Open Journal of Discrete Mathematics, 2013, 3, 60-69
http://dx.doi.org/10.4236/ojdm.2013.31013 Published Online January 2013 (http://www.scirp.org/journal/ojdm)
Dominating Sets and Domination Polynomials
of Square of Paths
A. Vijayan1, K. Lal Gipson2
1
Department of Mathematics, Nesamony Memorial Christian College,
Marthandam, India
2
Department of Mathematics, Mar Ephraem College of Engineering and Technology,
Kanayakumari District, India
Email: [email protected], [email protected]
Received August 30, 2012; revised September 30, 2012; accepted October 8, 2012
ABSTRACT
Let G = (V, E) be a simple graph. A set S  V is a dominating set of G, if every vertex in V-S is adjacent to at least one
vertex in S. Let Pn2 be the square of the Path Pn and let D Pn2 , i denote the family of all dominating sets of Pn2








with cardinality i. Let d Pn2 , i  D Pn2 , i . In this paper, we obtain a recursive formula for d Pn2 , i . Using this re-


cursive formula, we construct the polynomial, D Pn2 , x  
n
n
i 
5


d Pn2 , i xi , which we call domination polynomial of
Pn2 and obtain some properties of this polynomial.
Keywords: Domination Set; Domination Number; Domination Polynomials
G1 , , Gm , then
1. Introduction
Let G = (V, E) be a simple graph of order |V| = n. For any
vertex v  V, the open neighbourhood of  is the set
N  v   u  V uv  E and the closed neighbourhood
of  is the set N  v   N  v   v . For a set S  V, the
open neighbourhood of S is N  S   U nS N  v  and the
closed neighbourhood of S is N  S   N  S   S . A set S
 V is a dominating set of G, if N[S] = V, or equivalently,
every vertex in V-S is adjacent to at least one vertex in S.
The domination number of a graph G is defined as the
minimum size of a dominating set of vertices in G and it
is denoted as   G  . A simple path is a path in which all
its internal vertices have degree two, and the end vertices
have degree one and is denoted by Pn .
Definition 1.1.: Let D  G , i  be the family of dominating sets of a graph G with Cardinality i and let
d  G, i   D  G, i  .
Then the domination polynomial D  G, x  of G is
defined as
V G 
D  G , x    i   G  d  G , i  x i .
where   G  is the domination number of G.
Result 1.2. [1]: If a graph G consists of m components
Copyright © 2013 SciRes.
D  G, x   D  G1 , x  D  Gm , x  .
Result 1.3. [1]: Let G1 and G2 be graphs of order
n1 and n2 respectively. Then

 1  x 
D  G1  G2 , x   1  x  1  1
n
n2

1
 D  G1 , x   D  G2 , x  .
Result 1.4. [1]: Let G be a bipartite graph with bipartition (X,Y). Then G contains a matching that saturates
every vertex in X if and only if for all S  V N  S   S .
Result 1.5. [1]: Let G be a graph of order n. Then for
n
every 0  i  , we have d  G, i   d  G , i 1 .
2
Result 1.6. [1]: For any graph G of order n,
  G  K1   n .
Result 1.7. [1]: For any graph G of order n and
n  m  2n , we have
 n  2nm
.
d  G  K1 , m   
2
m  n
Hence D  G  K1 , x   x n  x  2  .
n
Definition 1.8: The Square of a graph with the same
set of vertices as G and an edge between two vertices if
OJDM
A. VIJAYAN, K. L. GIPSON
and only if there is a path of length at most 2 between
them. The second power of a graph is also called its
Square.
Let Pn2 be the square of the path Pn (2nd power )with
n vertices. Let D Pn2 , i be the family of dominating

2
n
sets of the graph P




d P ,i  D P ,i
2
n



.


D Pn2 , x  
n
i 
5




d Pn2 , i xi




Y  D P , i 1 ,
2
n 6
Y contains at least one vertex labeled n – 6, n – 7 or
n – 8.
If n  6  Y , then Y  D Pn25 , i  1 , a contradiction.
Hence n – 7 or n – 8  Y, but then in this case,
Y   x  D Pn2 , i , for any x [n], also a contradiction.
Lemma 2.5.:
1) If








D Pn21 , i  1  D Pn23 , i  1  

Let D Pn2 , i be the family of dominating sets of Pn2
with cardinality i. We investigate the dominating sets of
Pn2 . We need the following lemma to prove our main
results in this section.
n
Lemma 2.1. [3]:  Pn2    By Lemma 2.1 and
5
the definition of domination number, one has the following Lemma:
Lemma 2.2. D Pn2 , i   if and only if i  n or
n
i    . A simple path is a path in which all its internal
5
vertices have degree two, and the end vertices have degree one. The following Lemma follows from observation [2].
Lemma 2.3. [2]: If a graph G contains a simple path
of length 5k – 1, then every dominating set of G must
contain at least k vertices of the path.
In order to find a dominating set of Pn2 , with cardinality i, we only need to consider
 



