Spectral properties of the Cayley Graphs of split metacyclic groups
Kashyap Rajeevsarathy, Siddhartha Sarkar, S. Lakshmivarahan, Pawan Kumar Aurora
aa r X i v : . [ m a t h . C O ] M a y SPECTRAL PROPERTIES OF THE CAYLEY GRAPHS OFSPLIT METACYCLIC GROUPS
K.RAJEEVSARATHY, S. SARKAR, S. LAKSHMIVARAHAN, AND P. K. AURORA
Abstract.
Let Γ(
G, S ) denote the Cayley graph of a group G with respect toa set S ⊂ G . In this paper, we analyze the spectral properties of the Cayleygraphs T m,n,k = Γ( Z m ⋉ k Z n , { ( ± , , (0 , ± } ), where m, n ≥ k m ≡ n ). We show that the adjacency matrix of T m,n,k , upto relabeling, isa block circulant matrix, and we also obtain an explicit description of theseblocks. By extending a result due to Walker-Mieghem to Hermitian matrices,we show that T m,n,k is not Ramanujan, when either m >
8, or n ≥ Introduction
For a finite group G and a subset S of G , let Γ( G, S ) denote the Cayley Graph of G with respect to the set S . In this paper, we analyze the spectral properties of acollection of connected undirected Cayley graphs of degree 4 arising from metacyclicgroups, namely the graphs {T m,n,k } , where T m,n,k = Γ( Z m ⋉ k Z n , { ( ± , , (0 , ± } ) , m, n ≥ k m ≡ n ) . Our analysis is mainly driven by the fact that the properties of a subcollection ofgraphs known as supertoroids , have been widely studied [2, 3, 6, 13] clasically, andmore recently, the diamater [9] and the degree-diameter problem of this family ofgraphs have also been analyzed (see [4, 7, 10]).Let A ( X ) denote the adjacency matrix of a graph X , let Θ denote the zeromatrix of appropriate dimension, and let Diag m ( p , p , . . . , p τ ) denote the diagonalmatrix ( a ij ) m × m defined by a ii = 1, if i ∈ { p , . . . , p τ } , and a ii = 0, otherwise. Byusing a special labeling scheme for the vertices of T m,n,k and applying tools fromelementary number theory, we obtain the main result of the paper. Theorem 1.
Let α be the order of k in Z × n , and let C m denote the m -cycle graph.For ≤ i, j ≤ n − , the matrix A ( T m,n,k ) = ( A ij ) is block-circulant with m × m blocks such that for each i , A ii = A ( C m ) and for each j = i , A ij is a diagonalmatrix with or t , or t nonzero entries given by the following conditions.(i) If k s
6≡ − n ) , for any s , then A ij = ( Diag( ξ, ξ + α, . . . , ξ + ( t − α ) , if j ≡ ± k α − ξ , for ≤ ξ ≤ α − , and Θ , otherwise.(ii) If α is even and k α/ ≡ − n ) , then A ij = ( Diag( ξ, ξ + α , . . . , ξ + (2 t − α ) , if j ≡ ± k α − ξ , for ≤ ξ ≤ ( α/ − , and Θ , otherwise. By appealing to the theory of circulant matrices [1], we show that the matrix A ( T m,n,k ) is conjugate with a block diagonal matrix M = ( M ij ) n × n with Hermitian Mathematics Subject Classification.
Primary 68R10; Secondary 05C50.
Key words and phrases.
Cayley Graphs, Ramanujan Graphs, Cyclic groups, Semi-directproduct. × m blocks, one of whose blocks has the eigenvalues 2 + 2 cos(2 πk/m ), 0 ≤ k ≤ m −
1. Extending a result due to Walker-Mieghem [11] that bounds thelargest eigenvalue of a symmetric matrix, to the case of Hermitian matrices (seeCorollary 4.6), we derive a lower bound on the largest eigenvalue of M . As aconsequence of this spectral analysis of A ( T m,n,k ), we determine lower bounds on m and n for the nonexistence of Ramanujan graphs [8] in the family {T m,n,k } . Corollary 1.
The graph T m,n,k is not Ramanujan when either m > , or n ≥ . Finally, using software written for Mathematica [12], we list the collection of allpairs in { ( m, n ) : 3 ≤ m ≤ ≤ n < } for which there exist at least one k > T m,n,k is Ramanujan.This paper is organized in the following manner. In Section 2, we provide acomplete analysis of the structure of A ( T m,n,k ) and the main result of this paper,while the spectral properties of T m,n,k are analyzed in Section 3.2. Adjacency matrix of T m,n,k In this section, we obtain a complete description of the structure of A ( T m,n,k ).For integers m, n ≥
3, let the group Z m ⋉ k Z n be given by the presentation Z m ⋉ k Z n = h x, y : x m = 1 = y n , x − yx = y k i , where k ∈ Z × n is of order α > k m ≡ n ). Clearly, m = tα , andwe will call t as the p eriod of k . Denoting the vertex set of a graph X by V ( X )and its edge set by E ( X ), we have the following. Definition 2.1.
For i ∈ Z n we define the i th packet V i of V ( T m,n,k ) as V i := { x τα + ξ y ik ξ : 0 ≤ τ ≤ t − ≤ ξ ≤ α − } . By definition, it is clear that | V i | = tα = m , and V ( T m,n,k ) = ⊔ n − i =0 V i . We willnow define an an ordering on each packet. Definition 2.2.
Given x τα + ξ y ik ξ , x τ ′ α + ξ ′ y ik ξ ′ ∈ V i , we define x τα + ξ y ik ξ < x τ ′ α + ξ ′ y ik ξ ′ if, and only if, either τ < τ ′ , or τ = τ ′ and ξ < ξ ′ . We will now define an ordering on the vertices across the V i . Definition 2.3.
