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On Kirchhoff index and number of spanning trees of
linear pentagonal cylinder and Mobius chain graph
Md. Abdus Sahir* and Sk. Md. Abu Nayeemβ
Department of Mathematics and Statistics, Aliah University, Kolkata β 700 160, India.
Abstract
In this paper, we derive closed-form formulas for Kirchhoff index and Wiener index of
linear pentagonal cylinder graph and linear pentagonal Mobius chain graph. We also obtain
explicit formulas for finding total number of spanning trees for both the graphs.
MSC (2020): Primary: 05C09; Secondary: 05C50.
Keywords. Pentagonal cylinder, Mobius chain, Kirchhoff index, Wiener index, spanning tree.
1 Introduction
Let G = (V,E) be a connected molecular graph having vertex set V and edge set E. A topological
index of G is a numerical quantity involving different graph parameters such as number of vertices,
number of edges, degree, eccentricity, distance between vertices, etc. Winner index, one of the
oldest topological indices, was introduced by Harold Wiener [22] and was defined as W (G) =
βi< j
di j, where di j is the length of the shortest path between the vertices i and j. Motivated by the idea
of Wiener index, the idea of resistance distance and Kirchhoff index was introduced by Klein and
Randic [10]. Kirchhoff index, initially known as resistance index, was defined as K f (G) = βi< j
ri j,
where ri j is the effective resistance between vertex i and vertex j calculated using Ohmβs law
considering all the edges of G as unit resistors. Klein and Randic also proved that for any vertex
pair i, j in a graph G, ri j β€ di j and K f (G)β€W (G) with equality holds if and only if G is a tree.
In the last few decades, researchers have focused on various topological indices such as Wiener
index, Randic index [19], Kirchhoff index, Gutman index [6], Estrada index [3], Zagreb index
[8] etc. Especially Kirchhoff index is seeking a lot of attention of researchers as it has wide
applications in physics, chemistry, graph theory and various related subjects. Readers are referred
to [4, 16, 20, 21, 23, 26, 27] for some recent works. Many researchers have concentrated on finding
*Email: [email protected]β Corresponding author. Email: [email protected]
1
the Kirchhoff index and the number of spanning trees for many interesting graphs such as linear
hexagonal chain [24], linear pentagonal chain [20], Mobius hexagonal chain [21], periodic linear
chain [1], crossed hexagonal chain [17], linear octagonal chain [28], Mobius/ cylinder octagonal
chain [12] and many others [5, 13, 14, 18]. Normalized Laplacian spectrum and the number of
spanning trees of linear pentagonal chains have been obtained by He et al. [9]. Although the
Kirchhoff index of linear pentagonal chain was found long back in 2010 [20], but to the best of
our knowledge, Kirchhoff indices for linear pentagonal cylinder and Mobius chains have not been
obtained so far. In the present paper, we aim to obtain those.
The Laplacian matrix L(G) = (li j) of a simple connected graph G is defined as,
li j =
di, if i = j
β1, if i βΌ j
0, otherwise,
where di is the degree of the vertex i.
Since L(G) is a symmetric matrix, all of its eigenvalues are real. Moreover all are non-negative,
i.e., if the eigenvalues Ξ»i,s (i = 1,2, . . . ,n) are indexed in the increasing order of their values,
0 = Ξ»1 β€ Ξ»2 β€ Β·Β· Β· β€ Ξ»n. Since G is connected, Ξ»1 = 0 is a simple eigenvalue.
Gutman and Mohar [7] and Zhu et al. [29] obtained the following lemma.
Lemma 1 [7, 29] For a connected graph G with n-vertices, n β₯ 2,
K f (G) = nn
βk=2
1
Ξ»k
Β·
Like Wiener index, Kirchhoff index also gives description of the underlying structure of a
molecular graph [23]. Obtaining closed-form formulae for Kirchhoff index of general graphs is
not straight forward, but we can derive closed-form formulae for some special classes of graphs
like cycles [11], complete graphs [15], circulant graph, etc.
In this paper, we derive closed-form formulas for Kirchhoff index and Wiener index of linear
pentagonal cylinder graph Pn (Figure 1) and pentagonal Mobius chain graph Pβ²n (Figure 2) on 5n
(n β₯ 2) vertices. Also we present the formulas for the total number of spanning trees for those
graphs.
2
1
1β²
2
2β²
1
3
3β²
4
2
4β²
5
5β²
2nβ1
2nβ1β²
2n
n
2nβ²
Figure 1: Linear pentagonal cylinder graph Pn.
1
1β²
2
2β²
1
3
3β²
4
2
4β²
5
5β²
2nβ1
2nβ1β²
2n
n
2nβ²
Figure 2: Pentagonal Mobius chain graph Pβ²n.
