#### Abstract

Let be a simple graph. A set is a dominating set of , if every vertex in is adjacent to at least one vertex in . Let be the family of all dominating sets of a path with cardinality , and let . In this paper, we construct , and obtain a recursive formula for . Using this recursive formula, we consider the polynomial , which we call domination polynomial of paths and obtain some properties of this polynomial.

#### 1. Introduction

Let be a simple graph of order . For any vertex , the open neighborhood of is the set and the closed neighborhood of is the set . For a set , the open neighborhood of is and the closed neighborhood of is . A set is a dominating set of , if , or equivalently, every vertex in is adjacent to at least one vertex in . The domination number is the minimum cardinality of a dominating set in . A dominating set with cardinality is called a -set, and the family of -sets is denoted by . For a detailed treatment of this parameter, the reader is referred to [1]. It is well known and generally accepted that the problemof determining the dominating sets of an arbitrary graph is a difficult one (see [2]). A path is a connected graph in which two vertices have degree 1 and the remaining vertices have degree 2. Let be the path with vertices. Let be the family of dominating sets of a path with cardinality and let . We call the polynomial , the domination polynomial of the path . For a detailed treatment of the domination polynomial of a graph, the reader is referred to [3].

In the next section we construct the families of the dominating sets of paths by a recursive method. In Section 3, we use the results obtained in Section 2 to study the domination polynomial of paths.

As usual we use , for the smallest integer greater than or equal to . In this article we denote the set simply by .

#### 2. Dominating Sets of Paths

Let be the family of dominating sets of with cardinality . We will investigate dominating sets of path. We need the following lemmas to prove our main results in this article.

Lemma 2.1 (see [4, page 371]). *.*

By Lemma 2.1 and the definition of domination number, one has the following lemma.

Lemma 2.2. *, if and only if or .*

A *simple
path* is a path in which all its internal vertices have degree two. The
following lemma follows from observation.

Lemma 2.3. *If a graph contains a
simple path of length , then every dominating set of must contain
at least vertices of the path.*

To find a dominating set of with cardinality , we do not need to consider dominating sets of with cardinality . We show this in Lemma 2.4. Therefore, we only need to consider , and . The families of these dominating sets can be empty or otherwise. Thus, we have eight combinations of whether these three families are empty or not. Two of these combinations are not possible (see Lemma 2.5(i) and (ii)). Also, the combination that does not need to be considered because it implies (see Lemma 2.5(iii)). Thus we only need to consider five combinations or cases. We consider these cases in Theorem 2.7.

Lemma 2.4. *If , and there exists such that , then .*

*Proof. *Suppose that . Since , contains at
least one vertex labeled or . If , then , a contradiction. Hence, , but then in this case, , for any , also a contradiction.

Lemma 2.5. *
(i) If , then .**
(ii) If and , then .**
(iii) If , then .*

*Proof. *(i) Since , by Lemma 2.2, or . In either case we have .

(ii) Suppose that , so by Lemma 2.2, we have or . If , then , and hence, , a contradiction. Hence . So , and hence, , also a contradiction.

(iii)
Suppose that . Let . Then at least one vertex
labeled or is in . If , then by Lemma 2.3, at least one vertex labeled or is in . If or , then , a contradiction. If , then , a contradiction. Now suppose
that . Then by Lemma 2.3, at least one vertex labeled or is in . If or , then , a contradiction. If , then , a contradiction. Therefore .

Lemma 2.6. *If , then*

(i), and if and only if and for some (ii) and if and only if ;(iii) and if and only if and for some ;(iv) and if and only if ;(v) and if and only if .

*Proof. *(i) () Since , by Lemma 2.2, or . If , then , and by Lemma 2.2, , a contradiction. So , and since , together , which give us and for some .

() If and for some , then by Lemma 2.2, , and .

(ii)() Since , by Lemma 2.2, or . If , then , and hence , a contradiction. So . Also since , . Therefore . () If , then by Lemma 2.2, and .(iii)() Since , by Lemma 2.2, or . If , then and by Lemma 2.2, , a contradiction. So , but because . Hence, . Therefore and for some . () If and for some , then by Lemma 2.2, , , and . (iv)() Since , by Lemma 2.2, or . Since , by Lemma 2.2, . Therefore is not possible. Hence . Thus or , but because . So . () If , then by Lemma 2.2, , , and .(v) () Since , and , then by applying Lemma 2.2, , and So and
hence . () If ,thenthe result follows
from Lemma 2.2.

By Lemma 2.4, for the construction of , it's sufficient to consider , and . By Lemma 2.5, we need only to consider the following five cases.