Given v ∈ V i and v ∈ V j , for i = j , we define v < v if, and only if, i < j. The orderings in Definitions 2.2 and 2.3 together yield an ordering on V ( T m,n,k ).Hence, we have V ( T m,n,k ) = { v , v , . . . , v mn − } , where v i < v j , for i < j , and let A ( T m,n,k ) be the adjacency matrix associated with this labeling. It is apparentthat this ordering induces a block matrix structure on A ( T m,n,k ) A ( T m,n,k ) = A A · · · A n − A A · · · A n − ... ... . . . ... A n − A n − · · · A n − n − , where each block A ij is an m × m matrix that represents the adjacency betweenthe vertex packets V i and V j . Definition 2.4.
An edge ( v , v ) ∈ E ( T m,n,k ) is called an x -sibling pair (resp. y -sibling pair ) if v = v x ± (resp. v = v y ± ). Lemma 2.5.
Let ( v , v ) ∈ V i × V j , for i = j . Then:(i) ( v , v ) is an x -sibling pair if, and only if, i ≡ j (mod n ) , and ii) ( v , v ) is an y -sibling pair implies i j (mod n ) .Proof. Let v = x τα + ξ y ik ξ and v = x τ ′ α + ξ ′ y jk ξ ′ where 0 ≤ τ, τ ′ ≤ t − ≤ ξ, ξ ′ ≤ α −
1. Then v = v x ± if, and only if, τ ′ α + ξ ′ ≡ τ α + ξ ± m ) and jk ξ ′ ≡ ik ξ ± (mod n ) , and (i) follows.To show (ii), we assume on the contrary that i ≡ j (mod n ). Since by hypothesis jk ξ ′ ≡ ik ξ ± n ), our assumption would imply that 1 ≡ n ), which isimpossible. (cid:3) Corollary 2.6.
For ≤ i ≤ n − , the sub-blocks A ii of A ( T m,n,k ) is a circulantmatrix. Moreover, A ii = A jj , whenever i = j .Proof. From Lemma 2.5, it is apparent that the nonzero entries in A ( T m,n,k ) con-tributed by the edges in E ( T m,n,k ) forming x -sibling pairs appear in the diagonalblocks A ii , while those contributed by the y -sibling pairs appear in the off diagonalblocks A ij , for i = j . Given a vertex v = x τα + ξ y ik ξ ∈ V i , since vx = x τα + ξ +1 y ik ξ +1 ,the ordering defined on V ( T m,n,k ) makes vx an immediate successor of v , exceptwhen ( τ, α ) = ( t − , α − v and vx (resp.) represent the lastand first vertices (resp.) of V i (under the ordering). Hence, for each i , we have A ii = ( b rs ) m × m , where b rs = 1, if | r − s | >
1, and b rs = 0, otherwise. (cid:3) Each diagonal block A ii of A ( T m,n,k ) described in Corollary 2.6 is, in fact, theadjacency matrix A ( C m ) of the m -cycle graph C m . We will now turn our attentionto the A ij , for i = j . In this regard, it would be convenient to further partitioneach packet V i = ⊔ t − τ =0 V i,τ into ( i, τ ) -subpackets , where V i,τ := { x τα + ξ y ik ξ : 0 ≤ ξ ≤ α − } . This partition induces a block diagonal structure on each off-diagonalblock A ij of the form A ij = B ij B ij · · · B ij ,t − B ij B ij · · · B ij ,t − ... ... . . . ... B ijt − , B ijt − , · · · B ijt − ,t − (*)where B ijτ,µ arise from the adjacencies between the vertices in the packets V i,τ , V j,µ .The leads us to the following lemma, which describes the structure of the blocks B ijτ,µ . Lemma 2.7.
Consider the structure of each off-diagonal block A ij of A ( T m,n,k ) ,as described in (*) above. Then:(i) B ijτ,µ is a null matrix, if τ = µ , and(ii) B ijτ,τ = B ijµ,µ , for each ≤ τ, µ ≤ t − .Proof. As i j (mod n ), the nonzero entries of the matrix A ij correspond to theedges that form y -sibling pairs. Now if v = x τα + ξ y ik ξ , v = x µα + ξ ′ y jk ξ ′ , then thecorresponding matrix entry is 1 if, and only if, v = v y ± . This implies τ = µ and ξ = ξ ′ , and (i) follows.For (ii), assume without loss of generality, that τ > µ . Then ( x τα + ξ y ik ξ , x τα + ξ ′ y jk ξ ′ ) ∈ E ( T m,n,k ), if, and only if, ( x µα + ξ y ik ξ , x µα + ξ ′ y jk ξ ′ ) ∈ E ( T m,n,k ), and the result fol-lows. (cid:3) y Lemma 2.7, we can see that when i = j , A ij has the form A ij = B ij Θ · · · Θ ΘΘ B ij · · · Θ Θ... ... . . . ...Θ Θ · · · B ij ΘΘ Θ · · · Θ B ij where Θ represents the null matrix, and so it remains to obtain a description of B ij . We begin by noting that ( x τα + ξ y ik ξ , x τα + ξ ′ y jk ξ ′ ) ∈ E ( T m,n,k ) if, and only if, ξ ≡ ξ ′ (mod α ) and jk ξ ′ ≡ ik ξ ± n ) , which further holds true if, and only if, j ≡ i ± k α − ξ (mod n ) . (1)The condition ξ ≡ ξ ′ (mod α ) shows that the matrix B ij is a diagonal matrix foreach τ . Hence, we have the following. Lemma 2.8. B ij is a diagonal matrix whose nonzero entries of B ij are contributedby the edges ( x τα + ξ y ik ξ , x τα + ξ y jk ξ ) and ( x τα + ξ ′ y ik ξ ′ , x τα + ξ ′ y jk ξ ′ ) , for ≤ ξ, ξ ′ ≤ α − , satisfying the conditions j ≡ i + k α − ξ and j ≡ i − k α − ξ ′ . (2) In particular, if i, j ∈ Z n such that neither i − j nor j − i is a power of the unit k ,then A ij is a null matrix. In the following lemma, we further analyze the B ij to show that it can have atmost two nonzero entries. Lemma 2.9.
