The average order of a connected vertex set in $K_m \times P_n$

The a v erage order of a connected v ertex set in 1 K m × P n 2 Mingyuan Ma ∗ , Han Ren † Arm y Engineering Univ ersit y of PLA, Nanjing,210001,China Sc ho ol of Mathematical Sciences,East China Normal Univ ersity ,Shanghai,200241,China 3 Abstract. Let G b e a connected graph. Let N ( G ) and S ( G ) b e the num b er of connected 4 sets of G and the sum of the orders of these connected sets of G , respectively . Then A ( G ) = S ( G ) N ( G ) 5 is called the a verage order of a connected set of G . In this pap er, w e deriv e a closed-form form ula 6 for A ( K m × P n ), where K m × P n is the Cartesian pro duct of the complete graph K m and the 7 path P n . 8 Keyw ords. connected sets, av erage order, densit y of connected sets 9 1 In tro duction. 10 All graphs in this pap er are simple and finite. Let G be a connected graph of n v ertices. A v ertex subset M ⊆ V ( G ) is called a c onne cte d set of G if the subgraph induced by M is connected. Let N ( G ) denote the num b er of connected sets in G . Let { M 1 , M 2 , . . . , M N ( G ) } b e the collection of all connected sets of G . Then w e denote S ( G ) = N ( G ) P i =1 | M i | as the sum of orders of the connected sets in G . F urthermore, let A ( G ) = S ( G ) N ( G ) and D ( G ) = A ( G ) n denote the a verage order of connected sets of G and the density of connected sets of v ertices, 11 resp ectiv ely . 12 There are numerous research concerning the av erage order and density of connected sets of a graph. In 1983, Jamison[6] prov ed that among all trees of order n, the path P n minimizes the a v erage order of a subtree. In 2010, Vince and W ang[16] prov ed that if T is a tree all of whose in ternal vertices hav e degree at least three, then 1 2 ≤ D ( T ) < 3 4 . In 2018, Kro eker, Mol and ∗ E-mail: 623mm y@sina.com. † E-mail: hren@math.ecn u.edu.cn 1 Oellermann[8] pro ved that n 2 ≤ A ( G ) ≤ n +1 2 for a connected graph of order n . F or ladders and circular ladders, Vince[14] ga ve explicit closed formulas for b oth the num b er and the a verage order of connected induced subgraphs of them in terms of the classic P ell num b ers in 2021. F or example, the av erage order of a connected set of the ladder L n is A ( L n ) = (32 − 45 ¯ β ( n ) − 32 β ( n )) + n (10 + 21 β ( n ) + 30 ¯ β ( n )) 2( β ( n + 3) − 4 n − 7) , (1 . 1) where β ( n ) denotes P ell-Lucas num b er and ¯ β ( n ) denotes Pell num b er. 1 Let H and Q b e t wo graphs with v ertex sets V ( H ) = { u 1 , u 2 , . . . , u s } and V ( Q ) = { v 1 , v 2 , . . . , v t } , 2 resp ectiv ely . The Cartesian pr o duct of H and Q , denoted by H × Q , is the graph with vertex set 3 V ( H ) × V ( Q ) and edge set { ( u i , v p )( u j , v q ) | u i u j ∈ E ( H ) and v p = v q , or v p v q ∈ E ( Q ) and u i = 4 u j } . Let P l b e the path of l v ertices. In fact, L n is exactly P n × P 2 . In 2021, Vince [14] 5 prop osed the following question: Find a form ula for N ( P n × P n ). So far as we know, there is 6 not any formula for N ( P n × P n ) for an arbitrary n . In 2025, Ma et al.