46)


Sb (4n + 4m minus 4)a 1113946b


2

middot 1113946b

0xdx + 4mna

21113946

b

0x3dx

Sb 12


+12


+ mna2b4

(47)

Now take Da

SbDa 12



+ 2mna2b4

(48)

Similarly


+12


+ 2mna2b4

(49)

Take Db



+ 8mna2b4

(50)



Put the values

SS D(G) 12




(52)


SS D(G) 18mn + 2m + 2n minus 2 (53)



H(G) 53

(m + n minus 1) +73

mn (54)

Proof suppose



times a2b2

+(4mn) times a2b4

(55)




+ 8mna6

(56)

Take Sa




middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References









































middot 1113946a

0x3dt + 8mn 1113946

a

0x5dx

SaJf(a b) 43

(m + n minus 1)a3

+12


+23

mna6

(57)


H(G) 2SaJf(a b)|a1

243

(m + n minus 1) +12


mn1113876 1113877

H(G) 243

minus12

1113874 1113875m +43

minus12

1113874 1113875n +12

minus43

1113874 1113875 +43

+12

1113874 11138751113876 1113877mn

H(G) 256

m +56

n +76

mn minus56

1113876 1113877

(58)

Now the result is

H(G) 53

(m + n minus 1) +73

mn (59)



SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (60)

Proof suppose



times a2b2

+(4mn) times a2b4

(61)


6040

200

5

4

3

2

1

0

times104

010

2030

4050

(a)

6040

200

5

4

3

2

1

0

times104

010

2030

4050

(b)


50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(a)

50

0

ndash50 ndash50

0

50

5

0

ndash5

times104

(b)





+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References









































+ 32mna2b4

(62)Take Da



+ 64mna2b4

(63)Take Jf(a b)



+ 64mnx6

(64)

Take S(a)

SaJDaDbf(a b) 83

(m + n minus 1)a3


+323

mna6



(m + n minus 1)


mn

83

minus 41113874 1113875m +83

minus 41113874 1113875n

+323

+ 41113874 1113875mn + 4 minus83

1113874 1113875

(66)

50

0

ndash50 ndash50

0

50

600040002000


(a)

50

0

ndash50 ndash50

0

50

600040002000


(b)


50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(a)

50

0

ndash50 ndash50

0

50

4

2

0

ndash2

ndash4

times104

(b)




SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References








































SaJDaDb(f(a b)) 443

mn minus43

(m + n minus 1) (67)


Data Availability


Disclosure




References







































m-polynomials and degree-based topological indices of the

Documents