Consider i, j, ∈ Z n satisfying equation (1).(i) If k η
6≡ − n ) for any η ∈ { , . . . , α − } , then B ij has exactly onenonzero entry. In particular, if α is odd, then B ij exactly one nonzero entry.(ii) if k η ≡ − n ) , for some η ∈ { , . . . , α − } , then α is even and B ij hasexactly two nonzero entries contributed by the edges ( x τα + ξ y ik ξ , x τα + ξ y jk ξ ) and ( x τα + ξ + α/ y ik ξ + α/ , x τα + ξ + α/ y jk ξ + α/ ) .Proof. Clearly, when k η
6≡ − n ), there is exactly one solution to equa-tion (1), and so (i) follows. For (ii), it suffices to show that α is even. To see this,we first note that if equation 2 holds, then ξ ξ ′ (mod α ). Assuming withoutloss of generality that ξ < ξ ′ , this would imply k ξ ′ − ξ + 1 ≡ n ), and so α | ξ ′ − ξ ). If α is odd, then α | ( ξ ′ − ξ ), which would contradict the fact that ξ ξ ′ (mod α ). Thus, we have α is even, and ξ ′ − ξ ≡ α/ α ), and (ii)follows. (cid:3) Let Diag m ( p , p , . . . , p τ ) denote the diagonal matrix ( a ij ) m × m defined by a ii =1, if i ∈ { p , . . . , p τ } , and a ii = 0, otherwise. Putting together the results inCorollary 2.6, and Lemmas 2.7 - 2.9, and the fact that A ij = A i +1 ,j +1 , we have themain the result of the paper, which gives a complete description of A ( T m,n,k ). Theorem 2.10 (Main result) . The matrix A ( T m,n,k ) = ( A ij ) n × n is block-circulantwith m × m blocks such that for each i , A ii = A ( C m ) and for each j = i , A ij is adiagonal matrix with or t , or t nonzero entries given by the following conditions.(i) If k s
6≡ − n ) , for any s , then A ij = ( Diag( ξ, ξ + α, . . . , ξ + ( t − α ) , if j ≡ ± k α − ξ , for ≤ ξ ≤ α − , and Θ , otherwise. ii) If α is even and k α/ ≡ − n ) , then A ij = ( Diag( ξ, ξ + α , . . . , ξ + (2 t − α ) , if j ≡ ± k α − ξ , for ≤ ξ ≤ ( α/ − , and Θ , otherwise. On the spectrum of T m,n,k The central goal of this section is to analyze the spectrum of A ( T m,n,k ). FromTheorem 2.10, we know that all succsessive rows of blocks of A ( T m,n,k ) = ( A ij )are obtained by cyclically permuting the first row of blocks ( A , . . . , A m − ). Itis well known [1] that a block circulant matrix of type A ( T m,n,k ) can be block-diagonalized matrix by conjugating with the matrix F n ⊗ F m , where F n = ( f ij ) n × n is the Fourier matrix is defined by f ij = ω ijn , ω n being the primitive n th root ofunity. A direct computation yields the following lemma. Lemma 3.1.
For ≤ i, j ≤ n − , let M = ( M ij ) = ( F n ⊗ F m ) A ( T m,n,k )( F n ⊗ F m ) ∗ .Then:(i) for ≤ i ≤ m − , M ii = n − X s =0 ω isn F m A s F ∗ m , (ii) for ≤ i, j ≤ m − , ( F m A F ∗ m ) i,j = m − m − X τ =0 ( ω i − m + ω i − m ) ω τ ( i − j ) m , and(iii) for ≤ ℓ ≤ n − and ≤ i, j ≤ m − , the matrix F m A ℓ F ∗ m is given by thefollowing conditions.(a) If k s
6≡ − (mod n ) for any s , then we have ( F m A ℓ F ∗ m ) ij = m t − X s =0 ω ( ξ + sα )( i − j ) m , if ℓ ≡ ± k α − ξ , for ≤ ξ ≤ α − , and , otherwise.(b) If α is even and k α/ ≡ − (mod n ), then we have ( F m A l F ∗ m ) ij = m t − X s =0 ω ( ξ + sα/ i − j ) m , if ℓ ≡ ± k α − ξ , for ≤ ξ ≤ ( α/ − , and , otherwise. This leads us to the following proposition, which describes the elements of thediagonal blocks M ii . Proposition 3.2.
For ≤ i ≤ n − and ≤ a, b ≤ m − , the matrix M ii is givenby the following conditions.(i) If k τ
6≡ − n ) , for some τ , then ( M ii ) a,b = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m , if t | a − b, a = b, ( ω a − m + ω a − m ) + tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) , if a = b, and , otherwise. ii) If α is even and k α/ ≡ − n ) , then ( M ii ) a,b = tm ( α/ − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m , if t | a − b, a = b, ( ω a − m + ω a − m ) + 2 tm ( α/ − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) , if a = b, and , otherwise.Proof. Be Lemma 3.1 (i), we have M ii = n − X s =0 ω isn F m A s F ∗ m . Suppose that k τ
6≡ − n ), for some τ . Then M ii = F m A F ∗ m + α − X ξ =0 ( ω ik α − ξ n + ω − ik α − ξ n )( F m A ,k α − ξ F ∗ m ) . Thus, by Lemma 3.1 (ii) - (iii), for 0 ≤ a, b ≤ m −
1, we have( M ii ) ab = ( F m A F ∗ m ) ab + α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n )( F m A ,k α − ξ F ∗ m ) ab = 1 m m − X τ =0 ( ω a − m + ω a − m ) ω τ ( a − b ) m + 1 m α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) t − X s =0 ω ( ξ + sα )( a − b ) m . Replacing τ by ξ + sα , we get( M ii ) a,b = 1 m α − X ξ =0 t − X s =0 ω ( ξ + sα )( a − b ) m (cid:2) ( ω a − m + ω a − m ) + ( ω ik α − ξ n + ω ik α − ξ n ) (cid:3) = 1 m α − X ξ =0 (cid:2) ( ω a − m + ω a − m ) + ( ω ik α − ξ n + ω ik α − ξ n ) (cid:3) ω ξ ( a − b ) m t − X s =0 ( ω α ( a − b ) m ) s . As ω α ( a − b ) m is a t th root of unity, we have t − X s =0 ( ω α ( a − b ) m ) s = ( t ∤ a − bt otherwise,which implies that( M ii ) a,b = tm − α − X ξ =0 (cid:2) ( ω a − m + ω a − m ) + ( ω ik α − ξ n + ω ik α − ξ n ) (cid:3) ω ξ ( a − b ) m , if t | a − b , and is 0 otherwise. When t | a − b , ω a − b is an α th root of unity, and so α − X ξ =0 ω ξ ( a − b ) m = ( α, if a = b, and0 otherwise,which proves (i). The result in (ii) follows from a similar argument. (cid:3) An immediate consequence of Proposition 3.2 is the following corollary.
