On the 2 Domination Number of Complete Grid Graphs

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http://www.scirp.org/journal/ojdm ISSN Online: 2161-7643 ISSN Print: 2161-7635

On the 2-Domination Number of Complete Grid

Graphs

Ramy Shaheen, Suhail Mahfud, Khames Almanea

Department of Mathematics, Faculty of Science, Tishreen University, Lattakia, Syria

Abstract

A set D of vertices of a graph G = (V, E) is called k-dominating if every vertex v V∈ −D is adjacent to some k vertices of D. The k-domination number of a graph G, γk

( )

G , is the order of a smallest k-dominating set of G. In this paper we calculate the k-domination number (for k = 2) of the product of two paths Pm × Pn

for m = 1, 2, 3, 4, 5 and arbitrary n. These results were shown an error in the paper

[1].

Keywords

k-Dominating Set, k-Domination Number, 2-Dominating Set, 2-Domination Number, Cartesian Product Graphs, Paths

1. Introduction

Let G=

(

V E,

)

be a graph. A subset of vertices D⊆V is called a 2-dominating set of

G if for every v V∈ , either v∈D or v is adjacent to at least two vertices of D. The 2-domination number γ2

( )

G is equal to min

{

D D: is a 2−dominating set of G

}

.

The Cartesian product G ×_{H of two graphs G and H is the graph with vertex set}

(

)

( )

V G×H =V G ×V H , where two vertices

(

v v₁, ₂

)

,

(

u u₁, ₂

)

∈ ×G H are adjacent

if and only if either v u1 1∈E G

( )

and v2=u2 or v u2 2∈E H

( )

and v1=u1.

Let G be a path of order n with vertex set V G

( ) {

= 1, 2,,n

}

. Then for two paths of

order m and n respectively, we have P_m×P_n =

{

( )

i j, :1≤ ≤i m,1≤ ≤j n

}

. The jth col-umn of Pm×Pn is Kj=

{

( )

i j, :i=1,,m

}

. If D is a 2-dominating set for Pm×Pn,

then we put Wj= ∩D Kj. Let sj=Wj . The sequence

(

s s1, 2,,sn

)

is called a 2-dominating sequence corresponding to D. For a graph G, we refer to minimum and maximum degrees by δ

( )

G and ∆

( )

G , and for simplicity denoted those by δ and Δ,

respectively. Also, we denote by V and E to order and size of graph G,

respec-tively.

How to cite this paper: Shaheen, R., Mah- fud, S. and Almanea, K. (2017) On the 2-Domination Number of Complete Grid Graphs. Open Journal of Discrete Mathe-matics, 7, 32-50.

http://dx.doi.org/10.4236/ojdm.2017.71004

Received: November 22, 2016 Accepted: January 20, 2017 Published: January 23, 2017

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2. Notation and Terminology

Fink and Jacobson [2][3] in 1985 to introduced the concept of multiple domination. A subset D⊆V is k-dominating in G if every vertex of V–D has at least k neighbors in D. The cardinality of a minimum k-dominating set is called the k-domination num-ber γk

( )

G of G. Clearly, g G1

( )

=g G

( )

. Naturally, every k-dominating set of a graph

G contains all vertices of degree less than k. Of course, every

(

k+1

)

-dominating set is

also a k-dominating set and so γk

( )

G ≤γk+1

( )

G . Moreover, the vertex set V is the only

(

∆ +1

)

-dominating set but evidently it is not a minimum ∆-dominating set. Thus

every graph G satisfies

( )

1

( )

1

( )

k G k G G G V

γ ≤γ ₊ ≤≤γ_∆ <γ_∆+ = .

For a comprehensive treatment of domination in graphs, see the monographs by Haynes et al. [4]. Also, for more information see [5][6]. Fink and Jacobson [2], intro-duced the following theorems:

Theorem 2.1 [2]. If k≥2, is an integer and G is a graph with k≤ ∆

( )

G , then

( )

2

k G G k

γ ≥γ + − _.

Theorem 2.2 [2]. If T is a tree, then 2

( )

1 2 T T

γ ≥ + _.

In [6], Hansberg and Volkmann, proved the following theorem.

Theorem 2.4 [6]. Let G=

(

V E,

)

be a graph of order n and minimum degree δ and

let k∈N. If

(

1

)

2

ln 1 k

δ δ

+ ≥

+ , then

( )

(

)

₍

₎

1

1 0

ln 1

1 ! 1

i k

k k

i V

G k

i δ

γ δ

δ _δ

−

− =

 

 

≤ _ + + _

+ _

∑

+ _.

Cockayne, et al. [7], established an upper bound for the k-domination number of a graph G has minimum degree k, they gave the following result.

Theorem 2.3 [7]. Let G be a graph with minimum degree at least k, then

( ) ( )

1 k

k V G

k

γ ≤

+ .

Blidia, et al. [8], studied the k-domination number. They introduced the following results.

Theorem 2.5 [8]. Let G be a bipartite graph and S is the set of all vertices of degree at most k– 1, then

( )

2 k

V S

G

γ ≤ + _.

Favaron, et al. [9], gave new upper bounds of γk

( )

G .

Corollary 2.6 [9]. Let G be a graph of order n and minimum degree δ. If k≤δ is

an integer, then

( )

.

2 1

k G V

k δ γ

δ

≤

+ −

In [4], Haynes et al. showed that the 2-domination number is bounded from below by the total domination number for every nontrivial tree.

Theorem 2.7 [4]. For every nontrivial tree, γ2

( )

T ≥γt

( )

T . Also, Volkmann [10] gave the important following result.

Theorem 2.8 [10]. Let G be a graph with minimum degree δ ≥ +k 1, then

( )

1 .

2 k k

V G

G γ

γ ₊ ≤ +

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an upper and lower bounds. Also, in [12], he introduced the following results. Theorem 2.9 [12].

1) γ2

( )

Cn = n2.

2) γ2

(

C3×Cn

)

=n n: ≡0 mod 3

(

)

,

(

)

(

)

2 C3 Cn n 1:n 1, 2 mod 3

γ × = + ≡ _.

3) γ2

(

C4×Cn

)

= +n n 2 : n≡0, 3, 5 mod 8

(

)

,

(

)

(

)

2 C4 Cn n n 2 1:n 1, 2, 4, 6, 7 mod14

γ × = +_ _+ ≡ .

4) γ2

(

C5×Cn

)

=2n.

5) γ2

(

C6×Cn

)

=2 :n n≡0 mod 3

(

)

,

(

)

(

)

2 C6 Cn 2n 2 :n 1, 2 mod 3

γ × = + ≡ _.

