On Jacobian group and complexity of the generalized Petersen graph GP(n,k) through Chebyshev polynomials
aa r X i v : . [ m a t h . C O ] D ec On Jacobian group and complexity of the generalized Petersengraph GP ( n, k ) through Chebyshev polynomials Y. S. Kwon, A. D. Mednykh, I. A. Mednykh, Abstract
In the present paper we find a simple algorithm for counting Jacobian group of the gener-alized Petersen graph GP ( n, k ) . Also,we obtain a closed formula for the number of spanningtrees of this graph in terms of Chebyshev polynomials.
Key Words: spanning tree, Jacobian group, Petersen graph, Chebyshev polynomial
AMS Mathematics Subject Classification:
The notion of the Jacobian group of a graph, which is also known as the Picard group, thecritical group, and the dollar or sandpile group, was independently introduced by many authors([1], [2], [3], [4]). This notion arises as a discrete version of the Jacobian in the classical theoryof Riemann surfaces. It also admits a natural interpretation in various areas of physics, codingtheory, and financial mathematics. The Jacobian group is an important algebraic invariant of afinite graph. In particular, its order coincides with the number of spanning trees of the graph,which is known for some simplest graphs, such as the wheel, fan, prism, ladder, and M¨obiusladder [5], grids [14], lattices [16], Sierpinski gaskets [6, 7], 3-prism and 3-anti-prism [17]. At thesame time, the structure of the Jacobian is known only in particular cases [1], [3], [10], [19], [20],[21] and [22]. We mention that the number of spanning trees for circulant graphs is expressed isterms of the Chebyshev polynomials; it was found in [11], [12], and [13]. We show that similarresults are also true for the generalized Petersen graph GP ( n, k ) . The generalized Petersen graph GP ( n, k ) has vertex set and edge set given by V ( P ( n, k )) = { u i , v i | i = 1 , , . . . , n } E ( P ( n, k )) = { u i u i +1 , u i v i , v i v i + k | i = 1 , , . . . , n } , where the subscripts are expressed as integers modulo n . The classical Petersen graph is P (5 , GP ( n, k ) by their result. In this paper we find a closed formulafor the number of spanning trees for GP ( n, k ) through Chebyshev polynomials. Also, we suggestan effective algorithm for calculating Jacobian of GP ( n, k ) . Consider a connected finite graph G, allowed to have multiple edges but without loops. Weendow each edge of G with the two possible directions. Since G has no loops, this operation Department of Mathematics, Yeungnam University, Korea Sobolev Institute of Mathematics, Novosibirsk State University, Siberian Federal University Sobolev Institute of Mathematics, Novosibirsk State University, Siberian Federal University
1s well defined. Let O = O ( G ) be the set of directed edges of G. Given e ∈ O ( G ) , we denoteits initial and terminal vertices by s ( e ) and t ( e ) , respectively. Recall that a closed directedpath in G is a sequence of directed edges e i ∈ O ( G ) , i = 1 , . . . , n such that t ( e i ) = s ( e i +1 ) for i = 1 , . . . , n − t ( e n ) = s ( e ) . Following [2] and [4], the Jacobian group , or simply
Jacobian
J ac ( G ) of a graph G is definedas the (maximal) abelian group generated by flows ω ( e ) , e ∈ O ( G ) , obeying the following twoKirchhoff laws: K : the flow through each vertex of G vanishes, that is P e ∈ O,t ( e )= x ω ( e ) = 0 for all x ∈ V ( G ); K : the flow along each closed directed path W in G vanishes, that is P e ∈ W ω ( e ) = 0 . Equivalent definitions of the group