Corollary 3.3.
The block M = ( m ij ) m × m of M is a diagonal matrix given by m kk = 2 + 2 cos(2 πk/m ) , for ≤ k ≤ m − . . Spectral expanders in {T m,n,k } Since solvable groups of bounded derived length do not yield expander familiesof graphs, the family {T m,n,k } as a whole is not expander. However, it is aninteresting pursuit to determine the graphs in {T m,n,k } with the largest possiblespectral gaps (i.e. Ramanujan graphs), as these graphs are expected to be the bestspectral expanders. To fix notation, let k = λ ( X ) ≥ λ ( X ) ≥ . . . ≥ λ n − ( X ) ≥ − k be the spectrum of a k -regular graph X , and let λ ( X ) = ( max { λ ( X ) | , | λ n − ( X ) |} , if X is bipartite, andmax { λ ( X ) | , | λ n − ( X ) |} , otherwise. . Such a graph X is said to be Ramanujan , if λ ( X ) ≤ √ k −
1. Clearly, a nec-essary condition condition for T m,n,k to be Ramanujan λ ( T m,n,k ) ≤ √
3. FromCorollary 3.3, it is apparent that λ ( T m,n,k ) ≥ π/m ), which implies that λ ( T m,n,k ) > √
3, when m >
8, and so we have the following corollary.
Corollary 4.1.
The graph T m,n,k is not Ramanujan when m > . For a fixed 2 ≤ m ≤
8, we can find significantly large n yielding Ramanujan( m, n, k )-supertoroids. For example, our computations show that T , , is Ra-manujan. So, in order to derive an upper bound on n , we will require a positivelower bound for λ ( T m,n,k ) that involves n .4.1. A lower bound for λ ( T m,n,k ) . In order to derive such a bound, we will usethe following extension of a result due to Walker-Mieghem [11], which gives a lowerbound for the largest eigenvalue of a Hermitian matrix. Though the proof of thisresult is analogous to the proof in [11], we include it for completion.
Theorem 4.2.
Let A = ( a ij ) m × m be a Hermitian matrix, and let ˜ λ = sup ≤ i ≤ m − | λ i ( A ) | .Suppose that f ( t ) = P ∞ k =0 f k t k is an analytic function with radius of convergence R f such that f k is real for each k , and f ( t ) is increasing for all real t . Then for all t > ˜ λ/R f , the largest eigenvalue of A can be bounded from below by tf − m ∞ X k =0 f k N k t − k ! , where N k = u T A k u = P mi =1 P mj =1 ( A k ) ij and N = m , u = [1 , , . . . , T is an m × vector.Proof. Let A = [ a ij ] ≤ i,j ≤ m be an m × m Hermitian matrix with eigenvalues λ min ( A ) = λ ≤ λ ≤ . . . ≤ λ m = λ max ( A ) . Using Rayleigh’s quotient theorem (see [5, Theorem 4.2.2]), we have λ max ( A ) = max x ∈ C m \{ } x T Axx T x . Setting x = u = [1 , . . . , T , we have λ max ( A ) ≥ m m X i,j =1 a ij . lso, note that x T Ax ∈ R , for any x ∈ C m , which follows from the spetral theoremfor Hermitian matrices. Now, consider a complex valued function f ( z ) with radiusof convergence R f > f ( z ) = ∞ X k =0 a k z k , which satisfies: (i) a k ∈ R for all k ∈ N , (ii) f is increasing on the real line. Wenow apply this to the series A t = ∞ X k =0 a k A k t − k , where t > P ∞ k =0 b k A k is convergent if, andonly if, || A || < R g for any matrix norm || . || , where g ( z ) = P ∞ k =0 b k z k is a Taylorexpansion with radius of convergence R g > A t we require || A || /t < R f , that is, t > || A || /R f . This ispossible by choosing t > t >
0, for an eigenvalue λ ( ∈ R ) of A corresponding toan eigenvector v ∈ C m , we have A t v = ∞ X k =0 a k A k t − k v = ∞ X k =0 a k λ k t − k v = f (cid:16) λt (cid:17) v, where to make sense of convergence of the series we need to ensure | λ/t | < R f .This can be done by ensuring t > λ /R f , where λ = max ≤ j ≤ m | λ j | .Since f is increasing, using spectral mapping theorem for normal operators itfollows that for tR f > max (cid:8) || A || , λ (cid:9) , and so we have λ max ( A t ) = f (cid:16) λ max ( A ) t (cid:17) . (3)For t > A t , we get λ max ( A t ) ≥ u T A t um = 1 m ∞ X k =1 f k N k t − k , (4)where N k = u T A k u , for each k ≥
0. Moreover, applying Rayleigh’s bound to A k wehave N k ≤ mλ max ( A k ). Since λ max ( A k ) ≤ λ k , we have N k ≤ mλ k , and hence theseries on the right of (3) converges, if λ /t < R f , that is, for t > λ /R f . Finally,since f − ( x ) is increasing on the real line, the assertion follows from (4). (cid:3) In particular, we consider the case when f ( x ) = e ax (see [11, Section 2.1]) forderiving our bound on n , for which the lower bound in Theorem 4.2 takes the form N m + 12 (cid:18) N m − N m (cid:19) at + (cid:26) a (cid:18) N m − N m (cid:19) + (cid:18) N m − N N m (cid:19)(cid:27) a t + O ( t − ) , (5)for t > a e T √ m/ log(2) , where e T √ m = max ≤ j ≤ m { P mi =1 | a ij |} . In our case, wetake e T √ m = 4, as this is an upper bound for the absolute values of the eigenvaluesof the M ii . For notational consistency, we shall denote the parameters associatedwith our blocks M ii , for 1 ≤ i ≤ n −
1, by N ( i )1 , N ( i )2 , and N ( i )3 . To further simplifynotation, we define k to be regular if k s ≡ − n ), for all s , and we define k to be irregular , if 2 | α and k α/ ≡ − n ). Furthermore, we fix the notation,Ω ni,a,b = tm P α − ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m , if k is regular and t | a − b, tm P ( α/ − ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m , if k if irregular and 2 t | a − b, and0 , otherwise. hen a = b , as Ω ni,a,b is independent of a , we simply denote it by Ω ni . We will alsoneed the following two technical lemmas that involve several tedious computations. Lemma 4.3.