6) γ2

(

C7×Cn

)

=5n2 : n≡0, 3,11 mod14

(

)

,

(

)

(

)

2 C7 Cn 5n2 1:n 5, 6, 7, 8, 9,10 mod14

γ × = + ≡ ,

(

)

(

)

2 C7 Cn 5n 2 2 :n 1, 2, 4,12,13 mod14

γ × =_ _+ ≡ .

In this paper we calculate the k-domination number (for k = 2) of the product of two paths Pm×Pn for m = 1, 2, 3, 4, 5 and arbitrary n. These results were shown an error

in the paper [1]. We believe that these results were wrong. In our paper we will provide improved and corrected her, especially for m = 3, 4, 5.

The following formulas appeared in [1],

( ) (

)

(

)

(

)

(

)

2 Pn n 1 2 2 P2 Pn n 2 P3 Pn 2n n 2 2 P4 Pn 2 .n

γ = + ⋅γ × = ⋅γ × = − ⋅γ × =

(

)

(

) (

)

2 P5 Pn 3n n2 2 P2k1 Pn k 1 n n2 .

γ × = − ⋅γ + × = + − 

(

)

(

)

(

)

(

)

2

2 2 : 1 mod 2 ,

2 : 0 mod 2 .

m n

P P m n n m

P P m n m

γ γ

× =_ _ −_ _ ≡

× =_ _ ≡

In this paper, we correct the results in [1] and proves the following:

( ) (

)

(

)

(

)

2 Pn n 1 2 2 P2 Pn n 2 P3 Pn n n 3 .

γ =_ + _⋅γ × = ⋅γ × = +_ _

(

)

(

)

(

)

(

)

2 4

2 4 : 3, 7 mod 8 ,

2 4 1: 0,1, 2, 4, 5, 6 mod 8 .

n

P P n n n

γ γ

× = −_ _ ≡

× = −_ _+ ≡

(

)

(

)

(

)

(

)

2 5

2 7 : 1, 2, 3, 5 mod 7 ,

2 7 1: 0, 4, 6 mod 7 .

n

P P n n n

γ γ

× = +_ _ ≡

× = +_ _+ ≡

3. Main Results

Our main results here are to establish the domination number of Cartesian product of two paths Pm and Pn for m = 1, 2, 3, 4, 5 and arbitrary n. We study 2-dominating

sets in complete grid graphs using one technique: by given a minimum of upper 2-dominating set D of Pm×Pn and then we establish that D is a minimum 2-domi-

nating set of Pm×Pn for several values of m and arbitrary n. Definitely we have

(

)

2 Pm Pn D

γ × = _.

Let G be a path of order n with vertex set V G

( ) {

= 1, 2,,n

}

. For two paths of order

m and n respectively is:

( )

{

, :1 ,1

}

m n

P ×P = i j ≤ ≤i m ≤ ≤j n . The jth column P_m×P_n is

( )

{

, : 1, ,

}

j

K = i j i=  m .

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sequence

(

s s1, 2,,sn

)

is called a 2-dominating sequence corresponding to D. Always we have s s1, n≥ m 3. Suppose that sj =0 for some j (where j ≠ 1 or n). The

vertic-es of the jth column can only be 2-dominated by verticvertic-es of the (j − 1)st columns and (j + 1)st columns. Thus we have sj−1+sj+1=2m, then sj−1=sj+1=m. In general

1 4 1 2

j j j

s− + s +s+ ≥ m.

Notice 3.1.

1) The study of 2-dominating sequence

(

s s1, 2,,sn

)

is the same as the study of the 2-dominating sequence

(

s sn, n−1,,s1

)

.

2) If subsequence

(

s sj, j+1,,sj k+

)

is not possible, then its reverse

(

sj k+ ,,sj+1,sj

)

is not possible.

3) We say that two subsequences

(

sj,,sj q+

) (

, sj q+ +1,,sj r+

)

are equivalent, if the sequence

(

sj,,sj q+ ,sj q+ +1,,sj r+

)

is possible.

We need the useful following lemma.

Lemma 3.1. There is a minimum 2-dominating set for Pm×Pn with 2-dominating

sequence

(

s s1, 2−,,sn

)

such that, for all j=1, 2,,n, is m 4≤sj≤3m 4. Proof. Let D be a minimum 2-dominating set for Pm×Pn with 2-dominating

se-quence

(

s s1, 2−,,sn

)

. Assume that for some j, sj is large. Then we modify D by mov-ing two vertices from column j, one to column j − 1 and another one to column j + 1, such that the resulting set is still 2-dominating set for Pm×Pn. For 1≤ ≤i m and 1≤ ≤j n, let W = ∩D

{

( ) (

i j, , i+1,j

) (

, i+2,j

) (

, i+3,j

)

}

. If W =4, then we define

(

) ( ) (

{

) (

)

}

1 – , , 1, 1 , 2, 1 , 3,

D = D W ∪ i j i+ j− i+ j+ i+ j , see Figure 1. We repeat this

process if necessary eventually leads to a 2-dominating set with required properties. Also, we get D1 is a 2-dominating set for Pm×Pn with D = D1. Thus, we can assume

that every four consecutive vertices of the jth column include at most three vertices of D. This implies that s_j≤ _3m 4_, for all 1 ≤ j ≤ n.

To prove the lower bound, we suppose that Kj∩D is be a maximum, i.e.,

3 4

j

s = _ m _. Then for each

( )

i j, ∉D, we have

(

) (

)

{

i−1,j+1 , ,i j+1 , i+1,j+1

}

∩D ≥1. When s_j= _3m 4_, there at must

3 4 4

m−_ m  _{ }= m _ vertices does not in K_j∩D. This implies that s_j₊₁≥ _m 4_. So,

the same as for sj−1≥ m 4. □

By Lemma 3.1, always we have a minimum 2-dominating set D with 2-dominating sequence

(

s s1, 2,,sn

)

, such that m 4≤sj≤3m 4, for all j=1, 2,,n.

Lemma 3.2. 2

( )

1 2

n

n P

γ _{= } + _

 .

(5)

Proof. Let

(

2 1 ;1

)

2 n D= k− ≤ ≤k  _{ }

 

 .