3
2 Preliminaries
Let G be a graph with vertex set V (G). A permutation Ο of V (G) is called an automorphism if u and
v are adjacent in G if and only if Ο(u) and Ο(v) are also adjacent in G. Suppose V0 = {1, 2, . . . , p},
V1 = {1,2, . . . ,q} and V2 = {1β²,2β², . . . ,qβ²} are the vertex partition for the automorphism Ο such that
Ο(i) = i for all i βV0,Ο(i) = iβ² for all i βV1 and Ο(iβ²) = i for all iβ² βV2. It is easy to follow that Ο
can be decomposed as product of disjoint 1-cycles and transpositions, i.e.,
Ο = (1)(2) Β· Β· Β·(p)(1,1β²)(2,2β²) Β· Β· Β·(q,qβ²),
where p+2q = |V (G)|. Then by suitable arrangement of vertices, the Laplacian matrix L(G) of G
can be expressed into the block matrix form β
L(G) =
LV0V0LV0V1
LV0V2
LV1V0LV1V1
LV1V2
LV2V0LV2V1
LV2V2
where the submatrix LVrVscorrespond to the vertices of Vr and Vs,r,s = 0,1,2 respectively.
Let
LA(G) =
[
LV0V0
β2LV0V1β
2LV1V0LV1V1
+LV1V2
]
and
LS(G) = LV1V1βLV1V2
.
Yang and Yu [25] and many others like Yang and Zhang [24] have used the Laplacian decomposi-
tion formula to find the Kirchhoff indices of certain classes of graphs where some automorphisms
are found. We describe it in the form of the following lemma.
Lemma 2 The characteristic polynomial of L(G) is equal to the product of that of LA(G) and
LS(G), i.e.,
det(L(G)βΞ» I) = det(LA(G)βΞ» I) Β·det(LS(G)βΞ» I).
Let Ξ»i (i = 1,2, . . . ,3n) and Β΅ j ( j = 1,2, . . . ,2n) are the eigenvalues of LA(G) and LS(G) ar-
ranged in ascending order of their values. Then by Lemma 2, the spectrum of L(G) is given by
{0 = Ξ»1 β€ Ξ»2 β€ Β·Β· Β· β€ Ξ»3n}β
{Β΅1 β€ Β΅2 β€ Β·Β· Β· β€ Β΅2n}.
To avoid confusion, we denote LA(Pn) and LS(Pn) by LA and LS respectively and LA(Pβ²n) and
LS(Pβ²n) by Lβ²
A and Lβ²S respectively. Also we denote the block matrices constituting the Laplacian
4
matrix of linear pentagonal cylinder graph Pn by LVrVs,r,s = 0,1,2 and those of pentagonal Mobius
chain graph Pβ²n by Lβ²
VrVs,r,s = 0,1,2 respectively. Then,
LV0V0= Lβ²
V0V0=
2 0 Β· Β· Β· 0
0 2 Β· Β· Β· 0...
.... . .
...
0 0 Β· Β· Β· 2
nΓn
, LV0V1= Lβ²
V0V1=
0 β1 0 0 Β· Β· Β· 0
0 0 0 β1 Β· Β· Β· 0...
......
.... . .
...
0 0 0 0 Β· Β· Β· β1
nΓ2n
LV1V1=
3 β1 0 Β· Β· Β· 0 β1
β1 3 β1 Β· Β· Β· 0 0
0 β1 3 Β· Β· Β· 0 0...
......
. . ....
...
0 0 0 Β· Β· Β· 3 β1
β1 0 0 Β· Β· Β· β1 3
2nΓ2n
, LV1V2=
β1 0 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 0
0 0 β1 Β· Β· Β· 0 0...
......
. . ....
...
0 0 0 Β· Β· Β· β1 0
0 0 0 Β· Β· Β· 0 0
2nΓ2n
, and
Lβ²V1V1
=
3 β1 0 Β· Β· Β· 0 0
β1 3 β1 Β· Β· Β· 0 0
0 β1 3 Β· Β· Β· 0 0...
......
. . ....
...
0 0 0 Β· Β· Β· 3 β1
0 0 0 Β· Β· Β· β1 3
2nΓ2n
, Lβ²V1V2
=
β1 0 0 Β· Β· Β· 0 β1
0 0 0 Β· Β· Β· 0 0
0 0 β1 Β· Β· Β· 0 0...
......
. . ....
...
0 0 0 Β· Β· Β· β1 0
β1 0 0 Β· Β· Β· 0 0
2nΓ2n
.