With Ω ni,a,b and Ω ni defined as above, we have:(i) m − X a =0 ( M ii ) a,a = m Ω ni . (ii) m X ≤ a,b ≤ m − ( ω a − m + ω a − m )( ω b − m + ω b − m )Ω ni,a,b = ( ω ikn + ω ikn + ω ik α − n + ω ik α − n , if k is regular ,ω ik α − n + ω ik α − n , if k is irregular.(iii) Let δ = 1 , if k is regular, and δ = 2 , otherwise. Then the following sumsvanish(a) X ≤ a = b ≤ m − ,δt | a − b Ω ni,a,b ω a − m , X ≤ a = b ≤ m − ,δt | a − b Ω ni,a,b ω a − m , (b) X ≤ a = b ≤ m − ,δt | a − b Ω ni,a,b ω b − m , X ≤ a = b ≤ m − ,δt | a − b Ω ni,a,b ω b − m , and(c) m X ≤ a,b ≤ m − h ω b − m (cid:0) Ω ni (1+ k − ) ,a,b +Ω ni (1 − k − ) ,a,b (cid:1) + ω b − m (cid:0) Ω ni (1+ k ) ,a,b +Ω ni (1 − k ) ,a,b (cid:1)i ,and(d) m X ≤ a = b ≤ m − ,δt | a − b Ω ni,a,b ( ω a − m + ω a − m + ω b − m + ω b − m ) . Proof.
From Proposition 3.2, we have( M ii ) a,b = Ω ni,a,b + τ mi,a,b , where τ mi,a,b = ( ( ω a − m + ω a − m ) , if a = b, and0 , otherwise.Hence, we have m − X a =0 ( M ii ) a,a = m − X a =0 ( ω a − m + ω a − m ) + m Ω ni . Since ω m is an m th root of unity, the first sum vanishes, and (i) follows.To show (ii), we first consider the case when k is regular. Upon reparametrizingthe entries of the sum as( a, a + ηt ) : 0 ≤ a ≤ m − , ≤ η ≤ α − , and rearranging, the sum on the left becomes1 m α − X η =0 Ω ni,a,a + ηt m − X a =0 ( ω a − m + ω a − m )( ω a + ηt − m + ω a + ηt − m ) . (6)Since m ≥
3, the inner sum above can be rewritten as ω ηtm m − X a =0 ω a − m + ω ηtm m − X a =0 ω a − m + m ( ω ηtm + ω ηtm ) = m ( ω ηtm + ω ηtm )Consequently, after rearranging, (6) yields α − X η =0 Ω ni,a,a + ηt ( ω ηtm + ω ηtm ) = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) α − X η =0 ( ω ( ξ +1) ηtm + ω ( ξ − ηtm ) . The inner sum on the right can be further broken into two parts, where the twoparts survive if, and only if, ξ = α − ξ = 1, respectively. Further, when theseparts are nonzero, each assumes the value α , and hence the result follows for thecase when k is regular. The argument for the case when k is irregular is similar. onsider the first sum in (iii)(a). While k is regular, we subdivide this suminto parts with a fixed a , and allowing b to run over all elements a + ηt (mod m )(1 ≤ η ≤ α − ni,a,a + ηt depends only on η , the sum equals m − X a =0 α − X η =1 Ω ni,a,a + ηt ω a − m = α − X η =1 Ω ni,a,a + ηt m − X a =0 ω a − m = 0 . On the other hand, when k is irregular, the sum runs over all b = a + η (2 t ) (mod m ) (1 ≤ η ≤ α − δt | a − b . Thus, it suffices to verify that1 m m − X a =0 ω a − m (cid:0) Ω ni (1+ k − ) + Ω ni (1 − k − ) (cid:1) = 0 = 1 m m − X a =0 ω a − m (cid:0) Ω ni (1+ k ) + Ω ni (1 − k ) (cid:1) . But this follows from the fact that the terms within the parantheses in the twosums above are independent of a . (cid:3) Lemma 4.4.
For ≤ a, b ≤ m − , let M ii = ( u ab ) and M ii = ( v ab ) , and let δ = 1 ,if k is regular, and δ = 2 , otherwise. Then:(i) u ab = Ω ni,a,b ( ω a − m + ω a − m + ω b − m + ω b − m ) + Ω n i,a,b , if a = b, δt | a − b ( ω a − m + ω a − m ) + 2 + 2Ω ni ( ω a − m + ω a − m ) + Ω n i , if a = b, and , otherwise.(ii) v ab = Ω ni,a,b h ( ω a − m + ω a − m ) + ( ω b − m + ω b − m ) + ( ω a − m + ω a − m )( ω b − m + ω b − m ) + 3 i +Ω n i,a,b h ( ω a − m + ω a − m ) + ( ω b − m + ω b − m ) i + Ω n i,a,b + ω b − m (cid:0) Ω ni (1+ k − ) ,a,b + Ω ni (1 − k − ) ,a,b (cid:1) + ω b − m (cid:0) Ω ni (1+ k ) ,a,b + Ω ni (1 − k ) ,a,b (cid:1) , if a = b and δt | a − b, ( ω a − m + ω a − m ) + 4( ω a − m + ω a − m ) + 3Ω ni h ( ω a − m + ω a − m ) + 1 i +2Ω n i ( ω a − m + ω a − m ) + Ω n i + ω a − m (cid:0) Ω ni (1+ k − ) + Ω ni (1 − k − ) (cid:1) + ω a − m (cid:0) Ω ni (1+ k ) + Ω ni (1 − k ) (cid:1) , if a = b, and , if δt ∤ a − b. Proof.