 

D  P , i  1 , D  P , i  1
, i  1 . The families of these dominating sets
D Pn21 , i  1 , D Pn2 2 , i  1 ,
2
n 3
2
n4
and D Pn25
can be empty or otherwise. Thus, we have eight combinations, whether these five families are empty or not.
Two of these combinations are not possible (Lemma
2.5(i), & (ii)). Also, the combination that


Proof: Suppose that Y  D Pn25 , i  1 . Since
2. Main Result



or equal to x and  x  for the smallest integer greater
than or equal to x. Also, we denote the set 1, 2, , n
by [n], throughout this paper.



Y  D Pn25 , i  1 .
the domination polynomial of the graph Pn2 [2].
In the next section, we construct the families of the
dominating sets of the square of paths by recursive
method.
As usual we use  x  for the largest integer less than


does not need to be considered because it implies that
D Pn2 , i   (See Lemma 2.5 (iii)). Thus we only need
to consider five combinations or cases. We consider
those cases in Theorem 2.7.
Lemma 2.4.: If Y  D Pn2 6 , i  1 , and there exists x
 [n] such that Y   x  D Pn2 , i , then
We call the polynomial
n

D Pn2 4 , i  1  D Pn25 , i  1  

with cardinality i and let
2
n
61




D Pn21 , i  1  D Pn2 2 , i  1  D Pn23 , i  1  
Copyright © 2013 SciRes.
then






D Pn2 2 , i  1  
2) If


D Pn21 , i  1   , D Pn23 , i  1  
then
D Pn2 2 , i  1   ,
3) If

DP
DP



, i  1   , D  P , i  1   ,
, i  1   , then D  P , i   
D Pn21 , i  1   , D Pn2 2 , i  1   ,
2
n 3
2
n 5
2
n4
2
n
Proof:
1) Since D Pn21 , i  1  D Pn23 , i  1   , by Lemma
 n  3
2.2, i  1  n  1 (or) i  1  
 . In either case, we
 5 
have D Pn2 2 , i  1   .
2) Suppose that D Pn2 2 , i  1   , so by Lemma 2.2,
n  2
we have i  1  n  2 or i  1  
 If i  1  n  2 ,
 5 
then i  1  n  3 . Therefore D Pn23 , i  1   , a contradiction.
3) Suppose that D Pn2 , i   , Let Y  D Pn2 , i .
Thus, at least one vertex labeled n or n – 1 or n – 2 is in Y.
If n  Y, then by Lemma (2.3), at least one vertex labeled
n – 1, n – 2, n – 3, n – 4 or n – 5 is in Y. If n – 1  Y or














OJDM
A. VIJAYAN, K. L. GIPSON
62


n – 2  Y, then Y – n  D Pn21 , i  1 , a contradiction.
If n – 3  Y, then Y – n  D Pn2 2 ,i  1 , a contra-


diction. Now, suppose that n – 1  Y. Then, by Lemma
2.3, at least one vertex labeled n – 2, n – 3, n – 4, n – 5 or
n – 6 is in Y. If n – 2  Y or n – 3  Y, then
Y – n  1  D Pn2 2 , i  1 a contradiction. If n – 4  Y,




then Y – n  1  D P , i  1 , a contradiction.
If n – 5  Y, then Y – n  1  D Pn2 4 , i  1 , a con2
n 3




D Pn2 , i   , a contradiction.
n  2
2
So i  
  1 , and since D Pn , i   , we have
 5 
n
n  2
 5   i   5   1 , which implies that n = 5k and i = k
 


for some kN.
() If n = 5k and i = k for some kN, then by Lemma
2.2,













 D P , i  1   and D P
2
n 5

, i  1  




DP





, i  1   , if and only if n = 5k + 2 and
D Pn23 , i  1   , D Pn2 4 , i  1   ,
2
n 5
 5k  2 
i
 for some kN;
 5 




4) D Pn21 , i  1   , D Pn2 2 , i  1   ,

DP



, i  1   if and only if i = n – 3.
5) D  P , i  1   , D  P , i  1   ,
D  P , i  1   , D  P , i  1   , and
D  P , i  1   if and only if
D Pn23 , i  1   , D Pn2 4 , i  1   , and
2
n 5
2
n 1
2
n2
2
n 3

,

2
n4
2
n 5
 n  1
 5  1  i  n  4 .