So,
LA = Lβ²A =
2 0 Β· Β· Β· 0 0 ββ
2 0 0 Β· Β· Β· 0
0 2 Β· Β· Β· 0 0 0 0 ββ
2 Β· Β· Β· 0...
.... . .
......
......
.... . .
...
0 0 Β· Β· Β· 2 0 0 0 0 Β· Β· Β· ββ
2
0 0 Β· Β· Β· 0 2 β1 0 0 Β· Β· Β· β1
ββ
2 0 Β· Β· Β· 0 β1 3 β1 0 Β· Β· Β· 0
0 0 Β· Β· Β· 0 0 β1 2 β1 Β· Β· Β· 0
0 ββ
2 Β· Β· Β· 0 0 0 β1 3 Β· Β· Β· 0...
.... . .
......
......
.... . .
...
0 0 Β· Β· Β· ββ
2 β1 0 0 0 Β· Β· Β· 3
3nΓ3n
,
5
LS =
4 β1 0 0 Β· Β· Β· 0 0 β1
β1 3 β1 0 Β· Β· Β· 0 0 0
0 β1 4 β1 Β· Β· Β· 0 0 0
0 0 β1 3 Β· Β· Β· 0 0 0...
......
.... . .
......
...
0 0 0 0 Β· Β· Β· 3 β1 0
0 0 0 0 Β· Β· Β· β1 4 β1
β1 0 0 0 Β· Β· Β· 0 β1 3
2nΓ2n
,
and Lβ²S =
4 β1 0 0 Β· Β· Β· 0 0 1
β1 3 β1 0 Β· Β· Β· 0 0 0
0 β1 4 β1 Β· Β· Β· 0 0 0
0 0 β1 3 Β· Β· Β· 0 0 0...
......
.... . .
......
...
0 0 0 0 Β· Β· Β· 3 β1 0
0 0 0 0 Β· Β· Β· β1 4 β1
1 0 0 0 Β· Β· Β· 0 β1 3
2nΓ2n
.
For our convenience, we denote the matrix
4 β1 0 0 Β· Β· Β· 0 0 0
β1 3 β1 0 Β· Β· Β· 0 0 0
0 β1 4 β1 Β· Β· Β· 0 0 0
0 0 β1 3 Β· Β· Β· 0 0 0...
......
.... . .
......
...
0 0 0 0 Β· Β· Β· 3 β1 0
0 0 0 0 Β· Β· Β· β1 4 β1
0 0 0 0 Β· Β· Β· 0 β1 3
2nΓ2n
by L0S, so that LS = L0
Sβe1eT2n and Lβ²
S = L0S+e1eT
2n where ei is the unit column vector of compatible
size with all its components 0 except the ith component which has the value 1.
In this paper, we shall use the following lemma, known as the matrix-determinant lemma to
compute the determinant of a matrix with rank one perturbation if the determinant of the original
matrix is known.
Lemma 3 Let M be an nΓ n matrix. Then det(M + uvT ) = det(M)+ vT adj(M)u, where u,v are
6
nΓ1 column vectors.
3 Kirchhoff index of pentagonal cylinder and Mobius chain
From Lemma 2, we have that the Kirchhoff index of linear pentagonal cylinder Pn is
K f (Pn) = 5n
(
3n
βi=2
1
Οi+
2n
βj=1
1
Β΅ j
)
,n β₯ 2
where Οi, i = 1,2, . . . ,3n and Β΅ j, j = 1,2, . . . ,2n are the eigenvalues of LA and LS respectively.
Let
det(xI3n βLA) = x3n +Ξ±1x3nβ1 + Β· Β· Β·+Ξ±3nβ2x2 +Ξ±3nβ1x,(since Ο1 = 0) (1)
and
det(xI2n βLS) = x2n +Ξ²1x2nβ1 + Β· Β· Β·+Ξ²2nβ2x2 +Ξ²2nβ1x+Ξ²2n. (2)
From Vietaβs formula, we have3n
βi=2
1Οi=βΞ±3nβ2
Ξ±3nβ1and
2n
βj=1
1Β΅i= Ξ²2nβ1
Ξ²2n= Ξ²2nβ1
det(LS)Β·
Hence
K f (Pn) = 5n
(
βΞ±3nβ2
Ξ±3nβ1+
Ξ²2nβ1
det(LS)
)
Β·
By similar argument,
K f (Pβ²n) = 5n
(
3n
βi=2
1
Οi+
2n
βj=1
1
Β΅ β²j
)
= 5n
(
βΞ±3nβ2
Ξ±3nβ1+
Ξ² β²2nβ1
det(Lβ²S)
)
,
where Β΅ β²j, j = 1,2, . . . ,2n are the eigenvalues of Lβ²
S and Ξ² β²2nβ1 is the coefficient of the first degree
term in the characteristic polyomial of Lβ²S respectively.