By definition, we have u ab = P m − τ =0 ( M ii ) a,τ ( M ii ) τ,b . First, we consider thecase when k is regular. If t ∤ a − b , clearly, by definition the u ab = 0, since either( M ii ) a,τ = 0, or ( M ii ) τ,b = 0. When a = b , from the definition we have u ab equalsΩ ni,a,b ( ω a − m + ω a − m + ω b − m + ω b − m ) + m − X τ =0 Ω ni,a,τ Ω ni,τ,b . (7)In the expression for u ab in (7) above, the terms Ω ni,a,b , Ω ni,a,τ , and Ω ni,τ,b are nonzero,only when t | a − τ, τ − b . Now the last term in (7) is by definition t m α − X ξ =0 α − X ξ ′ =0 ( ω ik α − ξ n + ω ik α − ξ n )( ω ik α − ξ ′ n + ω ik α − ξ ′ n ) X ≤ τ ≤ m − ,t | a − τ,τ − b ω ξ ( a − τ )+ ξ ′ ( τ − b ) m , where the inner sum runs on the indices τ = a, a + t, a +2 t, . . . , a +( α − t (mod m ),and so the inner sum is α − X η =0 ω ( − ξ ) ηt + ξ ′ [( a − b )+ ηt ] m = ω ξ ′ ( a − b ) m α − X η =0 ( ω tm ) ( ξ ′ − ξ ) η . ince ω tm is an α th root of unity and α | ξ ′ − ξ if, and only if, ξ ′ = ξ , the innersum is nonzero, only when ξ ′ = ξ , and while it is nonzero, it is equal to αω ξ ( a − b ) m .Thus, the last sum in (7) reduces to tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m + 2 tm α − X ξ =0 ω ξ ( a − b ) m , from which the result follows for the case a = b . When a = b , u ab equals( ω a − m + ω a − m ) + 2Ω ni ( ω a − m + ω a − m ) + X ≤ τ ≤ m − ,t | a − τ Ω ni,a,τ Ω ni,τ,a . (8)As before, the last sum in (8) above equals t m α − X ξ =0 α − X ξ ′ =0 ( ω ik α − ξ n + ω ik α − ξ n )( ω ik α − ξ ′ n + ω ik α − ξ ′ n ) X ≤ τ ≤ m − ,t | a − τ ω ξ ( a − τ )+ ξ ′ ( τ − a ) m , where the inner sum equals α − X η =0 ( ω tm ) ( ξ ′ − ξ ) η . Thus, the last sum (8) reduces to tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) = Ω n i + 2 , and the result follows for this case. The argument for the case when k is irregularis analogous.For showing (ii), we will first establish the following claim: Claim.
For 0 ≤ a, b ≤ m −
1, the sum m − X τ =0 ( ω τ − m + ω τ − m )Ω ni,a,τ Ω ni,τ,b is given by= ω b − m (cid:0) Ω ni (1+ k − ) ,a,b + Ω ni (1 − k − ) ,a,b (cid:1) + ω b − m (cid:0) Ω ni (1+ k ) ,a,b + Ω ni (1 − k ) ,a,b (cid:1) , if δt | a − b, α = 4 , and k is regular , ( ω b − m + ω b − m ) (cid:0) Ω ni (1+ k ) ,a,b + Ω ni (1 − k ) ,a,b (cid:1) , if 2 t | a − b, α = 4 , and k is irregular, and0 if δt ∤ a − b. Proof of claim.
While δt ∤ a − b , the sum is clearly 0. We first consider the casewhen a = b and k is regular, and so the sum becomes t m α − X ξ =0 α − X ξ ′ =0 ( ω ik α − ξ n + ω ik α − ξ n )( ω ik α − ξ ′ n + ω ik α − ξ ′ n ) X ≤ τ ≤ m − ,t | a − τ ( ω τ − m + ω τ − m ) ω ξ ( a − τ )+ ξ ′ ( τ − b ) m . (9)As this runs over all indices τ = a + ηt (mod m ) , η = 0 , , . . . , α − , the inner sum equals ω ξ ′ ( a − b ) m α − X η =0 ( ω a + ηt − m + ω a + ηt − m )( ω tm ) ( ξ ′ − ξ ) η . (10) reaking (10) further into two parts, we obtain ω ξ ′ ( a − b ) m ω a − m α − X η =0 ω ( ξ ′ − ξ − ηtm + ω ξ ′ ( a − b ) m ω a − m α − X η =0 ω ( ξ ′ − ξ +1) ηtm . Note that first part is nonzero, only when ξ ′ = ξ + 1, in which case, it takes thevalue αω ( ξ +1)( a − b ) m ω a − m , while the second part survives, only when ξ ′ = ξ −
1, whereit assumes the value αω ( ξ − a − b ) m ω a − m . As α ≥
3, the indices ξ + 1 and ξ − α . Plugging these values in (9), we get tm ω a − bm ω a − m α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n )( ω ik α − ξ − n + ω ik α − ξ − n ) ω ξ ( a − b ) m + tm ω a − bm ω a − m α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n )( ω ik α − ξ +1 n + ω ik α − ξ +1 n ) ω ξ ( a − b ) m , from which the result follows for this case. The arguments for the case when k is irregular and the case a = b is similar, and so the claim follows. We will nowproceed to prove (ii). When a = b , we have v ab = m − X τ =0 ( M ii ) a,τ ( M ii ) τ,b = ( M ii ) a,a ( M ii ) a,b + ( M ii ) a,b ( M ii ) b,b + X τ = a,b ( M ii ) a,τ ( M ii ) τ,b = (cid:2) ( ω a − m + ω a − m ) + 2 + 2Ω ni ( ω a − m + ω a − m ) + Ω n i (cid:3) Ω ni,a,b + (cid:2) Ω ni,a,b ( ω a − m + ω a − m + ω b − m + ω b − m ) + Ω n i,a,b (cid:3) ( ω b − m + ω b − m + Ω ni )+ X τ = a,b (cid:2) Ω ni,a,τ ( ω a − m + ω a − m + ω τ − m + ω τ − m ) + Ω n i,a,τ (cid:3) Ω ni,τ,b = m − X τ =0 ( ω τ − m + ω τ − m )Ω ni,a,τ Ω ni,τ,b + ( ω a − m + ω a − m ) m − X τ =0 Ω ni,a,τ Ω ni,τ,b + m − X τ =0 Ω n i,a,τ Ω ni,τ,b + (cid:2) ( ω a − m + ω a − m ) + ( ω b − m + ω b − m ) + ( ω a − m + ω a − m )( ω b − m + ω b − m ) + 2 (cid:3) Ω ni,a,b +( ω b − m + ω b − m )Ω n i,a,b . (11)From (11) above, it is clear that v ab = 0, if δt ∤ a − b . Observing that the third sumis Ω n i,a,b + Ω ni,a,b , we further simplify (11) using the claim we just proved and theproof of (i), to obtain the desired result for the case a = b . The argument for thecase when a = b is analogous. (cid:3) We will now derive an expression for N ik , for 1 ≤ k ≤ Proposition 4.5.