J ac ( G ) can be found in papers [1], [2], [3], [4], [10], [15], [18].We denote the vertex and edge set of G by V ( G ) and E ( G ) , respectively. Given u, v ∈ V ( G ) , we set a uv to be equal to the number of edges between vertices u and v. The matrix A = A ( G ) = { a uv } u,v ∈ V ( G ) , called the adjacency matrix of the graph G. The degree d ( v ) of a vertex v ∈ V ( G )is defined by d ( v ) = P u a uv . Let D = D ( G ) be the diagonal matrix indexed by the elementsof V ( G ) with d vv = d ( v ) . Matrix L = L ( G ) = D ( G ) − A ( G ) is called the Laplacian matrix , orsimply Laplacian , of the graph G. Recall [10] the following useful relation between the structure of the Laplacian matrix andthe Jacobian of a graph G. Consider the Laplacian L ( G ) as a homomorphism Z | V | → Z | V | , where | V | = | V ( G ) | is the number of vertices in G. The cokernel coker ( L ( G )) = Z | V | / im ( L ( G )) — isan abelian group. Let coker ( L ( G )) ∼ = Z d ⊕ Z d ⊕ · · · ⊕ Z d | V | be its Smith normal form satisfying the conditions d i (cid:12)(cid:12) d i +1 , (1 ≤ i ≤ | V | ) . If the graph is con-nected, then the groups Z d , Z d , . . . , Z d | V |− — are finite, and Z d | V | = Z . In this case,
J ac ( G ) ∼ = Z t ⊕ Z t ⊕ · · · ⊕ Z d | V |− is the Jacobian of the graph G. In other words,
J ac ( G ) is isomorphic to the torsion subgroup ofthe cokernel coker ( L ( G )) . Let M be an integer n × n matrix, then we can interpret M as a homomorphism from Z n to Z n . In this interpretation M has a kernel ker M, an image im M, and a cokernel coker M = Z n / im M. We emphasize that coker M of the matrix M coincides with its Smith normal form.In what follows, by I n we denote the identity matrix of order n. We call an n × n matrix circulant, and denote it by circ ( a , a , . . . , a n ) if it is of the form circ ( a , a , . . . , a n − ) = a a a . . . a n − a n − a a . . . a n − ... . . . ... a a a . . . a . Recall [23] that the eigenvalues of matrix C = circ ( a , a , . . . , a n − ) are given by the followingsimple formulas λ j = p ( ε jn ) , where p ( x ) = a + a x + . . . + a n − x n − and ε n is the order n primitiveroot of the unity. Moreover, the circulant matrix C = p ( T ) , where T = circ (0 , , , . . . ,
0) is thematrix representation of the shift operator T : ( x , x , . . . , x n − , x n − ) → ( x , x , . . . , x n − , x ) . Also, we note that all circulant n × n matrices share the same set of eigenvectors.2y ([8], lemma 2.1) the 2 n × n adjacency matrix of the generalized Petersen graph GP ( n, k )has the following block form A ( GP ( n, k )) = (cid:18) C kn I n I n C n (cid:19) , where C kn is the n × n circulant matrix of the form C kn = circ (0 , . . . , | {z } k times , , , . . . , , , , . . . , | {z } k − ) . Denote by L = L ( GP ( n, k )) the Laplacian of GP ( n, k ) . Since the graph GP ( n, k ) is three-valent, we have L = 3 I n − A ( GP ( n, k )) = (cid:18) I n − C kn − I n − I n I n − C n (cid:19) . Let P ( z ) be a bimonic integer Laurent polynomial. That is P ( z ) = z p + a z p +1 + . . . + a s − z p + s − + z p + s for some integers p, a , a , . . . , a s − and some positive integer s. Introduce the followingcompanion matrix A for the polynomial P ( z ) : A = (cid:18) I s − − , − a , . . . , − a s − (cid:19) , where I s − isthe identity ( s − × ( s −
1) matrix. We will use the following properties of A . Note thatdet A = ( − s − . Hence A is invertible and inverse matrix A − is also integer matrix. Thecharacteristic polynomial of A coincides with z − p P ( z ) . Let R be a nonzero commutative ring (integral domain). In most cases, we deal with the case R = Z . Denote by R Z the set of bi-infinite sequences ( x j ) j ∈ Z = ( . . . , x − , x , x , x , . . . ) where x j ∈ R for all j ∈ Z . This set is naturally endowed by the structure of Z − module. Define theshift operator T : R Z → R Z by the formula T ( x j ) j ∈ Z = ( x j +1 ) j ∈ Z . For an arbitrary integer ℓ wehave T ℓ ( x j ) j ∈ Z = ( x j + ℓ ) j ∈ Z . Simplifying notation we will write T ℓ x j = x j + ℓ . We set to be theidentity operator in R Z . For the sake of simplicity for any integer n , we will write n instead of n . We will use the following notation for the infinitely generated abelian group. Let A ξ , ξ ∈ Ξ , bea family of Z − linear operators in the space R Z . Then by h x | A ξ x = 0 , ξ ∈ Ξ i , where x = ( x j ) j ∈ Z we denote the abelian group generated by x j , j ∈ Z satisfying the set of relations A ξ x = 0 , ξ ∈ Ξ . We will use the following lemma.