Lemma 4 Let Rn =
β2 1 0 Β· Β· Β· 0 0
1 β2 1 Β· Β· Β· 0 0
0 1 β2 Β· Β· Β· 0 0...
......
. . ....
...
0 0 0 Β· Β· Β· β2 1
0 0 0 Β· Β· Β· 1 β2
nΓn
, then det(Rn) = (β1)n(1+n).
7
Proof. Here det(R1) =β2, det(R2) = 3 and det(Rn) =β2det(Rnβ1)βdet(Rnβ2), n β₯ 3. Solving
the recurrence relation, we get det(Rn) = (β1)n(1+n). οΏ½
Lemma 5 Let
Rn,m =
β2 1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
1 β2 1 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
0 1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· β2 1 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 1 β3 1 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 1 β2
nΓn
,
where β3 is at (m,m) position (m β€ n). Then det(Rn,m) = (β1)n(1+n+m+mnβm2).
Proof. Observe that Rn,m = Rn β ememT . By Lemma 3, we have
det(Rn,m) = det(Rn)β eTm adj(Rn)e
m
= det(Rn)β cofactor of the entry at (m,m) position of Rn,m
= det(Rn)β (β1)m+m det(Rmβ1) Β·det(Rnβm)
= (β1)n(1+n)β (β1)mβ1m Β· (β1)nβm(1+nβm) (by Lemma 4)
= (β1)n(1+n+m+mnβm2).
οΏ½
Lemma 6 For n β₯ 2,3n
βi=2
1Οi=βΞ±3nβ2
Ξ±3nβ1= 25n2+30nβ13
60Β·
Proof. Let M be a square matrix. By M{i}, we denote the submatrix of M, obtained by deleting the
ith row and ith column of M. With this notation, we have from (1) that Ξ±3nβ1 =3n
βi=1
det(βLA{i}).
Now, for 1 β€ i β€ n, det(βLA{i}) =β£
β£
β£
β£
β£
β2Inβ1 S
ST Q
β£
β£
β£
β£
β£
, where S =ββ
2LV0V1{i} and Q =βLV1V1
β
LV1V2. Using Schur complement, we have for 1β€ iβ€ n, det(βLA{i})= det(β2Inβ1) Β·det
(
Q+ 12ST S)
.
8
Now,
Q+1
2ST S =
β2 1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 1
1 β2 1 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
0 1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· β2 1 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 1 β3 1 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1
1 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 1 β2
2nΓ2n
,
with β3 at (2i,2i) position.
Thus, Q+ 12ST S = R2n,2i + e1eT
2n + e2neT1 .
Let R1 = R2n,2i + e1eT2n. Then,
det(
R1)
= det(R2n,2i)+ eT2n adj(R2n,2i)e1
= det(R2n,2i)+ cofactor of the entry at (1,2n) position of R2n,2i
= det(R2n,2i)+(β1)2n+1 Β·1= 1+2n+2i+4niβ4i2β1 (by Lemma 5)
= 2n+2i+4niβ4i2.
9
Thus,
det
(
Q+1
2ST S
)
= det(
R1)
+ eT1 adj(R1)e2n
= det(
R1)
+ cofactor of the entry at (2n,1) position of R1
= det(R1)+(β1)2n+1 Β·det
1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 1
β2 1 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
.... . .
......
.... . .
......
0 0 Β· Β· Β· β2 1 0 Β· Β· Β· 0 0
0 0 Β· Β· Β· 1 β3 1 Β· Β· Β· 0 0
0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0...
.... . .
......
.... . .
......
0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1
2nβ1Γ2nβ1
= det(R1)β [1+det(R2nβ2,2iβ1)]
= 2n+2i+4niβ4i2β [1+1+(2nβ2)+(2iβ1)+(4niβ2nβ4i+2)
β(4i2 β4i+1)]
(by Lemma 5)
= 2n.
Hence det(βLA{i}) = (β2)nβ1 Β·2n = (β1)nβ1n Β·2n for i = 1,2, . . . ,n.
Again, for n+ 1 β€ i β€ 3n, det(βLA{i}) =β£
β£
β£
β£
β£
β2In U
UT Q{iβn}
β£
β£
β£
β£
β£
, where U = ββ
2LV0V1and Q =
βLV1V1βLV1V2
. Using Schur complement, we have for n+1 β€ i β€ 3n,det(βLA{i}) = det(β2In) Β·det(
Q{iβn}+ 12UTU
)
.
10
Now,
Q{iβn}+ 1
2UTU =
β2 1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 1
1 β2 1 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
0 1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· β2 0 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 β2 1 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1
1 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 1 β2
(2nβ1)Γ(2nβ1)
,
(
the diagonal blocks are of size (iβnβ1)Γ (iβnβ1)
and (3nβ i)Γ (3nβ i) respectively.)