We have D is a 2-dominating set of Pn for n≡1 mod 2

(

)

with 1 2 n D =  + _

 , also

( )

{ }

D∪ n is a 2-dominating set of Pn for n≡0 mod 2

(

)

with

{ }

( )

1 2 n D∪ n =  + _

 . Let D1 be a minimum 2-dominating set for Pn with V P

( ) {

n = x x1, 2,,xn

}

. Since

( )

1 n n

x x ∉E P , we need to x x₁, n∈D₁, also if xj∉D1 then xj−1,xj+1 are belong to D1,

this implies that x2j−1∈D1 for 2

2 n j  

≤ ≤  _{ }. Thus implies that

1

2 1

2 2

n n

D ≥ + _{ }− =_ + _

   . We result that 2

( )

1 2

n

n P

γ _{= } + _

 . □

Theorem 3.1. γ2

(

P2×Pn

)

=n.

Proof. Let a set

(

1, 2 1 :1

)

(

2, 2

)

:1

2 2

n n

D=_ k− ≤ ≤k _{ }  _{ }∪ k ≤ ≤k  _{ }_

   

   .

It is clear that D =n. (1)

We can check that D is 2-dominating set for P2×Pn, see Figure 2. Let D1 be a minimum

2-dominating set for P2×Pn with dominating sequence

(

s1,,sn

)

. If si ≥1 for all

1, ,

j=  n, then 1 1

n

j j

D s n

=

∑

≥ _{. (2)}

Let s_j =0 for some j, then s_j₋₁=s_j₊₁=2, also we have s₁≥1 and sn≥1. Now we define a new sequence

(

s1′,,sn′

)

, (not necessarily a 2-dominating sequence) as

follows:

For s_j=2, if j=1 or n, we put s′ =_j s_j – 1, s₂′ =s₂+1 2 and sn′ =−1 sn−1+1 2.

If j≠1 or n, we put s′ =_j s_j – 1, s′ =_j₋₁ s_j₋₁+1 2 and s′ =_j₊₁ s_j₊₁+1 2. Otherwise s′ =j sj.

We get a sequence

(

s1′,,sn′

)

have property that each s′ ≥j 1 with

1 1

n n

j j

D s s n

= =

′

=

∑

=

∑

≥ . (3)

By (1), (2) and (3) is γ2

(

P2×Pn

)

=n. This completes the proof of the theorem. □

Theorem 3.2. 2

(

3

)

3

n

P P n

γ × _{= +  } 

 .

Proof. Let

(

)

(

)

(

) (

)

2, 3 2 :1 2, 3 :1

3 3

1

1, 3 1 , 3, 3 1 :1

3

n n

D k k k k

n

k k k

     

= − ≤ ≤_{ } ∪ ≤ ≤_{ }

   

   

  − 

∪_ − − ≤ ≤_ __

 

 

[image:5.595.189.552.94.343.2] [image:5.595.199.547.410.719.2]

.

(6)

(

) (

)

(

)

(

)

1

1, 3 2 , 3, 3 2 :1 2, 3 1 :1

3 3

2, 3 :1

3

n n

D k k k k k

n

k k

     − 

′ = − − ≤ ≤_{ } ∪ − ≤ ≤_ _

   

   

  

∪_ ≤ ≤_{ }_

 

 

.

We have

3 n D = +  n  

  and 3

n D′ = +  n  

 . (4) By definition D and D′ we note that

D is 2-dominating set for P3×Pn when n=0, 2 mod 3

(

)

, (see Figure 3, for P3 ×

P14).

D′ is 2-dominating set for P3×Pn when n=1 mod 3

(

)

, (see Figure 4, for P3 × P10).

Let D1 be a minimum 2-dominating set for P3×Pn with 2-dominating sequence

(

s1,,sn

)

we have s s1, n≥1 and

if s s1, n=1 then s s2, n−1≥2,

if s s1, n=2 then s s2, n−1≥1.

Also for 1< <j n, if s_j =0 then s_j₋₁=s_j₊₁=3, 1

j

s = then s_j₋₁+s_j₊₁≥3, 2

j

s = then s_j₋₁+s_j₊₁≥2,

If no one of s_j =0 for all j, then ₁

1 3

n

j j

n

D s n

=

 

= _{≥ +  }

 

∑

. (5)

Let s_j =0 ( j≠1 or n) for some j, we define a sequence

(

s1′,,sn′

)

, (not necessar-ily a 2-dominating sequence ) as follows:

If s_j =3, then we put s′ =_j s_j – 1, s′ =_j₋₁ s_j₋₁+1 2 and s′ =_j₊₁ s_j₊₁+1 2, otherwise

j j

s′ =s . We have \

1

1 1

n n

j j

D s s

= =

=

∑

=

∑

_{. We note that the sequence}

(

s₁′,,s_n′

)

have the

property if s′ =_j 1 then s′_j₋₁+s′_j₊₁≥3. Thus implies that

1

1 3

n

j j

n

D s n

=

  ′

= _{≥ +  }

 

∑

. (6)

From (4), (5) and (6) we get the required result. □

Theorem 3.3.

(

)

(

)

(

)

2 4

2 : 3, 7 mod 8 ,

4

2 1: 0,1, 2, 4, 5, 6 mod 8 .

4 n

n

n n

P P

n

n n

γ

 ₋  _≡

  

  

× _{= }

 

 ₋ ₊ _≡

 

 _{ }



[image:6.595.189.555.64.793.2]

Figure 3. A 2-dominating set for P3 × P14.

(7)

Proof. Let a set D defined as follows:

( ) ( )

{

}

(

) (

)

(

)

(

) (

)

(

)

(

)

1

2,1 , 3,1 1, 4 2 , 4, 4 2 ;1

4

2

2,8 5 ;1

8

3

1,8 4 , 3,8 4 , 4,8 4 ;1

8

4 6

2,8 3 ;1 3,8 1 ;1

8 8

n

D k k k

n

k k

n

k k k k

n n

k k k k

   − 

= ∪ − − ≤ ≤_ _

 

 



  − 

∪ − ≤ ≤_ _

 

 

  − 

∪ − − − ≤ ≤_ _

 

 

  −    − 

∪ − ≤ ≤_ _ ∪ − ≤ ≤_ _

   

  

(

) (

)

7

(

)

1

1,8 , 2,8 , 4,8 ;1 3,8 1 ;1

8 8

n n

k k k k k k

  



  −    − 

∪ ≤ ≤_ _ ∪ + ≤ ≤_ _

   

   

( )

{

2,

}

,

{

( )

3,

}

D′= n D′′= n .

We can check that the following sets are 2-dominating set for P4×Pn (see Figure 5,

for P4×P11) as indicated:

D is 2-dominating set for P4×Pn when n≡0, 4 mod 8

(

)

.