Lemma 3.1
Let T : R Z → R Z be the shift operator and R = Z . Consider two operators A and B given by the formulas A = P ( T ) , B = Q ( T ) , where P ( z ) and Q ( z ) are Laurent polynomialswith integer coefficients. Then h x | Ax = 0 , Bx = 0 i ∼ = coker A/ im ( B | coker A ) ∼ = coker coker A B. Proof.
Consider R Z as the abelian group Z ∞ of all bi-infnite integer sequences provided withthe natural addition. Then A and B can be considered as endomorphisms of Z ∞ . Their imagesim A and im B are subgroups in Z ∞ . Denote by h im A, im B i the subgroup generated by elementsof im A and im B. Since P ( z ) and Q ( z ) are Laurent polynomials the operators A = P ( T ) and B = Q ( T ) do commute. Hence, subgroup im A is invariant under endomorphism B. Indeed, let y ∈ im A then By = B ( Ax ) = A ( Bx ) ∈ im A. The later means that B : Z ∞ → Z ∞ induces anendomorphism of the group coker A = Z ∞ / im A. We denote this endomorphism by B | coker A . We3ote that the abelian group h x | Ax = 0 , Bx = 0 i is naturally isomorphic to Z ∞ / h im A, im B i . Sowe have Z ∞ / h im A, im B i ∼ = ( Z ∞ / im A ) / im ( B | coker A ) ∼ = coker A/ im( B | coker A ) ∼ = coker coker A B. The proof of the lemma is finished. GP ( n, k ) Theorem 4.1
Let L = L ( GP ( n, k )) be the Laplacian of the generalized Petersen graph GP ( n, k ) . Then coker L ∼ = coker ( A n − I ) , where A is k + 1) × k + 1) companion matrix for the Laurent polynomial (3 − z − − z )(3 − z − k − z k ) − . Proof.
Let L be the Laplacian matrix of the graph GP ( n, k ) . Then, as it was mentionedabove, L is a 2 n × n matrix of the form L = (cid:18) I n − C kn − I n − I n I n − C n (cid:19) , where C kn = circ (0 , . . . , | {z } k times , , , . . . , , , , . . . , | {z } k-1 times ) . Consider L as a Z − linear operator L : Z n → Z n . In this case, coker( L ) is the abeliangroup generated by elements x , x , . . . , x n , y , y , . . . , y n satisfying the linear system of equations L ( x , x , . . . , x n , y , y , . . . , y n ) t = 0 . By the property mentioned in Section 2, the Jacobian ofthe graph GP ( n, k ) is isomorphic to the finite part of cokernel of the operator L . So, it sufficesto show that cokernels of operators L : Z n → Z n and A n − I s k : Z s k → Z s k are isomorphic.To study the structure of coker( L ) we consider two bi-infinite sequences of elements ( x j ) j ∈ Z =( . . . , x − , x , x , x , . . . ) and ( y j ) j ∈ Z = ( . . . , y − , y , y , y , . . . ) . By circularity of the n × n blocksof matrix L we have the following representation for cokernel of L :coker( L ) = h x i , y i , i ∈ Z | x j − x j − k − x j + k − y j = 0 , y j − y j − − y j +1 − x j = 0 , x j + n = x j , y j + n = y j , j ∈ Z i . Consider the operator L ( T ) : R Z → R Z defined by L ( T ) = (3 − T − T − )(3 − T k − T − k ) − . Then by making use of the operator notation we can rewrite the cokernel of L in the followingway coker( L ) = h x, y | (3 − T k − T − k ) x = y, (3 − T − T − ) y = x, T n x = x, T n y = y i = h x | (3 − T − T − )(3 − T k − T − k ) x = x, T n x = x i = h x | ((3 − T − T − )(3 − T k − T − k ) − x = 0 , ( T n − x = 0 i = h x | L ( T ) x = 0 , ( T n − x = 0 i .