= R2 + e1eT2nβ1, where, R2 =
[
Riβnβ1 0
0 R3nβi
]
+ e2nβ1eT1 .
Clearly, det(
R2)
= det(Rnβiβ1) Β·det(R3nβi) and hence,
det
(
Q{iβn}+ 1
2UTU
)
= det(Riβnβ1)det(R3nβi)+ cofactor of the entry at the
(1,2nβ1) position of R2 (by Lemma 3)
= det(Riβnβ1)det(R3nβi)βdet(Riβnβ2)det(R3nβiβ1)
= β(nβ i)(3nβ i+1)+(iβnβ1)(3nβ i)
= β2n.
So, det(LA{i}) = (β1)n+1n Β·2n+1 for n+1 β€ i β€ 3n.
11
Hence,
Ξ±3nβ1 =3n
βi=1
det(βLA{i})
=n
βi=1
det(βLA{i})+3n
βi=n+1
det(βLA{i})
= n Β· (β1)n+1n Β·2n+2n Β· (β1)n+1n Β·2n+1
= (β1)n+12n Β·5n2. (3)
Again suppose M{i, j} denotes the principal submatrix of M obtained by deleting the ith row
and jth row and the corresponding columns. Then from (1), we have
Ξ±3nβ2 = β1β€i< jβ€3n
det(βLA{i, j})
= β1β€i< jβ€n
det(βLA{i, j})+ βn+1β€i< jβ€3n
det(βLA{i, j})+ β1β€i<n
n+1β€ jβ€3n
det(βLA{i, j}).
For 1 β€ i < j β€ n,
det(βLA{i, j}) =β£
β£
β£
β£
β£
β2Inβ2 0(nβ2)Γ2n
02nΓ(nβ2) E2nΓ2n
β£
β£
β£
β£
β£
,
where
E =
β2 1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 1
1 β2 1 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
0 1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· β2 1 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 1 β3 1 Β· Β· Β· 0 Β· Β· Β· 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1 0 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 Β· Β· Β· 0 Β· Β· Β· 1 β3 1 Β· Β· Β· 0 0
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 1 β2 Β· Β· Β· 0 0...
......
. . ....
......
. . ....
......
. . ....
...
0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· β2 1
1 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 0 0 0 Β· Β· Β· 1 β2
2nΓ2n
,
12
with β3 at the (2i,2i) and at (2 j,2 j) positions.
By repeated application of Lemma 3, we get det(E) = 4n+ 8i jβ 4n(iβ j)β 4(i2 + j2) and
hence det(βLA{i, j}) = (β1)nβ22nβ2[4n+8i jβ4n(iβ j)β4(i2+ j2)] for 1 β€ i < j β€ n.
For n+1 β€ i < j β€ 3n, let p = iβn and q = jβn. So, 1 β€ p < q β€ 2n.
Then, det(βLA{p,q})=β£
β£
β£
β£
β£
β2In 0
0 F(2nβ2)Γ(2nβ2)
β£
β£
β£
β£
β£
,where F =
Rpβ1 0 epβ1eT2nβqβ1
0 Rqβp 0
e2nβqβ1eTpβ1 0 R2nβqβ1
.
Proceeding as before, det(F) = 2n(qβ p)+2pqβ (p2+q2) and hence
det(βLA{p,q}) = (β1)n2n[2n(qβ p)+2pqβ (p2+q2)] for 1 β€ p < q β€ 2n.
Similarly, for 1 β€ i < n, n+1 β€ j β€ 3n, let q = jβn, i.e., for 1 β€ q β€ 2n,
det(βLA{i,q}) =β£
β£
β£
β£
β£
β2Inβ1 0
0 H(2nβ1)Γ(2nβ1)
β£
β£
β£
β£
β£
,
where det(H) =
βq2 +4qiβ4i2+2n+2n(qβ2i), if 2i β€ q
βq2 +4qiβ4i2+2nβ2n(qβ2i), if 2i > q .