 D Pn24 , i  1  D Pn25 , i  1  ,
by Lemma 2.2,
 n  3
i  1  n  2 or i  1< 
.
 5 
If
 n  1
 n  3
i 1  
, then i  1  


 5 
 5 


and hence D Pn2 2 , i  1   , a contradiction.
So i  1  n  2 . Also, D Pn21 , i  1   .


Therefore i  n  1 . Therefore i  n . But i  n .
Hence i = n.
() If i = n, then by lemma 2.2,




D Pn2 2 , i  1  D Pn23 , i  1

 

then D  P , i  1   .
3) () Since D  P ,i  1   , by Lemma 2.2,
 D P , i 1  D P , i 1  
2
n4
2
n 5
2
n 1
2
n 1
 n  1
i  1  n  1 or i  1  
.
 5 
If i  1  n  1 , then i  1  n  2 , by Lemma 2.2,

Proof:
1) ()
Since

D Pn2 2 , i  1  D Pn23 , i  1

if i = n;
3) D Pn21 , i  1   , D Pn2 2 , i  1   ,



 D Pn25 , i  1   then D Pn21 , i  1   if and only


2
n4
2) () Since
if and only
if n = 5k and i = k for some k  N;
2) D  Pn2 2 , i  1  D  Pn23 , i  1  D  Pn2 4 , i  1


2
n 3
and D Pn2 5 , i  1   .
1) D Pn21 , i  1  D Pn22 , i  1  D Pn23 , i  1
2
n4

 D P , i 1  D P , i 1  

tradiction. Therefore, D Pn2 , i   .
Lemma 2.6.: If D Pn2 , i   , then



D Pn21 , i  1  D Pn2 2 , i  1
tradiction
If n – 6  Y, then Y – n  1  D Pn25 , i  1 , a con-





D Pn2 2 , i  1  D Pn23 , i  1





D P , i 1  D P , i 1
2
n 1


2
n2

2
n4

by Lemma 2.2,
n  2
i  1  n  1 or i  1  

 5 
If i  1  n  1 , then i  n and by Lemma 2.2,
Copyright © 2013 SciRes.


a contradiction.
 D P , i  1  D P , i  1  ,
2
n 3

 D Pn24 , i  1  D Pn25 , i  1  ,
 n  1
Therefore i  
 1.
 5 
n  2
2
But i  1  
 , because D Pn  2 , i  1   .
 5 
n  2
Therefore i  
 1 .
 5 


OJDM
A. VIJAYAN, K. L. GIPSON
n  2
 n  1
Hence, 
1  i  

 1.
 5 
 5 
This holds only if n = 5k + 2 and
So
 n  1
 5   i 1  n  5


 5k  2 
i  k 1  
,
 5 
for some k  N.
 5k  2 
() If n = 5k + 2 and i  
 for some k  N,
 5 
then by Lemma 2.2,

DP


, i  1   , D  P

, i  1  
D Pn21 , i  1   , D Pn2 2 , i  1   ,
2
n 3
2
n4
and

 n  1
and hence 
 1  i  n  4 .
 5 
 n  1
() If 
  1  i  n  4 , then the result follows
 5 
from Lemma 2.2.
n
Theorem 2.7.: For every n  6 and i    ,
5
1) If




n  5
i  1  n  5 or i  1  
.
 5 

n  2
 5  1  i  n 1 .



, i  1  
D Pn21 ,i  1   , D Pn2 2 , i  1   ,
2
n 3
2
n4


, i  1   , D  P , i  1  
D  P , i  1   , then by ap-
D Pn25 , i  1   .
5) () Since


2
n 1
D P , i  1   , and
plying Lemma (2.2),
2
n 3
2
n 5









 D Pn2 4 , i  1  D Pn25 , i  1  
and


D Pn21 , i  1  
then


D Pn2 , i   n  .
3) If

DP


, i  1   , D  P

, i  1  
D Pn21 , i  1   , D Pn2 2 ,i  1   ,
2
n 3
2
n4
and


D Pn25 , i  1  
then

D Pn2 , i

 3,8, , n  4, n  2 , X 1  n , X 2  n  1 ,


X 3  n  2 X 1  D Pn23 , i  1 ,
 n  3
 n  3
 5   i 1  n  3 ,  5   i 1  n  4




Copyright © 2013 SciRes.

D Pn2 2 , i  1  D Pn23 , i  1
 n  1
n  2
 5   i 1  n 1 ,  5   i 1  n  2 ,




 n  3
 5   i 1  n  5 .