4o finish the proof, we apply Lemma 3.1 to the operators A = L ( T ) and B = Q ( T ) = T n − . By definition coker A is generated by the elements x j = e j + im A, j ∈ Z . Since the Laurentpolynomial P ( z ) = (3 − z − z − )(3 − z k − z − k ) − P ( z ) = z − k − + a z − k + . . . + a k +1 z k + z k +1 , where a , a , . . . , a k +1 are integers. Thenthe companion matrix A is (cid:18) I k +1 − , − a , . . . , − a k +1 (cid:19) . It is easy to see that det A = ( − k − and its inverse A − is also integer matrix.For convenience we set s = 2 k + 2 to be the size of matrix A . Let Z s be an abelian group gen-erated by the elements x , x , . . . , x s . Note that for any integer j ∈ Z , ( x j +1 , x j +2 , . . . , x j + s ) t = A j ( x , x , . . . , x s ) t , which implies that each element x j , j ∈ Z can be uniquely expressed as aninteger linear combination of the elements x , x , . . . , x s .Our present aim is to show that coker A ∼ = Z s . Then we describe the action of the endomor-phism B | coker A on the coker A . Setting x = ( x j ) j ∈ Z we can write coker A = h x | Ax = 0 i . So wehave the following representation of coker A .coker A = h x | Ax = 0 i == h x j , j ∈ Z | x ℓ + a x ℓ +1 + . . . + a s − x ℓ + s − + x ℓ + s = 0 , ℓ ∈ Z i = h x j , j ∈ Z | ( x ℓ +1 , x ℓ +2 , . . . , x ℓ + s ) t = A ( x ℓ , x ℓ +1 , . . . , x ℓ + s − ) t , ℓ ∈ Z i = h x j , j ∈ Z | ( x ℓ +1 , x ℓ +2 , . . . , x ℓ + s ) t = A ℓ ( x , x , . . . , x s ) t , ℓ ∈ Z i = h x , x , . . . , x s |∅i ∼ = Z s . Since the operators A = L ( T ) and T commute, the action T | coker A : x j → x j +1 , j ∈ Z onthe coker A is well defined. Now we describe the action of T | coker A on the set of generators x , x , . . . , x s . For any i = 1 , . . . , s −
1, we have T | coker ( x i ) = x i +1 and T | coker A ( x s ) = x s +1 = − x − a x − . . . − a s − x s − − a s − x s . Hence, the action of T | coker A on the coker A is given by thematrix A . Considering A as an endomorphism of the coker A, we can write T | coker A = A . Finally, B | coker A = Q ( T | coker A ) = Q ( A ) . Applying Lemma 3.1 we finish the proof of the theorem.