Hence,
Ξ±3nβ2 = (β1)nβ22nβ2 x4 +6x3 β7x2
3+(β1)n2n 4x4 β x2
3+(β1)n2nβ1 4n4 +12n3 βn2
3
= (β1)n2nβ2 25n4+30n3 β13n2
3Β·
Hence3n
βi=2
1Οi=βΞ±3nβ2
Ξ±3nβ1= (β1)n+12nβ2(25n4+30n3β13n2)
(β1)n+12nΒ·15n2 = 25n2+30nβ1360
Β· οΏ½
Lemma 7 Let, ri be the determinant of iΓ i submatrix of L0S formed by the first i rows and first
i columns of L0S. Then for 1 β€ i β€ 2n, ri = s1(
β2+
β3)i + s2(β
β2β
β3)i + s3(
β2β
β3)i +
s4(ββ
2+β
3)i where s1 =(β
2+β
3)(2+β
3)
4β
6, s2 =
(β
2+β
3)(β2+β
3)
4β
6, s3 =
(β
2ββ
3)(β2+β
3)
4β
6and s4 =
(β
2ββ
3)(2+β
3)
4β
6Β·
Proof. By observation r1 = 4, r2 = 11, r3 = 40. If we choose r0 = 1, then for 2 β€ i β€ 2n
ri =
3riβ1 β riβ2, when i is even
4riβ1 β riβ2, when i is odd.
13
For 1 β€ i β€ nβ1, let ci = r2i and di = r2i+1, then
ci = 3diβ1 β ciβ2, when i is even
di = 4ci βdiβ1, when i is odd.(4)
By substitution, we have
ri = 10riβ2β riβ4,4 β€ i β€ 2n.
The auxiliary equation for this recurrence relation is x4 β10x2 +1 = 0. Solving, we get
x =Β±(β
2+β
3), Β± (β
2ββ
3). (5)
Thus the general solution is ri = s1(β
2+β
3)i+s2(ββ
2ββ
3)i+s3(β
2ββ
3)i+s4(ββ
2+β3)i. Using the initial conditions r0 = 1, r1 = 4, r2 = 11 and r3 = 40 we get four linear equations
in s1,s2,s3 and s4. Solving those equations by Cramerβs rule, we get
s1 =(β
2+β
3)(2+β
3)
4β
6,
s2 =(β
2+β
3)(β2+β
3)
4β
6,
s3 =(β
2ββ
3)(β2+β
3)
4β
6,
s4 =(β
2ββ
3)(2+β
3)
4β
6Β·
οΏ½
Lemma 8 For n β₯ 2,3n
βi=2
1Β΅i= Ξ²2nβ1
det(LS)=
7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]+ 7
β3
48 [(4nβ2β
6n+1)(β
3ββ
2)2nβ1+(4n+2β
6n+1)(β
3+β
2)2nβ1](β
2+β
3)2n+(β
2ββ
3)2nβ2and
3n
βi=2
1Β΅ β²
i=
Ξ² β²2nβ1
det(Lβ²S)=
7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]+ 7
β3
48 [(4nβ2β
6n+1)(β
3ββ
2)2nβ1+(4n+2β
6n+1)(β
3+β
2)2nβ1](β
2+β
3)2n+(β
2ββ
3)2n+2.
Proof. Apply Lemma 3 on LS and Lβ²S to get,
det(Ls) = r2n β r2nβ2 β2
= (β
2+β
3)2n +(β
2ββ
3)2n β2 (6)
14
and
det(Lβ²s) = r2n β r2nβ2 +2
= (β
2+β
3)2n +(β
2ββ
3)2n +2 (7)
respectively.
It is not difficult to follow that Ξ²2nβ1 = Ξ² β²2nβ1 =
2nβ1
βi=0
rirβ²2nβ1βi where rβ²i is the determinant of
iΓ i submatrix formed by last i rows and last i columns of L0S for 1 β€ i < 2n. Note that rβ²1 = 4,
rβ²2 = 11 and rβ²3 = 30. For convenience, we also choose rβ²0 = 1.
Let f (x) =β
βi=0
rixi and g(x)=β
βi=0
rβ²ixi. Then Ξ²2nβ1 = coefficients of x2nβ1βcoefficients of x2nβ3
in f (x)g(x).
So,
f (x) =β
βi=0
rixi = 1+4x+11x2 +40x3 +β
βi=4
rixi
= 1+4x+11x2 +40x3 +β
βi=4
(10riβ2β riβ4)xi
= 1+4x+11x2 +40x3 +10x2β
βi=4
riβ2xiβ2 β x4β
βi=4
riβ4xiβ4
= 1+4x+11x2 +40x3 +10x2(
f (x)β1β4x)
β x4 f (x).
Hence, f (x) = x2+4x+1x4β10x2+1
Β·In a similar way, we can deduce that g(x) = x2+3x+1
x4β10x2+1Β·
Thus we have
f (x)g(x) =x4 +7x3 +14x2 +14x+1
(x4 β10x2 +1)2Β·
Expressing as partial fraction,
f (x)g(x) =12β
2β7β
6384
x+β
2+β
3+
β12β
2β7β
6384
xββ
2ββ
3+
12β
2+7β
6384
x+β
2ββ
3+
β12β
2+7β
6384
x+β
3ββ
2
+12β7
β3
192
(x+β
2+β
3)2+
12+7β
3192
(xββ
2ββ
3)2+
12+7β
3192
(xββ
3+β
2)2+
12β7β
3192
(x+β
3ββ
2)2.