[10] obtained a form ula 7 for N ( K m × P n ), where K m is the complete graph of m v ertices. Ho wev er, they do not giv e a 8 form ula for A ( K m × P n ). In this pap er we will establish a closed-form formula for A ( K m × P n ). 9 The arrangement of the paper is as follows: In Section 2, we introduce some notations. In 10 particular, we in tro duce the recurrence matrix A m . In Section 3, we develop a matrix method 11 for calculating the av erage order of connected subsets in K m × P n and provide a general form ula 12 together with its pro of. Subsequen tly , we give a form ula for the density of connected sets in 13 K m × P n . In Section 4, w e use the formulas from Section 3 to calculate A ( K 2 × P n ) and 14 D ( K 2 × P n ). Our results coincide with those of Vince[14]. Section 5 prop oses tw o questions on 15 K m × C n , where C n is the cycle of n vertices. 16 2 Notations and Preliminaries. 17 By the definition of K m × P n , it consists of n subgraphs, each of which is isomorphic to K m . 18 F or conv enience, we refer to eac h such subgraph as a layer . As shown in Figure 1, H = K 3 × P 3 19 has three lay ers, where the cycle v 1 , 1 v 1 , 2 v 1 , 3 v 1 , 1 forms the initial lay er and v 3 , 1 v 3 , 2 v 3 , 3 v 3 , 1 forms 20 the third lay er. 21 Let f ( m, k )(where 1 ≤ k ≤ n ) denote the n um b er of connected sets in K m × P n whic h 22 con tain at least one vertex of each la yer from the first to the k -th lay er of K m × P n . The 23 follo wing theorem follows directly . 24 Theorem 2.1[10]. N ( K m × P n ) = n P k =1 ( n + 1 − k ) f ( m, k ) . 25 T o compute f ( m, k ), we partition it into 2 m − 1 parts. This is b ecause every lay er of G 26 2 Figure 1: H . con tains m vertices, and the connected sets that con tain at least one v ertex from eac h of the first 1 k la yers induce a non-empt y v ertex subset in the k -th lay er. There are  m 1  +  m 2  + · · · +  m m  (= 2 2 m − 1) suc h nonempty vertex subsets in the k -th lay er. Denote these subsets b y { S k p } for 3 1 ≤ p ≤ 2 m − 1. 4 Let f S p ( m, k ) denote the num b er of connected subsets that contain at least one vertex of eac h lay er from the first to the ( k − 1)-th la yer with exactly the vertices of S k p in the k -th lay er. Then f ( m, k ) = 2 m − 1 X p =1 f S p ( m, k ) . Lemma 2.2. F or 1 ≤ p, q ≤ 2 m − 1 with p  = q , if | S k p | = | S k q | , then f S p ( m, k ) = f S q ( m, k ) . 5 The stateme n t follows directly from symmetry . 6 W e now in tro duce the follo wing definitions which will b e used in the form ula for A ( K m × P n ). Let f i ( m, k ) be the n umber of connected sets in K m × P n whic h contains at least one vertex of eac h of the first k − 1 la yer and i fixed vertices in the k -th la yer, where 1 ≤ i ≤ m . By Lemma 2.2, we hav e f ( m, k ) = m X i =1  m i  f i ( m, k ) . (2 . 1) 7 The following recurrence is tak en from [10]. 8 F or i = 1 , 2 , . . . , m , 9 f i ( m, k ) = m X j =1  m j  f j ( m, k − 1) − m − i X j =1  m − i j  f j ( m, k − 1) . 3 Since  m − i j  = 0 for j ≥ m − i + 1,the recurrence for eac h f i ( m, k ) can simplify to: f i ( m, k ) = m X j =1 [  m j  −  m − i j  ] f j ( m, k − 1) . Let 1 A m =            m 1  −  m − 1 1   m 2  −  m − 1 2  . . .  