D∪D′ is 2-dominating set for P4×Pn when n≡1, 2, 7 mod 8

(

)

.

D∪D′′ is 2-dominating set for P4×Pn when n≡3, 5, 6 mod 8

(

)

.

We have

(

)

(

)

(

)

2 1: 3, 7 mod 8 ,

4

2 : 1, 2, 5, 6 mod 8 , 4

2 1: 0, 4 mod 8 .

4 n

n n

n

D n n

n

n n

 ₋ ₋ _≡ 

  _{ } 

 

   

=_ −_{ } ≡ _

 

 

 _{ } 

− + ≡

 _{ } 

 

 

Let D1 be a minimum 2-dominating set for P4×Pn with 2-dominating sequence

(

s1,,sn

)

we shall show that

(

)

(

)

1

2 : 3, 7 mod 8 ,

4

2 1: 0,1, 2, 4, 5, 6 mod 8 .

4 n

n n

D

n

n n

 ₋  _≡ 

   

   

=  

 

 ₋ ₊ _≡ 

 

 _{ } 

 

By Lemma 3.1, we have 1≤s_j≤3. Thus If s_j =1 then s_j₋₁+s_j₊₁≥5.

If s_j=2 then s_j₋₁+s_j₊₁≥2. If s_j =3 then s_j₋₁+s_j₊₁≥2.

[image:7.595.210.536.87.258.2]

Also, we have s s1, n≥2. If s s1, n=2 then s s2, n−1≥2, and if s s1, n=3 then

(8)

2, n1 1

s s− ≥ .

We define a new set D1′ with sequence

(

s1′,,sn′

)

, (not necessarily a 2-dominating

sequence) as follows: if s_j ≥2, let 7

4

j j

M =s − . Now, for j=2 to j= −n 1, if

2 j

s ≥ , then we put

j j j

s′ = −s M , ₁ ₁

2 j

j j

M

s′ =− s− + and 1 1

2 j

j j

M s′ =+ s+ +

Thus, for 3≤ ≤ −j n 2, we have 7

4 j

s ≥ . Since if s_j ≥2 then 7 4 j

s′ ≥ and if

1 j

s = , then s_j₋₁+s_j₊₁=5 this implies that 1 1

14 6

5

4 4

j j

M − +M + = − = , which implies

that 1 1 3 7

1 .

2 2 4 4

j j

M M

s′ = +s − + + = + =

We have three cases:

Case 1: s s1, n≥2, then s s2, n−1≥2, these implies that 1 1

1 8

s′ ≥ +s and 1

8

n n

s′ ≥ +s also

(

)

1

1 1

1 1 2

7 2

1 1 7 3

2 2

8 8 4 4 4

n n n

j j n j

j j j

n n

D s s s s s

−

= = =

−

′ ′ ′ ′

=

∑

=

∑

= + +

∑

≥ + + + + = + .

Case 2: s s1, n =3 then s s2, n−1≥2. Thus implies that s s1′ ′ =, n 3 and

2 1

1

, 1 .

8 n

s s′ ′ ≥ +− Then

2

1 1 2 1

1 1 2

1 1 7 7 5

3 1 3 1

8 8 4 4 4

n n n

j j n n j

j j j

n

D s s s s s s s

−

= = =

′ ′ ′ ′ ′ ′

=

∑

=

∑

= + + + +

∑

≥ + + + + + + = +

Case 3: s1=2, 3sn = and s2≥2, sn−1≥1 or s1=3, sn=2 and s2≥1, 2sn−1≥ .

Two cases are similar by symmetry. We consider the first case:

1 2, 22

s = s ≥ and sn =3, 1sn−1≥ , this implies that

1 2 1

1 7 1

2 , , 3, 1

8 4 n n 8

s′= + s′= s′ = s′− = + and

(

)

2

1 1 2 1

1 1 3

1 7 1 7 7

2 3 1 4 1

8 4 8 4 4

n n n

j j n n j

j j j

n

D s s s s s s s n

−

= = =

′ ′ ′ ′ ′ ′

=

∑

=

∑

= + + + +

∑

≥ + + + + + + − = +

But, we have the 2-domination number is positive integer number, also we have

7 3

2

4 4 4

n n

n− _{ }= +

  for n≡3, 7 mod 8

(

)

,

(

)

(

)

(

)

7

1 For 0, 4 mod 8 ,

4

7 5

2 1 For 1, 5 mod 8 ,

4 4 4

7 6

For 2, 6 mod 8 ,

4 4

n

n n

n

 ₊ _≡

   

−_{ }+ = + ≡

  _

 ₊ _≡

 Thus implies that

(

)

(

)

1

2 ; 3, 7 mod 8 ,

4

2 1; 0,1, 2, 4, 5, 6 mod 8 ,

4 n

n n

D

n

n n

 ₋  _≡ 

   

   

≥  

 

 ₋ ₊ _≡ 

 

 _{ } 

(9)

Finally, we get

(

)

(

)

(

)

(

)

2 4

2 : 3, 7 mod 8 ,

4

2 1: 0,1, 2, 4, 5, 6 mod 8 , 4

n

P P n n

n

P P n n

γ

 

× = −_{ } ≡

   

× = −_{ }+ ≡

  This complete the proof of the theorem. □

Theorem 3.4.

(

)

(

)

(

)

2 5

2 : 1, 2, 3, 5 mod 7 ,

7

2 1: 0, 4, 6 mod 7 .

7 n

n

n n

P P

n

n n

γ

 ₊  _≡

  

× _{= }

 

 ₊ ₊ _≡

 

 _{ }

 Proof. Let a set D defined as follows:

( ) ( )

{

}

{

( ) ( ) ( )

(

)

}

{

( )

(

)

{

}

{

( ) ( ) ( )

(

)

}

( ) ( )

(

)

{

}

{

( ) ( )

(

)

}

( ) ( )

(

)

{

}

{

( ) ( )

(

)

}

2,1 , 4,1 1, , 2, , 5, : 2 mod 7

3, : 3 mod 7 1, , 4, , 5, : 4 mod 7

2, , 3, : 5 mod 7 2, , 5, : 6 mod 7

1, , 4, : 0 mod 7 3, , 4, : 1 mod 7 and 1

D j j j j

j j j j j j

j j j j j j j

= ∪ ≡

∪ ≡ ∪ ≡

∪ ≡ ∪ ≡ ≠

We can check that the following sets are 2-dominating set for P5×Pn (see Figure 6,

for P5×P23) as indicated:

{

}

{

}

{

( ) ( ) ( )

}

(

)

( )

{

}

(

)

(

)

{

}

{

}

{

( ) ( )

}

(

)

{

}

{

}

{

( ) ( ) ( )

}

(

)

2, , 3, , 5, : 1 mod 7 .