Corollary 4.2
Sandpile group Jac ( GP ( n, k )) of the generalized Petersen graph GP ( n, k ) is iso-morphic to the torsion subgroup of coker ( A n − I ) , where A is the companion matrix for theLaurent polynomial (3 − z − − z )(3 − z − k − z k ) − . The Corollary 4.2 gives a simple way to count Jacobian group Jac( GP ( n, k )) for small valuesof k and sufficiently large numbers n. The numerical results are presented in Tables 1 , . GP ( n, k ) Theorem 5.1
The number of spanning trees of the generalized Petersen graph GP ( n, k ) is givenby the formula τ ( GP ( n, k )) = ( − ( n − k − n k Y s =1 T n ( w s ) − w s − , where w s , s = 1 , , . . . , k are roots of the order k algebraic equation T k ( w ) − T k ( w ) − w − − , and T k ( w ) is the Chebyshev polynomial of the first kind. roof. By the celebrated Kirchhoff theorem, the number of spanning trees τ k ( n ) is equalto the product of nonzero eigenvalues of the Laplacian of a graph GP ( n, k ) divided by thenumber of its vertices 2 n. To investigate the spectrum of Laplacian matrix we note that matrix C kn = T − k + T k , where T = circ (0 , , . . . ,
0) is the n × n shift operator. The latter equality easilyfollows from the identity T n = I n . Hence, L = (cid:18) I n − T − k − T k − I n − I n I n − T − − T (cid:19) . The eigenvalues of circulant matrix T are ε jn , where ε n = e πin . Since all eigenvalues of T are distinct, the matrix T is conjugate to the diagonal matrix T = diag (1 , ε n , . . . , ε n − n ), wherediagonal entries of diag (1 , ε n , . . . , ε n − n ) are 1 , ε n , . . . , ε n − n . To find spectrum of L, without lossof generality, one can assume that T = T . Then the n × n blocks of L are diagonal matrices. Thisessentially simplifies the problem of finding eigenvalues of L. Indeed, let λ be an eigenvalue of L and ( x, y ) = ( x , . . . , x n , y , . . . , y n ) be the respective eigenvector. Then we have the followingsystem of equations (cid:26) (3 I n − T − k − T k ) x − y = λx − x + (3 I n − T − − T ) y = λy . From here we conclude that y = (3 I n − T − k − T k ) x − λx = (3 − λ − T − k − T k ) − x .Substituting y in the second equation, we have ((3 − λ − T − − T )(3 − λ − T − k − T k ) − x = 0.Recall the matrices under consideration are diagonal and the ( j + 1 , j + 1)-th entry of T is equal to ε jn . Therefore, we have ((3 − λ − ε − jn − ε jn )(3 − λ − ε − jkn − ε jkn ) − x j +1 = 0 and y j +1 = (3 − λ − ε − jkn − ε jkn ) x j +1 . So, for any j = 0 , . . . , n − L has two eigenvalues, say λ ,j and λ ,j satisfying thequadratic equation (3 − λ − ε − jn − ε jn )(3 − λ − ε − jkn − ε jkn ) − . The corresponding eigenvectorsare ( x, y ) , where x = e j +1 = (0 , . . . , |{z} ( j +1) − th , . . . ,
0) and y = (3 − λ − T − k − T k ) e j +1 . In particular,if j = 0 for λ , , λ , we have (1 − λ )(1 − λ ) − λ ( λ −
2) = 0 . That is, λ , = 0 and λ , = 2 . Since λ ,j and λ ,j are roots of the same quadratic equation, we obtain λ ,j λ ,j = P ( ε jn ) , where P ( z ) = (3 − z − − z )(3 − z − k − z k ) − . Now we have τ k ( n ) = 12 n λ , n − Y j =1 λ ,j λ ,j = 1 n n − Y j =1 λ ,j λ ,j = 1 n n − Y j =1 P ( ε jn ) . Lemma 5.2
The following identity holds (3 − z − − z )(3 − z − k − z k ) − w − T k ( w ) − T k ( w ) − w − − , where T k ( w ) is the Chebyshev polynomial of the first kind and w = ( z − + z ) . Proof.