m m  . . . . . . . . . . . .  m 1  −  m − i 1   m 2  −  m − i 2  . . .  m m  . . . . . . . . . . . .  m 1   m 2  . . .  m m            . Then [ f 1 ( m, k ) , f 2 ( m, k ) , . . . , f m ( m, k )] T = A m [ f 1 ( m, k − 1) , f 2 ( m, k − 1) , . . . , f m ( m, k − 1)] T . Th us 2 [ f 1 ( m, k ) , . . . , f m ( m, k )] T = A m [ f 1 ( m, k − 1) , . . . , f m ( m, k − 1)] T = A 2 m [ f 1 ( m, k − 2) , . . . , f m ( m, k − 2)] T = · · · = A k m [1] T , (2 . 2) where [1] T denote the column vector [1 , 1 , . . . , 1] T of length m . 3 In particular, f m ( m, k ) = m X i =1  m i  f i ( m, k − 1) = f ( m, k − 1) b y the equation (2.1). F or this reason, we call A m the r e curr enc e matrix . 4 Let v ( i, k ) denote the i -th column of A k − 1 m . Let ( A m ) T denote the transp ose of A m . The 5 follo wing theorem establishes the relationship b etw een v ( i, k ) and f i ( m, k ). 6 Theorem 2.3. [  m 1  ,  m 2  , . . . ,  m m  ] × v ( i, k ) =  m i  f i ( m, k ) . 4 Pro of. Let C m = diag (  m 1  ,  m 2  , . . . ,  m m  ) =         m 1  0 . . . 0 0  m 2  . . . . . . . . . . . . . . . 0 0 . . . 0  m m         . W e first state a useful claim. 1 Claim 1. F or every k ≥ 1, ( C m × A k m ) T = C m × A k m . Pro of of Claim 1. W e pro ceed by induction on k . The base case is k = 1. Using the iden tit y  m i  [  m j  −  m − i j  ] =  m i  m j  − m ! i ! j !( m − j − i )! =  m j  [  m i  −  m − j i  ] , w e observe that the ( i, j )-en try of C m A m equals the ( i, j )-entry of ( A m ) T C m . 2 Explicitly , 3 C m × A m =            m 1  [  m 1  −  m − 1 1  ] . . .  m 1  [  m i  −  m − 1 i  ] . . .  m 1  m m  . . . . . . . . . . . . . . .  m i  [  m 1  −  m − i 1  ] . . .  m i  [  m i  −  m − i i  ] . . .  m i  m m  . . . . . . . . . . . . . . .  m m  m 1  . . .  m m  m i  . . .  m m  m m            = ( A m ) T × C m . Assume ( C m A k − 1 m ) T = C m A k − 1 m . Then ( C m × A k m ) T = ( A m ) T × ( C m × A k − 1 m ) T = ( A m ) T × C m × A k − 1 m = C m × A k m , where w e used the iden tit y C m × A m = ( A m ) T × C m from the base case. This completes the 4 induction. 5 No w return to the proof of Theorem 2.3. 6 By Claim 1, the matrix C m A k − 1 m is symmetric. Hence its row sums equal its column sums. The vector of row sums is [1 , 1 , . . . , 1] × C m A k − 1 m = [  m 1  ,  m 2  , . . . ,  m m  ] A k − 1 m . 5 On the other hand, the vector of column sums is C m A k − 1 m × [1] T = [  m 1  f 1 ( m, k ) ,  m 2  f 2 ( m, k ) , . . . ,  m m  f m ( m, k )] T . Since A k − 1 m = [ v (1 , k ) , v (2 , k ) , . . . , v ( i, k ) , . . . , v ( m, k )], comparing the i -th en tries of the tw o 1 equal vectors gives [  m 1  ,  m 2  , . . . ,  m m  ] × v ( i, k ) =  m i  f i ( m, k ). 2 Need to say that we can obtain some by-products using the recurrence matrix. No w, let p m ( λ ) = | λE m − A m | , which is the c haracteristic p olynomial of A m . Then p m ( λ ) = λ m + c m,m λ m − 1 + · · · + c m, 1 . (2 . 3) Since the v alue of c m,m is related to the Fib onacci sequence, let us recall some concepts 3 ab out the Fib onacci sequence. W e denote the n -th term of the Fib onacci sequence by F ( n ), 4 where F ( n ) = 1 √ 5 [( 1+ √ 5 2 ) n − ( 1 − √ 5 2 ) n ] and F (0) = 0 , F (1) = 1 , F (2) = 1 , F (3) = 2. 