For ≤ i ≤ n − , we have:(i) N ( i )1 m = ω in + ω in ,(ii) N ( i )2 m = 4 + ω in + ω in , and(iii) N ( i )3 m = ( ω ikn + ω ikn + ω ik α − n + ω ik α − n + 7( ω in + ω in ) + ( ω in + ω in ) , if k is regular, and ω ik α − n + ω ik α − n + 7( ω in + ω in ) + ( ω in + ω in ) , if k is irregular. roof. By applying Lemma 4.3 (i), we get N ( i )1 m = 1 m X ≤ a,b ≤ m − ( M ii ) a,b = 1 m m − X a =0 ( M ii ) a,a + 1 m X ≤ a = b ≤ m − ( M ii ) a,b = Ω ni + 1 m X ≤ a = b ≤ m − Ω ni,a,b = 1 m X ≤ a,b ≤ m − Ω ni,a,b . Suppose that k is regular. Then N ( i )1 m = 1 m X ≤ a,b ≤ m − , t | a − b tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) ω ξ ( a − b ) m = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) X ≤ a,b ≤ m − , t | a − b ω ξ ( a − b ) m . For each 0 ≤ a ≤ m −
1, setting b = a + ηt , for 0 ≤ η ≤ α −
1, we get N ( i )1 m = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) m − X a =0 α − X η =0 ω ξ ( − ηt ) m = tm α − X ξ =0 ( ω ik α − ξ n + ω ik α − ξ n ) α − X η =0 ( ω tm ) ξη . Since ω tm is an α th root of unity, and so the last sum is 0 when ξ = 0, and is α ,when ξ = 0, from which (i) follows. The proof for the case when k is irregularfollows from similar arguments.To show (ii), we first apply Lemma 4.4 (i), to obtain N ( i )2 m = 1 m m − X a =0 h ( ω a − m + ω a − m ) + 2 + 2Ω ni ( ω a − m + ω a − m ) + Ω n i i + 1 m X ≤ a = b ≤ m − ,δt | a − b h Ω ni,a,b ( ω a − m + ω a − m + ω b − m + ω b − m ) + Ω n i,a,b i = 1 m m − X a =0 h ( ω a − m + ω a − m ) + 4 + Ω n i i + 1 m X ≤ a = b ≤ m − ,δt | a − b Ω n i,a,b . If m is odd, both ω a − m and ω a − m vary over all powers of ω m , and so we have1 m m − X a =0 ( ω a − m + ω a − m ) = 0Similarly, when m is even we have ω m is a primitive ( m/ th root of unity, andhence 1 m m − X a =0 ω a − m = 1 m m − X a =0 ω a − m + ( ω m ) m/ m m − X b =0 ω b − m = 0 . herefore, we have N ( i )2 m = 4 + 1 m m − X a =0 Ω n i + 1 m X ≤ a = b ≤ m − ,δt | a − b Ω n i,a,b = 4 + N (2 i )1 m , and(ii) follows.To show (iii), first suppose that α ≥ k is irregular. Using Lemma 4.4 (ii)and (i), we have N ( i )3 m = m − X a,b =0 Ω ni,a,b h ( ω a − m + ω a − m ) + ( ω b − m + ω b − m ) i + m − X a,b =0 Ω ni,a,b h ( ω a − m + ω a − m )( ω b − m + ω b − m ) i ++3 m − X a,b =0 Ω ni,a,b + m − X a,b =0 Ω n i,a,b h ( ω a − m ω a − m ) + ( ω b − m + ω b − m ) i + m − X a,b =0 Ω n i,a,b + m − X a =0 h ( ω a − m + ω a − m ) + 4( ω a − m + ω a − m ) i . By applying (i) and Lemma 4.3, and by observing that the first sum is 4 N ( i )1 /m ,the third sum is 3 N ( i )1 /m , the fifth sum is N (3 i )1 /m , the result follows. (cid:3) Plugging in the expressions for N ( i ) k from Proposition 4.5 in Equation (5), we obtainthe following. Corollary 4.6.
The largest eigenvalue of M ii is bounded below by B ( n, k, a, t, ǫ ) := ( ω n + ω n ) + at + (cid:16) ǫa − (cid:17) ( ω n + ω n ) (cid:16) a t (cid:17) + O ( t − ) , where ǫ = 2 , if k is regular, and ǫ = 3 , otherwise. Consequently, λ ( X ) ≥ B ( n, k, a, t, ǫ ) . It finally remains to obtain a lower bound for n beyond which the graph T m,n,k will not be Ramanujan.4.2. A lower bound for n.