Let us substitute z = e iϕ . It is easy to see that w = ( z − + z ) = cos ϕ, so wehave T k ( w ) = cos( k arccos w ) = cos( kϕ ). Then the statement of the lemma is equivalent to thefollowing elementary identity(3 − ϕ )(3 − kϕ )) − ϕ − kϕ ) − cos( kϕ ) − ϕ − − P ( z ) = 2( w − h k ( w ) , where w = ( z + z − ) and h k ( w ) = 2 T k ( w ) − ( T k ( w ) − / ( w − − k . Note that 2( w −
1) = ( z − z . Since h k (1) =2 T k (1) − T ′ k (1) − − − k = 0 , the Laurent polynomial P ( z ) has the root z = 1 with multiplicitytwo. Hence, the roots of P ( z ) are 1 , , z , /z , . . . , z k , /z k , where for all s = 1 , . . . , k, z s = 1 and w s = ( z s + z − s ) is a root of equation h k ( w ) = 0 . We set H ( z ) = k Q s =1 ( z − z s )( z − z − s ) . Then P ( z ) = z k +1 ( z − H ( z ) . Lemma 5.3
Let H ( z ) = k Q s =1 ( z − z s )( z − z − s ) and H (1) = 0 . Then n − Y j =1 H ( ε jn ) = k Y s =1 T n ( w s ) − w s − , where w s = ( z s + z − s ) , s = 1 , . . . , k and T n ( x ) is the Chebyshev polynomial of the first kind. Proof.
It is easy to check that n − Q j =1 ( z − ε jn ) = z n − z − if z = 1 . Also we note that ( z n + z − n ) = T n ( ( z + z − )) . By the substitution z = e i ϕ the latter follows from the evident identitycos( nϕ ) = T n (cos ϕ ) . Then we have n − Y j =1 H ( ε jn ) = n − Y j =1 k Y s =1 ( ε jn − z s )( ε jn − z − s )= k Y s =1 n − Y j =1 ( z s − ε jn )( z − s − ε jn )= k Y s =1 z ns − z s − z − ns − z − s − k Y s =1 T n ( w s ) − w s − . Note that n − Q j =1 (1 − ε jn ) = lim z → n − Q j =1 ( z − ε jn ) = lim z → z n − z − = n and n − Q j =1 ε jn = ( − n − . As a result,taking into account Lemma 5.3, we obtain τ k ( n ) = 1 n n − Y j =1 P ( ε jn ) = 1 n n − Y j =1 ( ε jn − ( ε jn ) k +1 H ( ε jn ) = ( − ( n − k +1) n n n − Y j =1 H ( ε jn )= ( − ( n − k − n k Y s =1 T n ( w s ) − w s − . orollary 5.4 τ ( GP ( n, k )) = n (cid:12)(cid:12)(cid:12)Q ks =1 U n − ( q w s ) (cid:12)(cid:12)(cid:12) , where w s , s = 1 , , . . . , k are the sameas in Theorem 5.1 and U n − ( w ) is the Chebyshev polynomial of the second kind. Proof.
Follows from the identity T n ( w ) − w − = U n − ( q w ) . GP ( n, . As a first consequence of theorem 5.1 we have the following formula for number of spanning treesof the n -prism graph GP ( n,
1) : τ ( n ) = n ( T n (2) − . This formula is well known and was independently obtained by many authors. For example,by J. Sedl´acˇek, J.W. Moon, N. Biggs and others ([5]). GP ( n, . Theorem 6.1
The number τ ( n ) of the spanning trees for the generalized Petersen graph GP ( n, is equal to ( − n n ( α − β ) , where the integers α and β are given by the equality: T n ( √ ) − α + β √ . Moreover, τ ( n ) = na ( n ) , where the integer sequence a ( n ) satisfies the following recursiverelation a ( n + 4) = a ( n + 3) + 3 a ( n + 2) − a ( n + 1) − a ( n ) ,a (0) = 0 , a (1) = 1 , a (2) = 1 , a (3) = 5 . Note a ( n ) is A192422 sequence in the On - Line Encyclopaedia of Integer Sequences. Proof.