D Pn2 , i  3,8, , n  7, n  2 .
2
n2
and

then
and

DP

D Pn25 , i  1  



, i  1   , D  P

2
n4
2) If
n  5
Therefore i  1  
 is not possible. Therefore
 5 
i  1  n  5 . Therefore i  n  4 .
Therefore i  n  3 .
Since D Pn2 3 , i  1   , i  1  n  4 .
Therefore i  n  3 . Hence i  n  3 .
() if i  n  3 then
By Lemma 2.2,

DP

2
n 3
and
Since D Pn2 2 , i  1   , by Lemma 2.2,


 D P , i 1  D P , i 1  
4) () Since D Pn25 , i  1   , by Lemma 2.2,


D Pn21 , i  1  D Pn2 2 , i  1
D Pn25 , i  1   .

63



DP
(2.1)

, i  1   ,
X 2  D Pn2 4 , i  1 , X 3  D Pn25 , i  1


4) If D Pn23 , i  1   ,


2
n2
D Pn21 , i  1   ,
OJDM
A. VIJAYAN, K. L. GIPSON
64
then
5) If


D Pn2 , i   n    x , x   n  .









D Pn21 , i  1   , D Pn2 2 , i  1   ,



D Pn2 3 , i  1   , D Pn2 4 , i  1   , D Pn2 5 , i  1  
then
 





D Pn2 , i  X 1  n , X 2  n  1 , X 3  n  2 X 1  D Pn21 , i  1 , X 2  D Pn2 2 , i  1 , X 3  D Pn23 , i  1 ,
n  2   X
4

and
Proof:
1) Since










2
n4





DP
2
n 3

By Lemma 2.6 (iii), n = 5k + 2 and
 5k  2 
i
  k 1
 5 
for some k  N.
Since
X  3,8, ,5k  2  D P52k , k ,



Also, if X  D P



2
n 5

Now let Y  D P52k  2 , k  1 . Then 5k + 2, or 5k + 1 is
in Y. If 5k + 2 Y, then by Lemma 3, at least one vertex
labeled 5k + 1, 5k or 5k – 1 is in Y. If 5k + 1 or 5k is in Y,
then
Y  5k  2  D P52k 1 , k ,


a contradiction; because D P52k 1 , k   .
Hence 5k – 1  Y, 5k  Y and 5k + 1  Y.
Therefore, Y  X  5k  2 for some X  D P52k , k ;

Copyright © 2013 SciRes.
2
5 k 1

,k

then


Therefore, we have

X 2  D P , i 1 , X 3  D P , i 1  D P


X  5k  2  D P52k  2 , k  1 .



D Pn25 , i  1   .
3,8 5k  2,5k  1 , X1  n , X 2  n  1 , X 3  n  2


, i  1  
X  5k  1  D P52k  2 , k  1


2
n4
and
 D Pn24 , i  1  D Pn25 , i  1  
2
n4


, i  1   , D  P
D Pn21 , i  1   , D Pn2 2 ,i  1   ,



3) We have
D Pn2 2 , i  1  D Pn23 , i  1


D Pn2 , i   n  .
and D P , i  1   by Lemma 2.6 i) n = 5k, and i =
k for some k  N.
n
elements.
Clearly the set 3,8, , n  7, n  2 has
5
By the definition of Pn2 , 3 has joining with 1, 2, 4, 5,
and 8 has joining with 6, 7, 9, 10. Therefore 3 and 8 covers all the vertices up to10for n = 10. Proceeding like this,
we obtain that 3,8, , n  7, n  2 covers all Vertices
n
are
up to n. The other sets with cardinality
5
2, 7,12, , n  3 , 4,9, , n  1 etc. In the first set,
n  3 does not cover the vertex n. The second set does
not cover the vertex 1and so on.
Therefore 3,8, , n  7, n  2 is the only dominatn
 k i.
ing set of cardinality
5
2) We have

(2.2)
By Lemma 2.6 (ii), we have i = n. So,
 D P , i 1  D P , i 1  
2
n 5


D Pn21 , i  1   .
D Pn21 , i  1  D Pn2 2 , i  1
2
n 3

 X 3  , n   X 4  X 3  , n  1   X 5  X 4  , X 4  D Pn2 4 , i  1 , X 5  D Pn25 , i  1

2
5k  2


X 1  D Pn23 , i  1 ,
(2.3)

, k 1
that is Y  3,8, ,5k  2,5k  1 . Now suppose that 5k
+ 1 Y and 5k + 2  Y. By Lemma 2.3 at least one vertex
labeled 5k, 5k – 1 or 5k – 2 is in Y. If 5k  Y then
Y  5k  1  D P52k , k  3,8, ,5k  2 ,




a contradiction because 5k  X for all X  D P52k , k .
Therefore 5k – 1, or 5k – 2 is in Y, but 5kY.
Thus Y  X  5k  1 for some X  D P52k 1 , k .