5 Theorem 2.4. If m ≥ 3 , then c m,m = F ( m + 1) − 2 m ; c m, 1 = 1 . In p articular, c 2 , 2 = 6 − 2 , c 2 , 1 = − 1 . 7 Pro of. It is a well-kno wn result that − c m,m equals the trail of the recurrence matrix A m . Hence, − c m,m = m X i =1  m i  − ⌊ m 2 ⌋ X i =1  m − i i  . Ob viously , the first part equals to 2 m − 1. By [7], the second part is ⌊ m 2 ⌋ P i =0  m − i i  −  m 0  = 8 F ( m + 1) − 1. Therefore, we ha v e c m,m = F ( m + 1) − 1 − 2 m + 1 = F ( m + 1) − 2 m . 9 The co efficient c m, 1 equals ( − 1) m | A m | . Ev ery element of the last column of A m is  m m  = 1. Multiplying this column by  m i  and subtracting the result from the i -th column for eac h 1 ≤ i ≤ m − 1, we obtain the follo wing determinant.                  −  m − 1 1  −  m − 1 2  . . . − 1 1 −  m − 2 1  . . . − 1 0 1 . . . . . . . . . . . . − 2 − 1 . . . 0 1 − 1 0 . . . 0 1 0 0 . . . 0 1                  , in which the upper-left ( m − 1) × ( m − 1) blo c k can b e reduced to an upp er-triangular matrix with -1 on ev ery diagonal entry . Since the reduction pro cess in volv es an o dd num b er of column 6 exc hanges (except when m = 2), w e obtain: c m, 1 = ( − 1) m | A m | = ( − 1) m × ( − 1) × ( − 1) m − 1 = ( − 1) 2 m = 1 . 1 Consequen tly , we obtain the following linear recurrence: 2 Theorem 2.5. F or every k ≥ m , f ( m, k ) = − c m,m f ( m, k − 1) − c m,m − 1 f ( m, k − 2) − · · · − 3 c m, 1 f ( m, k − m ) . 4 Pro of. Let 0 m,m denote the m × m zero matrix. By the Cayley-Hamilton Theorem[13], we ha v e p m ( A m ) = 0 m,m . It implies that A m m + c m,m A m − 1 m + · · · + c m, 1 E m = 0 m,m b y (2.3), where E m denotes the identit y matrix of order m . Hence A m m = − c m,m A m − 1 m − · · · − c m, 1 E m . Multiplying b oth sides by A k − m − 1 m × [1] T giv es 5 A k − 1 m × [1] T = − c m,m A k − 2 m × [1] T − · · · − c m, 1 A k − m − 1 m × [1] T [ f 1 ( m, k ) , . . . , f m ( m, k )] T = − c m,m [ f 1 ( m, k − 1) , . . . , f m ( m, k − 1)] T − . . . − c m, 1 [ f 1 ( m, k − m ) , . . . , f m ( m, k − m )] T . Multiplying b oth sides of the ab o ve equation by [  m 1  ,  m 2  , . . . ,  m m  ] yields 6 m X i =1  m i  f i ( m, k ) = − c m,m m X i =1  m i  f i ( m, k − 1) − · · · − c m, 1 m X i =1  m i  f i ( m, k − m − 1) . Applying (2.1) to each term, w e obtain 7 f ( m, k ) = − c m,m f ( m, k − 1) − c m,m − 1 f ( m, k − 2) − · · · − c m, 1 f ( m, k − m − 1) . 8 Remark 2.6. The or em 2.3 wil l b e use d in the pr o of of L emma 3.4 in Se ction 3. The or ems 9 2.4 and 2.5 wil l b e use d in Se ction 4. 10 7 3 The a v erage order of a connected set of K m × P n . 1 In this se ction, w e discuss the av erage order of connected sets of K m × P n . F or k = 1 , 2 , . . . , n , 2 let F ( m, k ) denote the collection of connected sets of K m × P k that contain at least one vertex 3 of each lay er, and let S ( F ( m, k )) denote the sum of orders of all connected sets in F ( m, k ). 4 Theorem 3.1. A ( K m × P n ) = n P k =1 ( n − k + 1) S ( F ( m, k )) n P k =1 ( n − k + 1) f ( m, k ) . 