2, : 0, 4 mod 7 .

: 2 mod 7 .

2, , 4, : 3, 5 mod 7 .

1, , 3, , 5, : 6 mod 7 .

n

D K D n n n n

D n n

D n

D K D n n n

D K D n n n n

− ∩ ∪ ≡

∪ ≡

≡

− ∩ ∪ ≡

We have 2

7 n D≤ n+   

  and

(

)

(

)

(

)

2 5

2 : 1, 2, 3, 5 mod 7 ,

7

2 1: 0, 4, 6 mod 7 .

7 n

n

n n

P P

n

n n

γ

 ₊  _≡

  

× _{≤ }

 

 ₊ ₊ _≡

 

 _{ }



This complete the proof of the theorem. □

Lemma 3.3. The following cases are not possible: 1) (1, 2, 3, 1).

[image:9.595.195.525.80.352.2]

2) (1, 2, 1). 3) (1, 4, 1, 1).

(10)

4) (1, 3, 1, 3, 1, 3). 5) (2, 1, 3). 6) (2, 2, 2, 2, 2, 2).

Proof. It follows directly from the drawing. Lemma 3.4.

1) There is one case for subsequence

(

s sj, j+1,sj+2,sj+3,sj+4

)

=

(

2, 2, 2, 2, 2

)

.

(

s sj, j+1,sj+2,sj+3

)

=

(

1, 3,1, 3

)

.

(

s sj, j+1,sj+2,sj+3,sj+4

)

=

(

1, 3,1, 3,1

)

.

(

s sj, j+1,sj+2,sj+3,sj+4

)

=

(

1, 2, 3, 2,1

)

.

Proof. It follows directly from the drawing (see Figure 7). Lemma 3.5.

1) 3 8

j

j j

s

+ ≥

∑

.

2) 5 12

j

j j

s

+ ≥

∑

.

3) 6 14

j

j j

s

+ ≥

∑

.

4) If s_j =3 then

6

15 j

j j

s

+ ≥

∑

.

5) If s_j=4 then

6

16 j

j j

s

+ ≥

∑

.

Proof. 1) By Lemma 3.3, imply that 3 8

j

j j

s

+ ≥

∑

.

2) By 1, we have 3 8

j

j j

s

+ ≥

∑

. If 3 8

j

j j

s

+ =

∑

, then we have the cases

(

s sj, j+1,sj+2,sj+3

)

=

(

1, 2, 3, 2 , 1, 3,1, 3 , 1, 3, 2, 2 , 1, 4,1, 2 , 2, 2, 2, 2

) (

)

.

From Lemma 3.3, we have sj+4+sj+5≥4, this implies that 5

12 j

j j

s

+ ≥

∑

.

If 4 9

j

j j

s

+ ≥

∑

then sj+4+sj+5≥3. This implies that 6

12 j

j j

s

+ ≥

∑

.

3) We have 2 5

j

j j

s

+ ≥

∑

and 6

4

5 j

j j

s

+

+ ≥

∑

. If 6

4

5 j

j j

s

+

+ =

∑

, then there is one case

(

sj+4,sj+5,sj+6

)

=

(

1, 3,1

)

(where the cases

(

1, 2,1 , 1, 2, 2

) (

)

are not possible). But the

case

(

1, 3,1

)

is not compatible with any of the cases when

3

8 j

j j

s

+ =

[image:10.595.194.547.51.774.2]

∑

, this implies that

(11)

3

9 j

j j

s

+ ≥

∑

. Then 6 14

j

j j

s

+ ≥

∑

(where the case

(

1, 3,1, 3,1, 3

)

is not possible). If

6

4

6 j

j j

s

+

+ ≥

∑

then 6 3 6

4

8 6 14

j j j

s s s

+ + +

+

= + ≥ + =

∑

.

4) We have s_j ≥3, then from 2 is

6

15 j

j j

s

+ ≥

∑

.

5) We have sj ≥4, then from 2 is

6

16 j

j j

s

+ ≥

∑

. This complete the proof of the Lemma. □

Lemma 3.6. If 6 14

j

j j

s

+ =

∑

, then s_j =1 or s_j₊₆ =1.

Proof. We suppose the contrary s sj, j+6≥2. From Lemma 3.5, s sj, j+6<3, else 6

15 j

j j

s

+ ≥

∑

. Now, we must study the case sj=sj+6=2. We have 5

2

10 j

j j

s

+

+ =

∑

, by Lem-

ma 3.3, the case

(

2, 2, 2, 2, 2, 2

)

is not possible, this implies that not all elements of the

subsequence

(

sj+1,,sj+5

)

are equal to the value 2. If sj+1,sj+2,sj+3,sj+4,sj+5≥2

where at least one of them is equal or greater than 3, then 6 15 j

j j

s

+ ≥

∑

, this is a

contra-diction with 6 14

j

j j

s

+ =

∑

. Now, we have 5 10

j

j j

s

+ =

∑

, where one of the subsequence ele-

ment

(

sj+1,,sj+5

)

is at most equal the value 1 (where 1≤sj ≤4). We consider the

cases s_j =1 for j+ ≤ ≤ +1 j j 5:

1) sj+1=1 or sj+5 =1 (where two cases are similar), we study the case sj+1=1

then sj+2=4, these implies that sj+sj+1+sj+2=7 . By Lemma 3.5, we have 6

3

8 j

j j

s

+

+ ≥

∑

then 6 15

j

j j

s

+ ≥

∑

, this is a contradiction.

2) sj+2=1 or sj+4=1 (where two cases are similar), we study the case sj+2=1

then sj+1≥3, (because the case

(

2, 2,1

)

is not possible). If sj+1=3 then sj+3≥3

and we have sj+6=2 then 5

4

4 j

j j

s

+

+ ≥

∑

(because two cases

(

1, 2, 2 , 2,1, 2

) (

)

are not

possible). Thus implies that 6 2 3 1 3 4 2 15

j

j j

s

+

≥ + + + + + =

∑

3) sj+3=1, then we have two subcases results from sj+2+sj+4 ≥6:

Subcase 1: sj+2 =sj+4 =3 then sj+1,sj+5≥2 (because two cases

(

s sj, j+1,sj+2

)

=

(

2,1, 3

)

and

(

sj+4,sj+5,sj+6

)

=

(

3,1, 2

)

are not possible). Thus implies

that 6 15

j

j j

s

+ ≥

∑

Subcase 2: If sj+2 =2, 4sj+4 = or conversely (two cases are similar in studying), so

we will study case sj+2=2, 4sj+4= then sj+5≥1, if sj+5≥2, then 6

15 j

j j

s

+ ≥

∑

,

be-cause sj+4+sj+5+sj+6≥8, we have 3

8 j

j j

s

+ ≥

∑

. Then 6 15

j

j j

s

+ ≥

∑

, this is a contradiction).