So
OJDM
A. VIJAYAN, K. L. GIPSON

 
, i  1 , X
65


D P52k  2 , k  1  3,8, ,5k  2,5k  1 , X 1  n , X 2  n  1 , X 3  n  2 X 1  D Pn23 , i  1 ,

X 2  D Pn24
3



(2.4)

 D Pn25 , i  1  D P52k  2 , k  1
From (2.3) and (2.4),



D Pn2 , i  3,8, , n  4, n  2 ,






X 1  n , X 2  n  1 , X 3  n  2 X 1  D Pn23 , i  1 , X 2  D Pn2 4 , i  1 , X 3  D Pn25 , i  1 .
4) If

DP
DP


, i  1   ,
, i  1   ,
D P , i 1   ,
2
n 3
2
n2
2
n 1


DP





2
n 5

Now let X 3  D P , i  1 then n – 3, n – 4 or n – 5
is in X3. If n – 3 or n – 4 or n – 5  X3, then
2
n 3


X 3  n  2  D Pn2 , i .






, i  1   .
X  D  P , i  1 , so at least one vertex labeled
Now let X 4  D Pn2 4 , i  1 , then n – 4, or n – 5 or n
– 6 is in X4.
If n – 4  X4, then X 4  n  D Pn2 , i . If n – 5  X4,
then
X 4  n  1  D Pn2 , i .

If n – 6  X4, then X 4  n  2  D Pn2 , i . Thus we
have
D P , i 1   , D P , i 1   ,
2
n 3

X 2  n  1  D Pn2 , i .
by Lemma 2.6 (iv) i = n – 1.
Therefore D Pn2 , i   n    x x   n  .
5) D Pn21 , i  1   , D Pn2 2 , i  1   ,


Let X 2  D Pn2 2 , i  1 , then n – 2 or n – 3 or n – 4 is
inX2.
If n – 2, n – 3 or n – 4  X2, then
2
n4

2
n2
Let 1
n – 1, n – 2 or n – 3 is in X1. If n – 1, n – 2 or n – 3 X1,
then
X 1  n  D Pn2 , i .


 X  n , X  n  1 , X  n  2 X  D  P , i  1 , X  D  P , i  1 , X  D  P , i  1
 n  2   X  X  , n   X  X  , n  1   X  X  X  D  P , i  1 , X  D  P
1
2
3
4
2
n 1
1
3
4
2
n2
2
3
5
4
Now suppose that n – 1 Y, n  Y then by Lemma 2.3,
at least one vertex labeled n – 2, n – 3 or in Y. If n – 2 
Y then Y   X 4  X 3   n  2 for some




X 4  D Pn2 4 , i  1 , X 3  D Pn2 3 , i  1





2
n 3
3
2
n4
4
2
n 5
5
, i  1  D Pn2 , i
(2.5)



 DP

, i  1 , X



 DP

, i  1
X 4  D Pn2 4 , i  1 , X 5  D Pn2 5 , i  1 ,
.
So


D Pn2 , i  X 1  n , X 2  n  1 , X 3  n  2 X 1  D Pn21 , i  1 , X 2  D Pn2 2 , i  1 , X 3  D Pn23 , i  1
4

Lemma 2.3 one vertex labeled n – 6, n – 7, in Y.
If n – 4  Y, then Y   X 5  X 4   n  1 for
Now suppose that, n – 5 Y & n – 4  Y, then by
  
 n  2   X

 X 3  , n   X 4  X 3  , n  1   X 5  X 4  X 4
2
n4
5
2
n 5
(2.6)
From (2.5) & (2.6)
  
 n  2   X



 DP

, i  1 , X

 DP

, i  1
D Pn2 , i  X 1  n , X 2  n  1 , X 3  n  2 X 1  D Pn21 , i  1 , X 2  D Pn2 2 , i  1 , X 3  D Pn23 , i  1
4
 X 3  , n   X 4  X 3  , n  1   X 5  X 4  X 4
3. Domination polynomial of Pn2


Let D P , x  
2
n
n
n
i 
5

2
n

d P ,i x
i
be the domination
polynomial of a path Pn2 . In this section, we derive the
Copyright © 2013 SciRes.
2
n4

5
2
n 5

expression for D Pn2 , x .
Theorem 3.1.
a) If D  Pn2 , i  is the family of dominating sets with
cardinality i of Pn2 , then
OJDM
A. VIJAYAN, K. L. GIPSON
66





 
 
 
 

d Pn2 , i  d Pn21 , i  1  d Pn2 2 , i  1  d Pn23 , i  1  d Pn24 , i  1  d Pn25 , i  1


where d Pn2 , i  D Pn2 , i .
b) For every n  6 ,







 