Observe that the coefficient of x2nβ1 in
1
(x+β
2+β
3)2= 1
(β
2+β
3)2
(
1+ xβ2+
β3
)β2
= (β
3ββ
2)2
(
β
βi=0
( β1β2+
β3)ixi
)2
is
15
(β
3ββ
2)2
(
2nβ1
βi=0
( β1β2+
β3)i( β1β
2+β
3)2nβ1βi
)
=β2n(β
3ββ
2)2n+1.
Considering the coefficients of x2nβ1 and x2nβ3 from each fraction obtained in similar manner,
we have Ξ²2nβ1 = Ξ² β²2nβ1 =
7β
6192
[
(4β2β
6)(5β2β
6)nβ1 β (4+2β
6)(5+2β
6)nβ1]
+7β
348
[
(4nβ2β
6n+1)(β
3ββ
2)2nβ1 +(4n+2β
6n+1)(β
3+β
2)2nβ1]
.
Hence3n
βi=2
1Β΅i
=7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]+ 7
β3
48 [(4nβ2β
6n+1)(β
3ββ
2)2nβ1+(4n+2β
6n+1)(β
3+β
2)2nβ1](β
2+β
3)2n+(β
2ββ
3)2nβ2and
3n
βi=2
1Β΅ β²
i
=7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]+ 7
β3
48 [(4nβ2β
6n+1)(β
3ββ
2)2nβ1+(4n+2β
6n+1)(β
3+β
2)2nβ1](β
2+β
3)2n+(β
2ββ
3)2n+2.οΏ½
Combining Lemma 6 and Lemma 8, we have the following theorem.
Theorem 1 For n β₯ 2,
K f (Pn) = 5n
(
βΞ±3nβ2
Ξ±3nβ1+
Ξ²2nβ1
det(LS)
)
= 5n
25n2 +30nβ13
60+
7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]
+ 7β
348 [(4nβ2
β6n+1)(
β3β
β2)2nβ1+(4n+2
β6n+1)(
β3+
β2)2nβ1]
(β
2+β
3)2n +(β
2ββ
3)2n β2
,
and
K f (Pβ²n) = 5n
(
βΞ±3nβ2
Ξ±3nβ1+
Ξ² β²2nβ1
det(Lβ²S)
)
= 5n
25n2 +30nβ13
60+
7β
6192 [(4β2
β6)(5β2
β6)nβ1β(4+2
β6)(5+2
β6)nβ1]
+ 7β
348 [(4nβ2
β6n+1)(
β3β
β2)2nβ1+(4n+2
β6n+1)(
β3+
β2)2nβ1]
(β
2+β
3)2n +(β
2ββ
3)2n +2
Β·
In 1997, Chung [2] showed that the number of spanning trees Ο(G) for a connected graph G is
equal to the product of non-zero Laplacian eigenvalues of G devided by the number of vertices of
G. From (3), (6) and (7) we have the following theorem.
16
Theorem 2 The total number of spanning trees of Pn is
Ο(Pn) =
3n
βi=2
Ξ»i
2n
βj=1
Β΅ j
5n
=(β1)3nβ1Ξ±3nβ1 det(Ls)
5n
= 2nn(
(β
2+β
3)2n +(β
2ββ
3)2n β2)
and the total number of spanning trees of Pβ²n is
Ο(Pβ²n) =
3n
βi=2
Ξ»i
2n
βj=1
Β΅ β²j
5n
=(β1)3nβ1Ξ±3nβ1 det(Lβ²
s)
5n
= 2nn(
(β
2+β
3)2n +(β
2ββ
3)2n +2)
.
4 Relation between Kirchhoff index and Wiener index
First we calculate Wiener index for Pn and Pβ²n.
Theorem 3 The Wiener index of Pn is
W (Pn) =
254
n3 +9n2, when n is even
254
n3 +9n2 β n4, when n is odd.
and the Wiener index of Pβ²n is
W (Pβ²n) =
254
n3 +9n2 β2n, when n is even
254
n3 +9n2 β 9n4, when n is odd.
Proof. There are three type of vertices of Pn,
1. a-type: vertex with degree 3 and non-adjacent to vertex of degree 2,
2. b-type: vertex with degree 3 and which is adjacent to a vertex of degree 2,
3. c-type: vertex of degree 2 (middle vertex) of Pn.
Now we observe the following.