If sj+5=1, then sj+4+sj+5+sj+6 =7 . We have 3

8 j

j j

s

+ ≥

(12)

6

15 j

j j

s

+ ≥

∑

this is a contradiction.

Finally, we get if 6 14

j

j j

s

+ =

∑

, then s_j =1 or s_j₊₆ =1. This completely the proof. □

Result 3.1. If 6 14

j

j j

s

+ =

∑

, then from Lemma 3.6, we have the cases for subsequence

(

s sj, j+1,sj+2,sj+3,sj+4,sj+5,sj+6

)

:

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 2 3

4 5 6

7 8 9

10

: 1, 2, 3, 2,1, 4,1 , : 1, 2, 3, 2, 2, 2, 2 , : 1, 2, 3, 2, 2, 3,1 ,

: 1, 2, 3, 3,1, 3,1 , : 1, 3,1, 3,1, 4,1 , : 1, 3,1, 3, 2, 2, 2 ,

: 1, 3,1, 3, 2, 3,1 , : 1, 3,1, 3, 3, 2,1 , : 1, 3,1, 4,1, 3,1 ,

: 1, 3, 2, 2, 2, 2

a a a

a

(

)

(

)

(

)

(

)

11

(

)

12

(

)

13 14 15

, 2 , : 1, 3, 2, 2, 2, 3,1 , : 1, 3, 2, 2, 3, 2,1 ,

: 1, 3, 2, 3,1, 3,1 , : 1, 4,1, 2, 3, 2,1 , : 1, 4,1, 3,1, 3,1 .

a a

a a a

It is 15 cases (where s_j =1 with

6

14 j

j j

s

+ =

∑

). We have three cases with sj+1=2,

1 3

j

s₊ = and s_j₊₁=4.

Case 1: sj+1=2 (including the cases sj=1 and sj+1=2 or sj+6=1 and sj+5=2).

We have these cases are a a a a a a1, 2, 3, 4, 8, 12,a14 and comes before these cases, sj−1=4

or comes after these cases sj+7=4, i.e., if sj=1, 2sj+1= then sj−1=4 and if

6 1, 25

j j

s₊ = s₊ = then s_j₊₇=4.

Case 2: sj+1=3, 4sj+1= and these are the 8 remaining cases. We will study these

cases after rejecting isomorphism cases when there is two cases or more, where

(

sj,,sj+6

) (

= sj+6,,sj

)

, then we will study only one case. We have 8 cases as

fol-lows:

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

5 6 7 9

10 11 13 15

: 1, 3,1, 3,1, 4,1 , : 1, 3,1, 3, 2, 2, 2 , : 1, 3,1, 3, 2, 3,1 , : 1, 3,1, 4,1, 3,1 ,

: 1, 3, 2, 2, 2, 2, 2 , : 1, 3, 2, 2, 2, 3,1 , : 1, 3, 2, 3,1, 3,1 , : 1, 4,1, 3,1, 3,1 .

a a a a

We note that two cases a a5, 15 are similar where one of them is contrary to the

oth-er one, so we study the case a5. Also, two cases a a7, 13 are similar, so we study the

case a7. Then we study these cases: a a a a a5, 6, ,7 9, 10,a11. □

Notice 3.2. We note that all the possible cases in Result 3.1, do not begin or end with 3 or 4 and it do not begin or end with sj+sj+1≥5 or sj+5+sj+6≥5 such that

2 j

s = or s_j₊₆=2, and s_j₊₁=3 or s_j₊₅=3. Thus implies that if s_j=2, 3s_j₊₁= ,

then 6 15

j

j j

s

+ ≥

∑

. Also, we note cases a a a5, 6, 7 are beginning with (1, 3, 1, 3), but from

Lemma 3.4, we get sj−1=4. Now, remains our three cases for studying by the

follow-ing lemma are:

(

)

(

)

(

)

9: 1, 3,1, 4,1, 3,1 , : 1, 3, 2, 2, 2, 2, 2 , : 1, 3, 2, 2, 2, 3,1 .10 11

a a a □

Result 3.2. If sj+1=3, 1sj= where kj∩ =s

{

( )

1,j

}

or kj∩ =s

{

( )

2,j

}

then

1 4

j

s₋ = , also for k_j∩ =s

{

( )

4,j

}

or k_j∩ =s

{

( )

5,j

}

because it are similar to two

cases kj∩ =s

{

( )

2,j

}

or kj∩ =s

{

( )

1,j

}

, respectively. □

Lemma 3.7. If 6 14

j

j j

s

+ =

∑

, such that sj+5=3, 1sj+6= , then 13

7

15 j

j j

s

+

+ ≥

(13)

if 13 7 15 j j j s + + =

∑

then 20

14 15 j j j s + + ≥

∑

.

Proof. By Result 3.2, if kj+6∩ =s

{

(

1,j+6

)

}

, kj+6∩ =s

{

(

2,j+6

)

}

,

(

)

{

}

6 4, 6

j

k₊ ∩ =s j+ or k_j₊₆∩ =s

{

(

5,j+6

)

}

then s_j₊₇=4. From Lemma 3.5, we get 13

7 16 j j j s + + ≥

∑

.Assume kj+6∩ =s

{

(

3,j+6

)

}

then we have two cases for kj+5∩s:

Case 1. kj+5∩ =s

{

(

1,j+5 , 3,

) (

j+5 , 5,

) (

j+5

)

}

. Then sj+7=4, by lemma 3.5, 13 7 16 j j j s + + ≥

∑

.

Case 2. kj+5∩ =s

{

(

1,j+5 , 2,

) (

j+5 , 5,

) (

j+5

)

}

or

(

) (

)

{

}

5 1, 5 , 4, 5 , 5, 5

j

k₊ ∩ =s j+ j+ j+ and both cases are similar, so we will consider

the first case. We have 3≤sj+7≤4 then by Lemma 3.5, 13 7 15 j j j s + + ≥

∑

. If sj+7=4 then 13 7 16 j j j s + + ≥

∑

. Assume sj+7 =3, if

13 7 16 j j j s + + ≥

∑

the proof is finish. Assume 13

7 15 j j j s + + =

∑

then we have cases sj+8=1, 2, 3 or 4.