D Pn2 , x  x  D Pn21 , x  D Pn2 2 , x  D Pn23 , x  D Pn24 , x  D Pn25 , x 

with the initial values
 
D  P , x  x
D  P , x  x
D  P , x  x


D P , x  x , D P , x  x  2x ,
2
1
2
2
2

 3x  3x ,
2
3
3
2
4
4
 4 x3  6 x 2  2 x ,
2
5
5
 5 x 4  10 x3  8 x 2  x .
2
 
2
n1





2
n 1
Since

2
n




2
n 5
2
n2
2
n 3




D Pn2 4 , i  1  D Pn2 5 , i  1  0




Therefore D Pn2 , i  D Pn21 , i  1

, i  1  
, i  1  0




Therefore d P , i  d P , i  1 .
Therefore, In this case d P , i  d Pn21 , i  1
2
n
2
n1
2
n





D Pn2 , i    n  1   X 2  , n  2   X 3  ,   X 4  X 3   n  2 ,  X 4  X 3   n ,

holds.
3) Suppose (iii) of Theorem 2.7 holds.
From (v), we have
2
n2
 X 5  X 4   n  1

2
n 3
2
n4
D Pn21 , i  1  D Pn2 2 , i  1   ,
2
n4

 

D  P , i  1  D  P , i  1  
D  P , i  1  D  P , i  1  
D  P , i  1  D  P , i  1  0

Since
2
n 3

D Pn2 , i    n   X 1  X 1  D Pn21 , i  1 .


 
, i  1  D  P
, i  1  D  P

Therefore, in this case d P , i  d Pn2 5 , i  1 holds.
2) Suppose (ii) of Theorem 2.7 holds.
From (v), we have
D Pn2 , i    n  1   X 5  X 5  D Pn25 , i  1 .

DP
DP

Therefore D Pn2 , i  D Pn25 , i  1 .
Therefore d Pn2 , i  d Pn2 5 , i  1 .
Proof:
a) Using (i), (ii), (iii), (iv) and (v) of Theorem 2.7, we
prove (a) part.
1) Suppose (i) of Theorem 2.7 holds.
From (v), we have


D Pn23 , i  1  D Pn2 4 , i  1  0







X 2  D Pn2 2 , i  1 , X 3  D Pn23 , i  1 , X 4  D Pn2 4 , i  1 and X 5  D Pn25 , i  1
.
Therefore
Since






d  P
d  P
2
n 3
Therefore







2
n 5
D Pn2 , i  D Pn2 2 , i  1  D Pn2 3 , i  1




 D Pn2 4 , i  1  D Pn2 5 , i  1


, i  1  d  P
, i  1 .
d Pn2 , i  d Pn2 2 , i  1
D Pn21 , i  1   , D Pn21 , i  1  0 .
.
2
n4

, i 1 .
4) Suppose (v) of Theorem 2.7 holds.
From (v), we have

D Pn2 , i  n   X 1  , n  1   X 2  , n  2   X 3  ,   X 4  X 3   n  2 ,  X 4  X 3   n ,
 X 5  X 4   n  1








X 1  D Pn21 , i  1 , X 2  D Pn2 2 , i  1 , X 3  D Pn23 , i  1 ,


X 4  D Pn2 4 , i  1 and X 5  D Pn25 , i  1 .
Therefore
Copyright © 2013 SciRes.
OJDM
A. VIJAYAN, K. L. GIPSON







67





D Pn2 , i  D Pn21 , i  1  D Pn2 2 , i  1  D Pn23 , i  1  D Pn24 , i  1  D Pn25 , i  1 .
Therefore



 
 
 
 

 d Pn2 , i  d Pn21 , i  1  d Pn2 2 , i  1  d Pn23 , i  1  d Pn2 4 , i  1  d Pn25 , i  1 .
Therefore, we have the theorem.
b)










 d  P , i  x   d  P , i  1 x   d  P , i  1 x   d  P , i  1 x   d  P
d  P , i  x  x  d  P , i  1 x  x  d  P , i  1 x  x  d  P , i  1 x
 x  d  P , i  1 x  x  d  P , i  1 x
 d  P , i  x  x   d  P , i  1 x   d  P , i  1 x   d  P , i  1 x
  d  P , i  1 x   d  P , i  1 x 
D  P , x   x  D  P , x   D  P , x   D  P , x   D  P , x   D  P , x  


d Pn2 , i x i  d Pn21 , i  1 xi  d Pn2 2 , i  1 xi  d Pn23 , i  1 xi  d Pn2 4 , i  1 xi  d Pn25 , i  1 xi
2
n
2
n
2
n 1
i
2
n 1
i 1
2
n2
i 1
2
n4
i 1
2
n 5
i 1
i
i
2
n
2
n2
i 1
2
n 1
i
2
n
2
n 1
2
n 3
 