17
1. Sum of the distances from an a-type vertex to
(a) all the upper vertices j : 2nβ1
βi=1
i+n = n2.
(b) all the lower vertices jβ² : 2n
βi=1
i+n = n2 +2n.
(c) all the middle vertices j:
i. when n is even: 4n/2
βi=1
i = n2(n+2).
ii. when n is odd: 4(n+1)/2
βi=1
iβ (n+1) = 12(n+1)2.
2. Sum of the distances from a b-type vertex to
(a) all the upper vertices j : 2nβ1
βi=1
i+n = n2.
(b) all the lower vertices jβ² : 2n
βi=1
i+(n+1) = (n+1)2.
(c) all the middle vertices j:
i. when n is even: 2n/2
βi=1
(2iβ1)+n = n2
2+n.
ii. when n is odd: 2(n+1)/2
βi=1
(2iβ1)β1 = (n+1)2β22
+n.
3. Sum of the distances from a c-type vertex to
(a) all the upper vertices j : 2n
βi=1
i+n = n2 +2n.
(b) all the lower vertices jβ² : 2n
βi=1
i+n = n2 +2n.
(c) all the middle vertices j:
i. when n is even: 4n/2
βi=1
i+nβ2 = n2+4nβ42
.
ii. when n is odd: 4(n+1)/2
βi=1
iβ4 =(n+1)(n+3)
2β4.
18
Hence the Wiener index for Pn is
W (Pn) =
2n
βj=1
(
2n2 +(n2 +2n)+(n+1)2+ n2(n+2)+ n2
2+n+n2 +2n
)
+2n
βj=1
(
n2+4nβ42
)
2
=2n(
2n2 +(n2 +2n)+(n+1)2+ n2(n+2)+ n2
2+n+n2 +2n
)
+n(
n2+4nβ42
)
2
=25
4n3 +9n2,
when n is even and
W (Pn) =
2n
βj=1
(
2n2 +(n2 +2n)+ 12(n+1)2+(n+1)2 + (n+1)2β2
2+n2 +2n
)
+n
βj=1
(
(n+1)(n+3)2
β4)
2
=2n(
2n2 +(n2 +2n)+ 12(n+1)2+(n+1)2 +
(n+1)2β22
+n2 +2n)
+n(
(n+1)(n+3)2
β4)
2
=25
4n3 +9n2 β n
4,
when n is odd.
By similar approach we get the Wiener index for Pβ²n as
W (Pβ²n) =
{
254
n3 +9n2 β2n, when n is even,254
n3 +9n2 β 9n4, when n is odd.
οΏ½
From Theorem 1 and Theorem 3 we have the following theorem.
Theorem 4
limnββ
W (Pn)
K f (Pn)= 3
and
limnββ
W (Pβ²n)
K f (Pβ²n)
= 3.
In Table 1, for different values of n, we list the values of Kirchhoff indices and Wiener indices
and the ratios of Wiener indices to Kirchhoff indices for both Pn and Pβ²n. It is evident from the table
that both the ratios are gradually approaching to 3.
19
n K f (Pn) W (Pn)W (Pn)K f (Pn)
K f (Pβ²n) W (Pβ²
n)W (Pβ²
n)K f (Pβ²
n)
2 39.083333 86 2.200426458 38.5 82 2.12987012987
3 107.715909 249 2.3116362505 107.583333 243 2.25871418206
4 226.166667 544 2.40530581812 226.142857 536 2.37018319796
5 406.806193 1005 2.47046386533 406.802434 995 2.44590473616
6 662.098485 1674 2.52832477029 662.097938 1662 2.51020265222
7 1004.536492 2583 2.57133515862 1004.536417 2569 2.55739857363
8 1446.619048 3776 2.61022416732 1446.619038 3760 2.5991639134
9 2000.845977 5283 2.64038314829 2000.845975 5265 2.63138695621
10 2679.717254 7150 2.66819194799 2679.717254 7130 2.660728474
20 19073.869017 53600 2.81012729783 19073.869017 53560 2.80803018791
99 2080862.36308 6152553 2.95673231885 2080862.36308 6152355 2.95663768188
Table 1: Kirchhoff index and Wiener index of pentagonal cylinder chain Pn and pentagonal Mobius
chain Pβ²n for different values of n β₯ 2.
5 Concluding remarks
In this paper, we have derived explicit formulas for Kirchhoff index and Wiener index of linear
pentagonal cylinder chain graph Pn of 5n vertices and linear pentagonal Mobius chain graph Pβ²n of
5n vertices. We have also established that for large values of n, Wiener index is almost three times
the Kirchhoff index for both the graphs.
Acknowledgement
University Grants Commission, India has provided a partial support through Senior Research Fel-
lowship to the first author for this work.
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