Subcase 2.1. If sj+8=4 then sj+9 ≥1. This implies that

13 13

7 10

3 4 1 8 8 16

j j j j j j s s + + + + ≥ + + + = + =

∑

{By Lemma 3.5, 3 8

j j j s + ≥

∑

}.

Subcase 2.2. If sj+8=3 then 13 9 9 j j j s + + ≥

∑

. If 13

9 9 j j j s + + >

∑

then 13

7 16 j j j s + + ≥

∑

. Assume

that 13

9 9 j j j s + + =

∑

then we have only one case

(

sj+9,,sj+13

)

=

(

1, 3,1, 3,1

)

or

(

sj+9,,sj+13

)

=

(

1, 2, 3, 2,1

)

. For any case we have sj+8=4. So, we get 13 9 9 j j j s + + >

∑

.

Which implies that 13

7 16 j j j s + + ≥

∑

.

Subcase 2.3. If sj+8 =1 then sj+9=4 {because the case

(

sj+5,sj+6,sj+7,sj+8,sj+9

)

=

(

3,1, 3,1, 3

)

is not possible, by Lemma 3.3}. Then

13 13

7 10

3 1 4 8 8 16

j j j j j j s s + + + + ≥ + + + ≥ + =

∑

.

Subcase 2.4. If sj+8=2 then sj+7=3, 2sj+8= , we have the following cases:

2.4.1. sj+9≥3 then

13 13

7 10

3 2 3 8 8 16

j j j j j j s s + + + + ≥ + + + ≥ + =

∑

.

2.4.2. sj+9≠1 {because there is only one case for

(

sj+7,sj+8,sj+9

)

=

(

3, 2,1

)

such

that

{

Kj+7∪Kj+8∪Kj+9

}

∩ =S 2,

{

(

j+7 , 3,

) (

j+7 , 4,

) (

j+7 , 1,

) (

j+8 , 5,

) (

j+8 , 3,

) (

j+9

)

}

But according to distribution vertices kj+5∩S and kj+6∩S we have

(

) (

)

{

52,7 7 , 3, 7 , 4, 7

}

j

k

j j j

+ ∩

≠ + + + .

(14)

(

sj+7,sj+8,sj+9

)

=

(

3, 2, 2

)

. We will study the cases that leads to 13 7 15 j j j s + + =

∑

, i.e.,

13 10 8 j j j s + + =

∑

, {because the cases which leads to 13

7 16 j j j s + + ≥

∑

the proof will be done}. Now,

we have the fixed case

(

sj+7,sj+8,sj+9

)

=

(

3, 2, 2

)

We will consider the vertices

10

j

k₊ ∩S which imply the following:

2.4.3.1. If sj+10=4 then

(

3, 2, 2, 4,sj+11,sj+12,sj+13

)

, this implies that

13 11 4 j j j s + + =

∑

and

(

sj+11,sj+12,sj+13

)

=

(

1, 2,1

)

is not possible.

2.4.3.2. If sj+10 =3 then

(

3, 2, 2, 3,sj+11,sj+12,sj+13

)

and

13 11 5 j j j s + + =

∑

which imply

that

(

sj+11,sj+12,sj+13

)

=

(

2,1, 2 , 2, 2,1 , 1, 2, 2

) (

)

or (1, 3, 1), and the only possible case

is (1, 3, 1). Thus implies that

(

sj+7,,sj+13

)

=

(

3, 2, 2, 3,1, 3,1

)

. By Lemma 3.4 and

Lemma 3.5 is sj+14 =4, these implies that 20 14 16 j j j s + + ≥

∑

.

2.4.3.3. If sj+10=2 then

(

3, 2, 2, 2,sj+11,sj+12,sj+13

)

, i.e.,

13 11 6 j j j s + + =

∑

. We have

11 1

j

s₊ ≠ {because the case

(

2, 2,1

)

is not possible}. Then we have the following cases

for sj+11,sj+12,sj+13:

1). If sj+11=4 then sj+12=1 and sj+13=1, but the case

(

4,1,1

)

is not possible.

2). If sj+11=3 and sj+12=1 then sj+13=2, also the case

(

3,1, 2

)

is not possible.

3). If sj+11=3, sj+12=2 and sj+13=1 then

(

sj,,sj+6

)

=

(

3, 2, 2, 2, 3, 2,1

)

which

gets sj+7=4 and 13 7 16 j j j s + + ≥

∑

.

4). If sj+11=2 and sj+12 =2 then sj+13=2, but the case

(

3, 2, 2, 2, 2, 2, 2

)

is not

possible. If sj+11=2, sj+12 =3 and sj+13=1 then we gets

(

sj,,sj+6

)

=

(

3, 2, 2, 2, 2, 3,1

)

During the proof of Lemma, we notice that if sj =3

and sj+1=1, then 8 2 15 j j j s + + ≥

∑

. This complete the proof. □

Result 3.3. Based on the Lemma 3.6, and the other Lemmas and results precede it.

We see that when we have case of 6 14

j j j s + =

∑

, then the only case that comes after it, is

13 7 15 j j j s + + =

∑

such that

(

sj+7,,sj+13

)

=

(

3, 2, 2, 2, 2, 3,1

)

which continues in the same

way or it is followed by 7 columns contain 16 vertices from S {by Lemma 3.6,

20 14 15 j j j s + + ≥

∑

, because sj+12 =3, sj+13=1}. When this case is repeated then 6 15 n j j n s = − ≥

∑

and then when the case 6 14

j j j s + =

∑

it is necessary, the case 6 1 7

6

16 j q r

j j q s + + − + + + ≥

∑

exists as

well {where j+ + − +6 q 1 7r≤n} these implies that

1 15 7 n j j n s =   ≥ _ _

∑

then

(

)

2 5 1 2 7 n n j j n

P P s n

γ =   × = ≥ _{+  }  

∑

. □

(15)

1) s1≥2 and s1+s2≥4

(

sn−1+sn≥4,sn≥2

)

.

2) If s1+s2 =4 then s1+s2+s3=8 (sn−1+sn=4 then sn−2+sn−1+sn =8).

3) s1+s2+s3≥6

(

sn−2+sn−1+sn≥6

)

.