5 Pro of. The n umerator is precisely the total sum of the orders of all non-empty connected 6 sets of K m × P n , while the denominator equals N ( K m × P n ) by Theorem 2.1, the total num b er 7 of such sets. Hence the formula giv es the av erage order. 8 Next, we intend to give a form ula for S ( F ( m, k )). 9 F or p = 1 , 2 , . . . , 2 m − 1 and k = 1 , 2 , . . . , n , let F S p ( m, k ) denote the collection of connected 10 subsets in K m × P n that contain at least one vertex from eac h lay er from the first to the k − 1-th 11 la y er and exactly all vertices of S k p in the k -th lay er. 12 Lemma 3.2. F or 1 ≤ p, q ≤ 2 m − 1 with p  = q , if | S k p | = | S k p | , then the total sum of the 13 or ders of c onne cte d subsets in F S p ( m, k ) e quals that in F S q ( m, k ) . 14 The stateme n t follows directly from symmetry . 15 No w, for i = 1 , 2 , . . . , m and k = 1 , 2 , . . . , n , let F i ( m, k ) denote the collection of connected sets of K m × P k that contain at least one vertex of eac h lay er with exactly i fixed vertices in the k -th la y er. Let s i ( m, k ) denote the total sum of the orders of connected sets in F i ( m, k ). Since ev ery connected set of K m × P 1 has order i , we ha v e [ s 1 ( m, 1) , s 2 ( m, 1) , . . . , s m ( m, 1)] T = [1 , 2 , . . . , m ] T . By Lemma 3.2, S ( F ( m, k )) = m X i =1  m i  s i ( m, k ) . T o compute s i ( m, k ), we introduce a new matrix B m and use a recursiv e m ultiplication sc heme in v olving matrices B m and A m . Let [1] T denote the column vector [1 , 1 , . . . , 1] T of 8 length m . Let B m = diag (1 , 2 , . . . , m ) =        1 0 . . . 0 0 2 . . . . . . . . . . . . . . . 0 0 . . . 0 m        . Let E m denote the identit y matrix of order m . Then w e hav e the follo wing lemma. 1 Lemma 3.3. [ s 1 ( m, k ) , s 2 ( m, k ) , . . . , s m ( m, k )] T = ( k − 1 X s =0 A k − s − 1 m B m A s m ) × [1] T , wher e A 0 m = E m . 2 Pro of. W e use induction on k . The base case is that k = 1. Since f i ( m, 1) = 1 for 3 i = 1 , 2 , . . . , m , we hav e 4 [ s 1 ( m, 1) , s 2 ( m, 1) , . . . , s m ( m, 1)] T = B m × [1] T = [1 f 1 ( m, 1) , 2 f 2 ( m, 1) , . . . , mf m ( m, 1)] T . So the statement holds for k = 1. Assume the statement true for k − 1 and consider the case k . Let [ s ′ 1 ( m, k ) , s ′ 2 ( m, k ) , . . . , s ′ m ( m, k )] T = A m [ s 1 ( m, k − 1) , s 2 ( m, k − 1) , . . . , s m ( m, k − 1)] T . By the definition of A m and the induction hypothesis, s ′ i ( m, k ) equals the sum of num b ers of 5 v ertices in F i ( m, k ) that lie in la y ers other than the k -th lay er. 6 On the other hand, the vector [1 f 1 ( m, k ) , 2 f 2 ( m, k ) , . . . , mf m ( m, k )] T giv es the sum of num b ers of v ertices in F i ( m, k ) that lie in the k -th lay er for i = 1 , 2 , . . . , m . By the equation (2.2), [ f 1 ( m, k ) , f 2 ( m, k ) , . . . , f m ( m, k )] T = A k − 1 m × [1] T . Hence [1 f 1 ( m, k ) , 2 f 2 ( m, k ) , . . . , mf m ( m, k )] T = B m × [ f 1 ( m, k ) , f 2 ( m, k ) , . . . , f m ( m, k )] T = B m A k − 1 m × [1] T . 