 
D  P , x   x  3x  3x
D  P , x  x  4x  6x  2x ,
D  P , x   x  5 x  10 x  8 x  x .
We obtain d  P , i  for 1 ≤ n ≤ 10 as shown in Table
D P , x  x, D P , x  x  2x ,
1.
2
2
2
3
3
2
2
4
4
3
2
5
5
4
2
2
3
2
2
n
Table 1. d  Pn2 , i  , the number of dominating set of Pn2 with
cardinality i.
i
1
2
3
4
5
6
7
8
9
i1
2
n 3
i1
2
n4
with the initial values
2
1
2
n 3



, i  1 x   d Pn25 , i  1 xi
i
i 1
2
n 5
2
n2
2
n 4
i
i 1
2
n2
i 1
2
n4
2
n 3
i
10
2
n 5
In the following Theorem, we obtain some properties
of d Pn2 , i
Theorem 3.2. The following properties hold for the
coefficients of D Pn2 , x ;
1) d P5n2 , n  1 for every n  N.



2)
3)

 
d  P , n   1 for every n  N
d  P , n  1  n for every n  2
2
n
2
n

  2  for every n  3
d  P , n  3  nC  2 for every n  4
d  P , n  4   nC   2n  6  for every n  5
d  P , n   n  1 for every n  N.
4) d Pn2 , n  2 
5)
6)
7)
n n 1
2
n
3
2
n
4
2
5 n 1
Proof:
1) Since D P52n , n  3,8, , 5k  2 , we have
d P5n2 , n  1 .
2) Since D Pn2 , n   n  , we have the result
n

1
1
2
2
1
3
3
3
1
4
2
6
4
1
5
1
8
10
5
1
6
0
9
18
15
6
1
7
0
8
27
33
21
7
1
8
0
6
34
60
54
28
8
1
9
0
3
37
93
114
82
36
9
1
10
0
1
34
126
206
196
118
45
10


Copyright © 2013 SciRes.
 
D  P , n  1   n    x
2
n
3) Since



x   n  , we have
d Pn2 , n  1  n .
4) By induction on n. The result is true for n = 3.
L.H.S.  d P32 ,1  3 (from table)


 3 2 
R.H.S  
3
 2 
1
Therefore, the result is true for n = 4.
Now suppose that the result is true for all numbers less
than “n” and we prove it for n.
By Theorem 3.1,
OJDM
A. VIJAYAN, K. L. GIPSON
68



 
, n  3  d  P
2
d Pn2 , n  2  d Pn21 , n  3  d Pn2
,n 3

 
 d Pn23 , n  3  d Pn24
2
n 5

,n 3
5) By induction on n. The result is true for n = 4.
L.H.S  d P42 ,1  2 (from table)


4  3 2
2
6
Therefore the result is true for n = 1. Now suppose the
result is for all natural numbers less than n.
By Theorem 3.1,

 n  1 n  2  



 

R.H.S  d P42 ,1 
 n  2  1
2
n2  3n  2  2n  4  2 n 2  n n  n  1



.
2
2
2


 
 
 
d Pn2 , n  3  d Pn21 , n  4  d Pn2 2 , n  4  d Pn23 , n  4  d Pn2 4 , n  4  d Pn25 , n  4

 n  1  n  2  n  3  2   n  1 n  2  
6
2

 n  3  1
1
 n  1 n  2  n  3  3  n  2  n  3 6  n  3  12  6 
6
1
1
  n  3  n  2  n  2   6   6    n  3  n 2  4  6   6 
6
6
1 3
1 3
2
2
  n  2n  3n  6  6    n  3n  2n  12 
6
6

6) By induction on n. Let n = 5
L.H.S  d P52 ,1  1 (from table)


R.H.S 
n  n  1 n  2  n  3
5 4  3 2
  2n  6  
  2  5  6  5  4  1
1 2  3  4
1 2  3  4
Therefore the result is true for n =1.
Now suppose that the result is true for all natural numbers less than or equal to n.