Theorem 2.7. *For every and .*

(i)If and , then .(ii)If , and , then .(iii) If and , then (iv)If and , then (v)If and , then

*Proof. *(i) , and . By Lemma 2.6(i), and for some . Therefore .

(ii), and . By Lemma 2.6(ii), we have . So .(iii) and . By Lemma 2.6(iii), and for some . Since , . Also, if , then . Therefore we
have Now let . Then or is in . If , then by Lemma 3, at least one vertex labeled or is in . If or is in , then , a contradiction because . Hence, , and . Therefore for some , that is . Now suppose that and . By Lemma 2.3, at least one vertex labeled , or is in Y. If , then , a contradiction because for all . Therefore or is in , but . Thus for some . So (iv), , and . By Lemma 2.6(iv), . Therefore .(v), and . Let , so at least one vertex
labeled or is in . If or , then .

Let , then or is in . If or , then .

Now let , then or is in . If , then , for . If , then . Therefore we have Now, let , then or . If , then by Lemma 2.3, at least one vertex labeled , or is in . If or , then for some . If , and , then for some . Now suppose that and , then by Lemma 3, at least one vertex labeled or is in . If or , then for some . If and , then for some . So

*Example 2.8. *Consider with . We use Theorem 2.7
to construct for . Since and , by Theorem 2.7, .

Since , and , we get .

Since , , and , then by Theorem
2.7,and, for the construction of , by Theorem 2.7,Finally, since , , and , then

#### 3. Domination Polynomial of a Path

Let be the domination polynomial of a path . In this section we study this polynomial.

Theorem 3.1. *
(i) If is the family of dominating
set with cardinality of , then**
(ii) For every ,**with the initial values , and .*

*Proof. *
(i) It follows from Theorem 2.7.

(ii) It follows from Part (i) and the
definition of the domination polynomial.

Using Theorem 3.1, we obtain for as shown in Table 1. There are interesting relationships between numbers in this table. In the following theorem, we obtain some properties of .

Theorem 3.2. *The following
properties hold for the coefficients of :*

(i)*, for every .*(ii)*, for every .*(iii)*, for every .*(iv)*, for every .*(v)*, for every .*(vi)*, for every .*(vii)*, for every .*(viii)*, for every .*(ix)*, for every .*(x)*.*(xi)*.*(xii)* If , then for every , with initial
values and .*(xiii)*For every , and , *(xiv)*For every , .*

*Proof. *(i) Since , we have .

(ii) Proof by induction on . Since , so . Therefore the result is true for . Now suppose that the result is true for all natural
numbers less than , and we prove it for . By part (i), Theorem 3.1, and the induction hypothesis we have .

(iii) Proof by induction on . The result is true for , because . Now suppose that the result is true for all natural
numbers less than , and we prove it for . By part (i), (ii), Theorem 3.1, and the induction hypothesis we have

(iv) Proof by induction on . Since , the result is true for . Now suppose that the result is true for all natural
numbers less than , and we prove it for . By Theorem 3.1, parts (ii), (iii), and the
induction hypothesis we have

(v) Since , we have the result.

(vi) Since , we have .

(vii) By induction on . The result is true for , because . Now suppose that the result is true for all numbers
less that , and we prove it for . By Theorem 3.1, induction hypothesis, part (v) and part (vi) we
have

(viii) By induction on . The result is true for , since . Now suppose that the result is true for all natural
numbers less than or equal and we prove it
for . By Theorem 3.1, induction hypothesis, parts (vii) and (vi) we
have

(ix) By induction on . Since , the result is true for . Now suppose that the result is true for all natural
numbers less than , and we prove it for . By Theorem 3.1, induction hypothesis, parts (viii) and (vii), we have

(x) We will prove that for
every , for , and for . We prove the first inequality by induction on . The result holds for . Suppose that result is true for all . Now we prove it for , that is for . By Theorem 3.1 and the induction hypothesis we
haveSimilarly, we have the other
inequality.

(xi) Proof by induction on . First, suppose that . Then . Now suppose that the result is true for every , and we prove for :

(xii) By Theorem 3.1, we
have

(xiii) Proof by induction on . Since , the theorem is true for . Now, suppose that the theorem is true for all
numbers less than , and we will prove it for . By Theorem 3.1 and the induction hypothesis, we can
write

(xiv) By
Theorem 3.1, we haveTherefore we have the result.

In the following theorem we use the generating function technique to find .

Theorem 3.3. *For every and , is the coefficient of in the
expansion of the function*

*Proof. * Set . By recursive formula for in Theorem 3.1 we can write in the
following form: By substituting the values from Table 1 (note that for all natural numbers and ), we haveTherefore we have the result.

#### Acknowledgments

Authors wish to thank the referee for his valuable comments and the research management center (RMC) of University Putra Malaysia for their partial financial support.