9 Therefore, [ s 1 ( m, k ) , s 2 ( m, k ) , . . . , s m ( m, k )] T 1 = [ s ′ 1 ( m, k ) , s ′ 2 ( m, k ) , . . . , s ′ m ( m, k )] T + [1 f 1 ( m, k ) , 2 f 2 ( m, k ) , . . . , mf m ( m, k )] T = A m × [ s 1 ( m, k − 1) , s 2 ( m, k − 1) , . . . , s m ( m, k − 1)] T × [1] T + B m × A k − 1 m × [1] T = A m × k − 2 X s =0 A k − s − 2 m B m A s m × [1] T + E m × B m × A k − 1 m × [1] T = k − 1 X s =0 A k − s − 1 m B m A s m × [1] T , whic h completes the induction. 2 Ho w ever, obtaining s i ( m, k ) requires ev aluating the matrix expression in Lemma 3.3. T o 3 simplify the computation we in tro duce an auxiliary result. 4 Let E i,i b e the m × m matrix whose ( i, i )-entry is 1 and all other en tries are 0. W e 5 note that B m = mE m − ( m − 1) E 1 , 1 − ( m − 2) E 2 , 2 − · · · − E m − 1 ,m − 1 . In order to compute 6 k − 1 P s =0 A k − s − 1 m B m A s m × [1] T , we first calculate k − 1 P s =0 A k − s − 1 m E i,i A s m × [1] T for i = 1 , 2 , . . . , m . 7 F or each j = 1 , 2 , . . . , m , we define [ x i, 1 ( k ) , x i, 2 ( k ) , . . . , x i,j ( k ) , . . . , x i,m ( k )] T = k − 1 X s =0 A k − s − 1 m E i,i A s m × [1] T . The v alues of s i ( m, k ) can then b e obtained from x i,j ( k ). 8 Lemma 3.4. m P j =1  m j  x i,j ( k ) = k P s =1  m i  f i ( m, s ) f i ( m, k + 1 − s ) . 9 Pro of. Since 10 E i,i =           0 · · · 0 · · · 0 . . . . . . . . . . . . 0 · · · 1 · · · 0 . . . . . . . . . . . . 0 · · · 0 · · · 0           , w e hav e 11 10 A k − s − 1 m E i,i A s m =        a 1 ,i b i, 1 a 1 ,i b i, 2 · · · a 1 ,i b i,m a 2 ,i b i, 1 a 2 ,i b i, 2 · · · a 2 ,i b i,m . . . . . . . . . . . . a m,i b i, 1 a m,i b i, 2 · · · a m,i b i,m        , where a j,i is the ( j, i )-entry of A k − s − 1 m and b i,j is the ( i, j )-entry of A s m . 1 Because A s m × [1] T = [ f 1 ( m, s + 1) , f 2 ( m, s + 1) , . . . , f m ( m, s + 1)] T , w e hav e m P j =1 b i,j = 2 f i ( m, s + 1). Hence the sum of the j -th ro w of A k − s − 1 m E i,i A s m is a j,i f i ( m, s + 1). 3 Since [ a 1 ,i , a 2 ,i , . . . , a m,i ] T is the i -th column of A k − s − 1 m , Theorem 2.3 implies [  m 1  ,  m 2  , . . . ,  m m  ] × [ a 1 ,i , a 2 ,i , . . . , a m,i ] T =  m i  f i ( m, k − s ) . Consequen tly , 4 [ x i, 1 ( k ) , x i, 2 ( k ) , . . . , x i,m ( k )] T = k − 1 X s =0 [ a 1 ,i f i ( m, s + 1) , a 2 ,i f i ( m, s + 1) , . . . , a m,i f i ( m, s + 1)] T Multiplying b oth sides of the ab o ve equation by [  m 1  ,  m 2  , . . . ,  m m  ] yields 5 m X j =1  m j  x i,j ( k ) = [  m 1  , . . . ,  m m  ] × k − 1 X s =0 [ a 1 ,i f i ( m, s + 1) , . . . , a m,i f i ( m, s + 1)] T = k − 1 X s =0  m i  f i ( m, k − s ) f i ( m, s + 1) = k X s =1  m i  f i ( m, s ) f i ( m, k + 1 − s ) . The lemma holds. 6 Theorem 3.5. S ( F ( m, k )) = mk f ( m, k ) − m − 1 P i =1 k P s =1  m i  ( m − i ) f i ( m, s ) f i ( m, k + 1 − s ) . 7 Pro of. Observe that 8 B m = mE m − m − 1 X i =1 ( m − i ) E i,i . 