Note that 2 T ( w ) − T ( w ) − w − − w − w −
7, and hence two roots of the equation2 T ( w ) − T ( w ) − w − − w = √ and w = −√ . By Theorem 5.1, we obtain τ ( n ) = ( − n − n Y s =1 T n ( w s ) − w s −
1= ( − n − n ( α + β √ α − β √ w − w −
1) = ( − n n ( α − β )20 . To prove the second statement, by Corollary 5.4, we have a ( n ) = | p ( n ) | , where p ( n ) = U n − ( θ ) U n − ( θ ) , θ = q w and θ = q w . Recall that Chebyshev polynomial U n ( θ ) satisfies the recursive relation U n +1 ( θ ) − θU n ( θ ) + U n − ( θ ) = 0 with initial data U ( θ ) = 1 and U ( θ ) = 2 θ. To find the recursive relation for thesequence p ( n ), we will use the following lemma.8 emma 6.2 Let P ( z ) and Q ( z ) are polynomials without multiple roots and let T u ( n ) = u ( n +1) be the shift operator. Suppose the sequences u ( n ) and v ( n ) satisfy the recursive relations P ( T ) u ( n ) = 0 and Q ( T ) v ( n ) = 0 respectively. Then the sequence p ( n ) = u ( n ) v ( n ) satisfies therecursive relations R ( T ) p ( n ) = 0 , where R ( z ) is the resultant of polynomials P ( ξ ) and ξ deg Q Q ( zξ ) with respect to ξ and deg Q is degree of Q ( z ) . Proof.
Let λ , λ , . . . , λ s and µ , µ , . . . , µ t be distinct roots of polynomials P ( z ) and Q ( z )respectively. Then each solution u ( n ) of the recursive equation P ( T ) u ( n ) = 0 is a linear combi-nation of the functions λ n , λ n , . . . , λ ns , while solution v ( n ) of the recursive equation Q ( T ) v ( n ) = 0is a linear combination of µ n , µ n , . . . , µ nt . Hence, their product p ( n ) is a linear combination of thefunctions λ nj µ nk , j = 1 , , . . . , s, k = 1 , , . . . , t. By definition of resultant, we have R ( λ j µ k ) = 0for all j, k and the proof of the lemma follows.Now we apply Lemma 6.2 to the polynomials P ( z ) = z − θ z + 1 and Q ( z ) = z − θ z + 1.Now the resultant of polynomials P ( ξ ) and ξ Q ( zξ ) is (cid:16) θ + p θ − − ( θ + p θ − z (cid:17) (cid:16) θ + p θ − − ( θ − p θ − z (cid:17) × (cid:16) θ − p θ − − ( θ + p θ − z (cid:17) (cid:16) θ − p θ − − ( θ − p θ − z (cid:17) . If we expand the equation, we get the equation z − iz + 3 z − iz + 1. Hence we have thefollowing recursive relation for p ( n ) = U n − ( θ ) U n − ( θ ) : p ( n ) − i p ( n + 1) + 3 p ( n + 2) − i p ( n + 3) + p ( n + 4) = 0 . Observing that a ( n ) = | p ( n ) | = ( − i ) n − p ( n ) and the initial values a (0) = 0 , a (1) = 1 , a (2) =1 , a (3) = 5, we get the result. GP ( n, . Theorem 6.3
The number τ ( n ) of the spanning trees for the generalized Petersen graph GP ( n, is given by the formula τ ( n ) = n Y s =1 T n ( w s ) − w s − , and w , w , w are roots of the equation w − w − w − . Moreover, τ (2 n ) = 12 n a ( n ) and τ (2 n + 1) = (2 n + 1) b ( n ) , where the integer sequences a ( n ) and b ( n ) satisfy the recursive relation u ( n ) − u ( n + 1) − u ( n + 2) − u ( n + 3) + 65 u ( n + 4) − u ( n + 5) − u ( n + 6) − u ( n + 7) + u ( n + 8) = 0 with the following initial data a (0) = 0 , a (1) = 1 , a (2) = 4 , a (3) = 9 , a (4) = 72 ,a (5) = 320 , u (6) = 1332 , a (7) = 68899 nd b (0) = 1 , b (1) = 1 , b (2) = 20 , b (3) = 83 , b (4) = 289 ,b (5) = 1693 , b (6) = 7775 , b (7) = 34820 . Proof.