 
 
 
 
d Pn2 , n  4  d Pn21 , n  5  d Pn2 2 , n  5  d Pn23 , n  5  d Pn2 4 , n  5  d Pn25 , n  5

n  n  1 n  2  n  3 n  4 
24
 2  n  1  6  
 n  2  n  3 n  4 
6
2
 n  3 n  4 
2

  n  4  1
1
 n  1 n  2  n  3 n  4   12  n  2  n  3 n  4   24  n  4   4  n  2  n  3 n  4   48n  192  48  24 
24 
1
 n  2  n  3 n  4   n  1  4  12  n  4   n  3  2  48n  168

24 
1

 n  2  n  3 n  4  n  3  12  n  4  n  1  48n  168
24 
1
 n  4   n  2  n  3 n  3  12  n  1  48n  168 

24 

1 
1
 n  4   n  2  n2  9  12n  12  48n     n  4  n3  9n  2n 2  18  36n  156
24 
24
1 4
1 4
 n  9n 2  2n3  18n  4n3  36n  8n 2  72  36n  156  
 n  6n3  11n 2  54n  144 

24
24 



7) From the table it is true.
Theorem 3.3.
1)
 i  n d  Pi 2 , n   5 i  2 d  Pi 2 , n  1
5n
n
2) For every j    ,
5
Copyright © 2013 SciRes.
5n 5



 


 
d Pn21 , j  1  d Pn2 , j  1  d Pn2 , j  d Pn25 , j
3) If Sn  
n
n
i 
5



d Pn2 , j , then for every n ≥ 6,
S n  Sn 1  Sn  2  Sn  3  Sn  4  Sn 5 with initial values S1
OJDM
A. VIJAYAN, K. L. GIPSON
= 1, S2 = 3, S3 = 7, S4 = 13 and S5 = 25.
Proof:
69
1) Proof by induction on n
First suppose that n = 2 then,
 i d  Pi 2 , 2   45  5 i 2 d  Pi 2 ,1
5k
5k
5k
5k
5k
5k
 i k d  Pi 2 , k    i k d  Pi 21 , k  1   i k d  Pi 22 , k  1   i k d  Pi 23 , k  1   i k d  Pi 2 4 , k  1   i k d  Pi25 , k  1
5 k 1
5 k 1
5 k 1
5 k 1
5 k 1
 5 i k 1 d  Pi 21 , k  2   5  i k 1 d  Pi 2 2 , k  2   5 i k 1 d  Pi 23 , k  2   5 i k 1 d  Pi 2 4 , k  2   5 i k 1 d  Pi 25 , k  2 
5 k 5
 5 i k 1 d  Pi 2 , k  1
10
5
We have the result.
2) By Theorem 3.1, We have

 

  
 
 

d  P , j   d  P , j   d  P , j   d  P , j   d  P
d  P , j  1  d  P , j  1  d  P , j   d  P , j 
d Pn21 , j  1  d Pn2 , j  1  d Pn2 , j  d Pn21 , j  d Pn2 2 , j  d Pn23 , j
2
n4
2
n 1
2
n
2
n 1
2
n
2
n 1
2
n 2
2
n 3
 
 
, j  d Pn2 4 , j  d Pn25 , j

2
n 5
Therefore we have the result
3) By theorem (3.1), we have
Sn  
Sn  

n
n
i 
5
n
n
i 
5

d Pn2 , j


 


 
 
 

 d Pn21 , j  1  d Pn2 2 , j  1  d Pn23 , j  1  d Pn2 4 , j  1  d Pn25 , j  1 


n 1
n
i    1
5
d Pn21 , j  1  
n 1
n
i    1
5


d Pn2 2 , j  1  
n 1
n
i    1
5


d Pn2 3 , j  1  
n 1
n
i    1
5


d Pn2 4 , j  1  
n 1
n
i    1
5


d Pn25 , j  1
Sn  Sn 1  Sn  2  Sn 3  Sn  4  Sn  5 .
.
4. Conclusions
In [2], the domination polynomial of path was studied
and obtained the very important property,



 
 

d Pn2 , i  d Pn21 , i  1  d Pn2 2 , i  1  d Pn2 3 , i  1 .
It is interesting that we have derived an analogues relation for the square of path of the form



 
, i  1  d  P
 
, i  1

d Pn2 , i  d Pn21 , i  1  d Pn2 2 , i  1  d Pn2 3 , i  1

 d Pn2 4
2
n 5
One can characterise the roots of the polynomial
D Pn2 , x and identify whether they are real or complex.

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Copyright © 2013 SciRes.
Another interesting character r to be investigated is
whether D Pn2 , x is log concave are not.
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REFERENCES
[1]
S. Alikhani and Y.-H. Peng, “Introduction to Domination
Polynomial of a Graph,” .arXiv:0905.2251v1[math.co],
2009.
[2]
S. Alikhani and Y.-H. Peng, “Domination Sets and
Domination Polynomials of Paths,” International Journal
of Mathematics and Mathematical Sciences, Vol. 2009,
2009, Article ID: 542040.
[3]
G. Chartand and P. Zhang, “Introduction to Graph Theory,” McGraw-Hill, Boston, 2005.
OJDM