11 Then 1 S ( F ( m, k )) = m X i =1  m i  s i ( m, k ) = [  m 1  ,  m 2  , . . . ,  m m  ] × [ s 1 ( m, k ) , s 2 ( m, k ) , . . . , s m ( m, k )] T = [  m 1  ,  m 2  , . . . ,  m m  ] × k − 1 X s =0 A k − s − 1 B m A s × [1] T . Applying Lemmas 3.4, we obtain 2 S ( F ( m, k )) = [  m 1  ,  m 2  , . . . ,  m m  ] × [ mk E m A k − 1 − m − 1 X i =1 ( m − i ) k − 1 X s =0 A k − s − 1 E i,i A s ] × [1] T = mk [  m 1  ,  m 2  , . . . ,  m m  ] × [ f 1 ( m, k ) , f 2 ( m, k ) , . . . , f m ( m, k )] T − m − 1 X i =1 ( m − i )[  m 1  ,  m 2  , . . . ,  m m  ] × [ x i, 1 ( k ) , x i, 2 ( k ) , . . . , x i,j ( k ) , . . . , x i,m ( k )] T = m X i =1  m i  f i ( m, k ) − m − 1 X i =1 m X j =1  m j  ( m − i ) x i,j ( k ) = mk f ( m, k ) − m − 1 X i =1 k X s =1  m i  ( m − i ) f i ( m, s ) f i ( m, k + 1 − s ) , whic h completes the pro of. 3 Com bining Theorems 3.1 and 3.5, we hav e the general formula for the av erage order and the 4 densit y of connected sets in K m × P n . 5 Theorem 3.6. A ( K m × P n ) = n P k =1 ( n − k + 1) { mk f ( m, k ) − m − 1 P i =1 k P s =1  m i  ( m − i ) f i ( m, s ) f i ( m, k + 1 − s ) } n P k =1 ( n − k + 1) f ( m, k ) . Theorem 3.7. D ( K m × P n ) = n P k =1 ( n − k + 1) { mk f ( m, k ) − m − 1 P i =1 k P s =1  m i  ( m − i ) f i ( m, s ) f i ( m, k + 1 − s ) } mn n P k =1 ( n − k + 1) f ( m, k ) . 12 1 4 The num b er and a v erage order of connected sets in K 2 × 2 P n . 3 In this section, w e apply the matrix method developed in Sections 2 and 3 to study the num b er 4 and the av erage order of connected sets in K 2 × P n . Need to sa y that Vince[14] has given a 5 form ula for A ( K 2 × P n )(cf. formula(1.1)). How ever, our metho d is different from that in [14]. 6 F or m = 2, the recurrence matrix is A 2 = " 1 1 2 1 # . By Theorem 2.4, the c haracteristic polynomial of A 2 is p 2 ( λ ) = λ 2 − 2 λ − 1. Solving p 2 ( λ ) = 0 giv es the tw o eigen v alues: λ 1 = 1 − √ 2 , λ 2 = 1 + √ 2 . By Theorem 2.5, we obtain the recurrence f (2 , k ) = 2 f (2 , k − 1) + f (2 , k − 2) , k ≥ 2 , with initial v alues f (2 , 1) = 3 and f (2 , 2) = 7. 7 Lemma 4.1. f (2 , k ) = (1 − √ 2) k +1 2 + (1 + √ 2) k +1 2 . Pro of. Assume that f (2 , k ) = C 1 (1 − √ 2) k + C 2 (1 + √ 2) k . Using the initial v alues f (2 , 1) = 3 8 and f (2 , 2) = 7, w e obtain: C 1 = 1 − √ 2 2 and C 2 = 1+ √ 2 2 . Substituting these co efficients gives 9 the stated formula. 10 One may note that f (2 , k ) = β ( k + 1), where β ( k ) denotes the k -th Pell-Lucas n um b er[7]. 11 Theorem 4.2. N ( K 2 × P n ) = f (2 , n + 2) − 4 n − 7 2 . Pro of. By Theorem 2.1, N ( K 2 × P n ) = n X k =1 ( n + 1 − k ) f (2 , k ) = ( n + 1) n X k =1 f (2 , k ) − n X k =1 k f (2 , k ) . 13 Applying formulas given in [7], w e hav e 1 n X k =1 f (2 , k ) = f (2 , n + 1) + f (2 , n ) − 4 2 . (4 . 1) Next we compute n P k =1 k f (2 , k ). By Lemma 4.1, n X k =1 k f (2 , k ) = n X k =1 k (1 − √ 2) k +1 2 + n X k =1 k (1 + √ 2) k +1 2 (4 . 2) Since n P k =1 k x k +1 = x 2 ( n P k =1 x k ) ′ and ( n P k =1 x k ) ′ = 1+ nx n +1 − ( n +1) x n (1 − x ) 2 , the right part of the equa- tion (4.2) can b e transfer to (1 − √ 2) 2 + n (1 − √ 2) n +3 − ( n + 1)(1 − √ 2) n +2 4 + (1 + √ 2) 2 + n (1 + √ 2) n +3 − ( n + 1)(1 + √ 2) n +2 4 . After simplification this equals 2 nf (2 , n + 2) − (2 n + 2) f (2 , n + 1) + 6 4 , whic h gives n X k =1 k f (2 , k ) = nf (2 , n + 2) − ( n + 1) f (2 , n + 1) + 3 2 . (4 . 3) No w substitute (4.1) and (4.3) into the expression for N ( K 2 × P n ): 2 N ( K 2 × P n ) = ( n + 1) f (2 , n + 1) + ( n + 1) f (2 , n ) − 4( n + 1) 2 − nf (2 , n + 2) − ( n + 1) f (2 , n + 1) + 3 2 = (2 n + 2) f (2 , n + 1) + ( n + 1) f (2 , n ) f (2 , n ) − nf (2 , n + 2) − 4 n − 7 2 = f (2 , n + 2) − 4 n − 7 2 , where the last equality follo ws from the recurrence f (2 , n + 2) = 2 f (2 , n + 1) + f (2 , n ). 