The first statement of the theorem directly follows from Theorem 5.1. To provethe second, by Corollary 5.4, we have τ ( n ) = n | c ( n ) | , where c ( n ) = Q s =1 U n − ( θ s ) and θ s = q w s , s = 1 , , . Then the sequence u ( n ) = U n − ( θ s ) satisfies the following recursive relation P s ( T ) u ( n ) = 0 , where P s ( z ) = z − θ s z + 1 . Suppose that λ, µ, ν are roots of the equations P ( λ ) = 0 , P ( µ ) = 0 , P ( ν ) = 0 . By applying Lemma 6.2 twice, we obtain that η = λ µ ν is aroot of the equation 1 + η + 11 η + η + η = √ η (1 + 2 η + 2 η + η ) . Denote by η , η , . . . , η distinct roots of the previous equation. Then the sequence c ( n ) = U n − ( θ ) U n − ( θ ) U n − ( θ ) can be written in the form c ( n ) = P j =1 c j η nj , where c , c , . . . , c aresuitable constants. Let ζ j = η j . Now c (2 n ) = P j =1 c j ζ nj and c (2 n + 1) = P j =1 ( c j η j ) ζ nj . Since(1 + η + 11 η + η + η ) = 6 η (1 + 2 η + 2 η + η ) , all ζ j , j = 1 , , . . . , ζ + 11 ζ + ζ + ζ ) − ζ (1 + 2 ζ + 2 ζ + ζ ) = 0 . Hence, 1 − ζ − ζ − ζ + 65 ζ − ζ − ζ − ζ + ζ = 0and both sequences c (2 n ) and c (2 n + 1) are solution of the difference equation u ( n ) − u ( n + 1) − u ( n + 2) − u ( n + 3) + 65 u ( n + 4) − u ( n + 5) − u ( n + 6) − u ( n + 7) + u ( n + 8) = 0 . We use the formulas a ( n ) = c (2 n ) / √ b ( n ) = c (2 n + 1) to calculate the initial elements ofsequences a ( n ) and b ( n ) directly. GP ( n, . Theorem 6.4
The number τ ( n ) of the spanning trees for the generalized Petersen graph GP ( n, is given by the formula τ ( n ) = ( − n − n Y s =1 T n ( w s ) − w s − , and w , w , w , w are roots of the equation w − w − w − . Moreover, τ ( n ) = n a ( n ) , where the integer sequences a ( n ) satisfies the recursive relation P ( T ) a ( n ) = 0 , where ( T ) = T − T − T − T + 10 T − T + 16 T + 50 T − T − T − T − T + 2 T + T + 1 and T a ( n ) = a ( n + 1) is the shift operator.The initial data of a ( n ) for n equal to − , − , − , − , − , − , − , , , , , , , , , are, respectively, − , , − , , − , , − , , , , , , , , , . Proof.
The first statement of the theorem directly follows from Theorem 5.1. ApplyingLemma 6.2 several times and using the same arguments as in the proof of Theorem 6.1 weconclude that the sequence a ( n ) = | p ( n ) | = ( − i ) n − p ( n ) , where p ( n ) = Q s =1 U n − ( q x s )satisfies the recursive relation P ( T ) a ( n ) = 0 . Theorem 4.1 is the first step to understand the structure of the Jacobian for GP ( n, k ) . Also, itgives a simple way for numerical calculations of Jac( GP ( n, k )) for small values of n and k. SeeTables 1 , , , . GP ( n, n Jac( GP ( n, τ ( n ) = | Jac( GP ( n, | Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z GP ( n, n Jac( GP ( n, τ ( n ) = | Jac( GP ( n, | Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z GP ( n, n Jac( GP ( n, τ ( n ) = | Jac( GP ( n, | Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z Z ⊕ Z ⊕ Z ⊕ Z Z ⊕ Z CKNOWLEDGMENTS
The first author was supported by the 2015 Yeungnam University Research Grant. The sec-ond and the third authors were supported by the Grant of the Russian Federation Govementat Siberian Federal University (grant no. 14.Y26.31.0006), by the Presidium of the RussianAcademy of Sciences (project no. 0314-2015-0011), and by the Russian Foundation for BasicResearch (projects no. 15-01-07906 and 16-31-00138).
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