4) 4

1 3

9 9

n

j j

j j n

s s = = −   ≥ _ ≥ _  

∑

.

5) 5

1 4

10 10

n

j j

j j n

s s

= = −

 

≥ _ ≥ _

 

∑

and if 5

1 10 j j s = =

∑

then 6

1 14 j j s = ≥

∑

, also if

4 10 n j j n s = − =

∑

then 5 14 n j j n s = − ≥

∑

6) 6

1 5

13 13

n

j j

j j n

s s = = −   ≥ _ ≥ _  

∑

.

7) 7

1 6

15 15

n

j j

j j n

s s = = −   ≥ _ ≥ _  

∑

.

8) If s1+s2 =5 then either 5 1 11 j j s = ≥

∑

or 6

1 14 j j s = ≥

∑

, also if sn−1+sn =5 then

ei-ther 4 11 n j j n s = − ≥

∑

or

5 14 n j j n s = − ≥

∑

.

Proof. The study of dominating sequence

(

s s1, 2,,sn

)

is the same as the study of the dominating sequence

(

s sn, n−1,,s1

)

, so we study one case

(

s s1, 2,,sn

)

. Also, the study of 1 r j j s =

∑

is the same as the study of

1

n

j j n r

s

= − +

∑

.

1) We have s1≥2, if s1=2 then s2≥3 thus, s1+s2≥5 if s1≥3 then

(

)

2 1 1 j 4

s ≥ ≤s ≤ these implies that s₁+s₂ ≥4.

2) If s1+s2 =4, then we have only one the case k1∩ =s

{

( ) ( ) ( )

1,1 , 3,1 , 5,1

}

these

implies that k2∩ =s

{

( )

3, 2

}

and s3 =4 then s1+s2+s3=8.

3) If s1+s2≥5, then 3 1 6 j j s = ≥

∑

{because 1≤s_j ≤4} and if s₁+s₂=4 then by 2,

is 3

1 8 j j s = =

∑

.

4) If s1+s2 =4 then 4 1 8 j j s = =

∑

these implies that 4

1 9 j j s = ≥

∑

and if s1+s2 ≥6

then 4

1 9 j j s = ≥

∑

{because s3+s4 ≥3}. Assume that s1+s2 =5, then we have three

cases:

4.1) s1=2, 3s2= then s3+s4≥4, because the case

(

s s s2, 3, 4

) (

= 3,1, 2

)

is not

possible. Also the case

(

s s s2, 3, 4

) (

= 3, 2,1

)

is not possible, else when

( ) ( ) ( )

{

}

2 2, 2 , 3, 2 , 4, 2

k ∩ =s and this is not possible.

4.2) s1=3, 2s2 = then s3+s4≥4 because the cases

(

s s s2, 3, 4

) (

= 2, 2,1

)

,

(

s s s2, 3, 4

) (

= 2,1, 2

)

are not possible.

4.3) s1=4, 1s2= then s3+s4≥4, because the cases

(

s s s s1, 2, 3, 4

) (

= 4,1, 2,1

)

,

(

s s s s1, 2, 3, 4

) (

= 4,1, 2, 2

)

are not possible. Thus implies that we have

(16)

5) By Lemma 3.4, we have two cases for 4 1 9 j j s = =

∑

and these two cases are

(

1, 2, 3, 2,1 , 1, 3,1, 3,1

) (

)

, furthermore these cannot be shown here because s₁≥2. Thus implies that we 5

1 10 j j s = ≥

∑

.

6). If s1+s2≥5 then

6 6

1 2

1 3

5 8 13

j j

s s s s

= =

= + + ≥ + =

∑

.

(where by Lemma 3.5, we have 3 8

j j j s + ≥

∑

). Let s1+s2 =4 then 3 1 8 j j s = =

∑

these

im-plies that 6 6

1 4 8 j j j j s s = = ≥ +

∑

. Thus implies that 6

1

8 5 13 j

j s

=

≥ + =

∑

{because 2 5

j j j s + ≥

∑

}.

7) If s1≥3 then from Lemma 3.5, 7 1 15 j j s = ≥

∑

. Let s1=2 {because s1>1} then

2 3

s ≥ . This implies that

7 1 15 j j s = ≥

∑

{by Notice 3.2}.

8) If s1+s2 =5 then either 5 1 11 j j s = ≥

∑

or 6

1 14 j j s = ≥

∑

. We have s1+s2 =5, then

we have three cases:

8.1) s1=4, 1s2= , then s3+ +s4 s5≥7 because the cases

(

s s s s s1, 2, 3, 4, 5

) (

= 4,1, 2, 2, 2 , 4,1, 3, 2,1 , 4,1, 2, 3,1 or 4,1, 3,1, 2

) (

)

(

)

are not possible.

Thus implies that 5

1 11 j j s = ≥

∑

.

8.2) s1=2, 3s2 = , then 5 1 10 j j s = ≥

∑

and if 5

1 10 j j s = =

∑

then

(

s s s s s1, 2, 3, 4, 5

) (

= 2, 3,1, 3,1

)

. By Lemma 3.4, s6 =4. Thus implies that 6 1 14 j j s = ≥

∑

.

8.3) s1=3, 2s2 = , then

(

s s s s s1, 2, 3, 4, 5

)

it has minimal numerals in the following

cases

(

s s s s s1, 2, 3, 4, 5

) (

= 3, 2, 2, 2, 2 , 3, 2,1, 4,1 or 3, 2, 3,1, 3

) (

)

(

)

and for the case

(

s s s3, 4, 5

) (

= 1, 3,1

)

is not compatible with the case

(

s s1, 2

) ( )

= 3, 2 . Thus implies that

5 1 11 j j s = ≥

∑

. This completes the proof. □

Theorem 3.5.

(

)

(

)

(

)

2 5

2 : 1, 2, 3, 5 mod 7 ,

7

2 1: 0, 4, 6 mod 7 .

7 n n n n P P n n n γ  ₊  _≡   _{ }  × _{= }    +_{ }+ ≡  _{ } 

Proof. By Result 3.3, we have 2

(

5

)

1 15 7 n n j j n

p p s

γ

=

 

× = _{≥ } _

 

∑

. By Theorem 3.4, we get

(

)

(

)

2 5 2 : 1, 2, 3, 5 mod 7 .

7

n

p p n n

γ × = + _{ } ≡

 

Now, for n≡0, 4, 6 mod 7

(

)

, by Theorem 3.4, we have ₂

(

₅

)

2 1

7

n

p p n

γ × ≤ + _{ }+