3 W e now turn to A ( K 2 × P n ). By Theorem 3.1, this requires the expression of S ( F (2 , k )). 4 By Theorem 3.5, S ( F (2 , k )) = 2 k f (2 , k ) − 2 k X i =1 f 1 (2 , i ) f 1 (2 , k + 1 − i ) . 14 Lemma 4.3. f 1 (2 , k ) = (1 + √ 2) k 2 √ 2 − (1 − √ 2) k 2 √ 2 . Pro of. Assume that f 1 (2 , k ) = C 3 (1 − √ 2) k + C 4 (1 + √ 2) k . Multiplying A 2 and A 2 2 b y 1 [1 , 1] T giv es the initial v alues f 1 (2 , 1) = 1 and f 1 (2 , 2) = 2. Solving for the co efficients yields: 2 C 3 = − 1 2 √ 2 and C 4 = 1 2 √ 2 . 3 Lemma 4.3 shows that f 1 (2 , k ) = ¯ β ( k ), where ¯ β ( k ) is the k -th Pell n um b er[7]. F rom [7, 4 p.212], we know that k P i =1 f 1 (2 , i ) f 1 (2 , k + 1 − i ) = ( k +2) f (2 ,k ) − f 1 (2 ,k +2) 4 . Therefore 5 S ( F (2 , k )) = 2 k f (2 , k ) − ( k + 2) f (2 , k ) − f 1 (2 , k + 2) 4 × 2 = (3 k − 2) f (2 , k ) + f 1 (2 , k + 2) 2 . Th us n X k =1 ( n − k + 1) S ( F ( k )) = n X k =1 ( n − k + 1) (3 k − 2) f (2 , k ) + f 1 (2 , k + 2) 2 . Using the identities given in [7], we hav e 6 n X k =1 f 1 (2 , k + 2) = f (2 , n + 3) − 7 2 . F ollowing the same method as in Theorem 4.2, we also obtain 7 n X k =1 k f 1 (2 , k + 2) = 1 2 [2( n − 1) f 1 (2 , n + 2) + (3 n − 1) f 1 (2 , n + 1) + nf 1 (2 , n ) + 5] , n X k =1 k 2 f (2 , k ) = 1 2 [(2 n 2 + 2 n + 1) f 1 (2 , n + 2) + (1 − 2 n ) f 1 (2 , n + 3) − 7] . F rom these w e derive 8 n X k =1 ( n − k + 1) S ( F ( k )) = (21 n − 44) f (2 , n ) + ( − 12 n + 26) f 1 (2 , n ) + 17 n + 44 4 + 12 f (2 , n ) − 7 f 1 (2 , n ) − 7 n − 12 4 = (21 n − 32) f (2 , n ) + ( − 12 n + 19) f 1 (2 , n ) + 10 n + 32 4 . 15 Com bining this with Theorem 4.2 gives the following closed-form expressions. 1 Theorem 4.4. A ( K 2 × P n ) = (21 n − 32) f (2 , n ) + ( − 12 n + 19) f 1 (2 , n ) + 10 n + 32 2 f (2 , n + 2) − 4 n − 7 . Theorem 4.5. D ( K 2 × P n ) = (21 n − 32) f (2 , n ) + ( − 12 n + 19) f 1 (2 , n ) + 10 n + 32 2 n [2 f (2 , n + 2) − 4 n − 7] . Remark 4.6. Sinc e f (2 , n ) = f (2 , n − 1) + 2 f 1 (2 , n ) , (21 n − 32) f (2 , n ) + ( − 12 n + 19) f 1 (2 , n ) + 10 n + 32 2 f (2 , n + 2) − 4 n − 7 = [32 − 45 f 1 (2 , n ) − 32 f (2 , n − 1)] + n [10 + 21 f (2 , n − 1) + 30 f 1 (2 , n )] 4 , which shows that our formulas c oincide with those obtaine d by Vinc e in [14]. 2 5 F urther w orks. 3 W e hav e discussed the num b er and av erage order of connected sets in K m × P n . A natural 4 follo w-up question is whether analogous formulas can b e obtained for K m × C n . Sp ecifically: 5 Question 1. Find a formula for N ( K m × C n ). 6 Question 2. Find a formula for A ( K m × C n ). 7 Exploring these questions w ould extend the results presented here and provide a more com- 8 plete picture of connected sets in pro ducts of complete graphs with simple linear structures. 9 References 10 [1] K. J. Balodis, L. Mol, O. R. Oellermann, M. E. Kro eker, On the mean order of connected 11 induced subgraphs of blo ck graphs, Australas. J. 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