The aim of this subsection is to establish that T m,n,k is not Ramanujan when n ≥ B ( n, k, a, t, ǫ ), by choosing a = 127 . ǫ = 2, t = 1000, and n ≥ ≥ .
476 (i.e ≥ √ ǫ = 3, choosing a = 111 .
5, and the same values of t and n as above, would bound the sum of thenon-error terms of B ( n, k, a, t, ǫ ) from below by 3 . B ( n, k, a, t, ǫ ) in Corollary 4.6, we use the general form(see [11, Appendix A]) of the Taylor expansion of f − ( m P ∞ k =0 f k N k z k ) around z = 0, which is given by f − ( h ( z )) = ∞ X k =1 c k z k where h ( z ) = m P ∞ k =0 f k N k z k and the characteristic coefficients c k = f k f (cid:18) N k m − (cid:16) N m (cid:17) k (cid:19) + 1 k (cid:16) N m (cid:17) k k − X j =2 ( − j (cid:18) k + j − j (cid:19) f − j s ∗ [ j, k − | f ( z ) + k − X n =2 n − X j =1 ( − j (cid:18) n + j − j (cid:19) f − n − j s ∗ [ j, n − | f ( z ) ! s [ n, k ] | h ( z ) n . (12) e define M ( n, t ) := X n + ... + n t = n, n i ≥ (cid:18) nn n . . . n t (cid:19) , if t ≥ , and1 , if t = 1 , where (cid:0) nn n ...n t (cid:1) = n ! n ! n ! ...n t ! . Notice that the quantity M ( n + j, j ) is the numberof surjective functions from a set of size n + j onto a set of size j . Further, we set S n := 1 n ! n X j =2 ( − j j ! M ( n + j, j ) , and the state a lemma, which we will use in our error estimates. Lemma 4.7.
For n ≥ , we have S n := 1 n ! n X j =2 ( − j j ! M ( n + j, j ) = ( − n + 1 n ! . Proof.
Multiplying both sides by n !, we need to prove that A ( n ) := n X j =1 ( − j j ! M ( n + j, j ) = ( − n n ! . Denoting I n = { , , . . . , n } , we first note that f ( n, j ) := M ( n + j,j ;2) j ! denotes thenumber of partitions of I n + j into j subsets, each of which contains at least twoelements. For any such partition of I n + j , if the element n + j belong to a subset(of this partition) of size ≥
3, then removing it will yield a partition of I n + j − into j subsets of the above type. Labeling the subsets by indices 1 , , . . . , j , this can bedone in j distinct ways.If the element n + j belongs to a subset of size 2, then removing this subset yieldsa partition of the set I n + j − \ { x } into j − x ∈ I n + j − . This canbe done in n + j − f ( n, j ) = jf ( n − , j ) + ( n + j − f ( n − , j − . Plugging this in the above sum for A ( n ), we obtain A ( n ) = n X j =1 ( − j jf ( n − , j ) + n X j =1 ( − j ( n + j − f ( n − , j − . Now changing the variable j j + 1 in the second sum, combining the coefficientsof f ( n − , k ) together, and using f ( n − , n ) = 0, we obtain A ( n ) = − nA ( n − n . (cid:3) We denote the three terms (in the order of their appearance) in the expressionfor c k in (12) for the case f ( x ) = e ax by R i,k , for 1 ≤ i ≤
3, where R ,k = a k − k ! (cid:18) N k m − (cid:16) N m (cid:17) k (cid:19) . (13)We will now derive simpler expressions for R ,k and R ,k . Proposition 4.8. (i) R ,k = a k − k (cid:16) N m (cid:17) k (cid:18) ( − n + 1 n ! (cid:19) .(ii) R ,k = a k − k − X n =2 ( − n − n X P ni =1 t i = k, t i ≥ (cid:0) N t m (cid:1) . . . (cid:0) N t n m (cid:1) . roof. The term s ∗ [ j, k − | f ( z ) of T is given by s ∗ [ j, k − | f ( z ) = X P ji =1 t i = k − ,t i ≥ j Y i =1 f t i +1 ( z )= X P ji =1 t i = k − ,t i ≥ a k − j ( t + 1)! . . . ( t j + 1)!= a k − j ( k − j )! X P ji =1 t i = k − ,t i ≥ ( k − j )!( t + 1)! . . . ( t j + 1)!= a k − j ( k − j )! M ( k − j, j ) . Consequently, we have R ,k = 1 k (cid:16) N m (cid:17) k k − X j =2 ( − j (cid:18) k + j − j (cid:19) f − j s ∗ [ j, k − | f ( z ) = 1 k (cid:16) N m (cid:17) k k − X j =2 ( − j (cid:18) k + j − j (cid:19) a k − ( k − j )! M ( k − j, j )= a k − k ! (cid:16) N m (cid:17) k k − X j =2 ( − j M ( k − j, j ; 2) j ! , from which (i) follows by a direct application of Lemma 4.7.To show (ii), we first compute s [ n, k ] | h ( z ) . We have s [ n, k ] | h ( z ) = X P ni =1 t i = k, t i ≥ n Y i =1 h t i ( z )= X P ni =1 t i = k, t i ≥ a t (cid:0) N t m (cid:1) . . . a t n (cid:0) N t n m (cid:1) = a k X P ni =1 t i = k, t i ≥ (cid:0) N t m (cid:1) . . . (cid:0) N t n m (cid:1) (14)From the computations for R ,k , we see that n − X j =1 ( − j n + j − j ! f − n − j s ∗ [ j, n − | f ( z ) = 1 a ( n − n − X j =1 ( − j M ( n − j, j ) j ! . (15) Finally, (ii) follows from (14), (15), and Lemma 4.7. (cid:3)
It now remains to obtain upper bounds on the error terms R i,k , whenever n ≥ R ,k and R ,k by plugging in the inequalities N k /m ≤ k and N /m = 2 cos(2 π/n ) ≥ (1 . k in (13). Moreover, we canobtain an upper bound for R ,k by applying the inequalities X P ni =1 t i = k, t i ≥ (cid:0) N t m (cid:1) . . . (cid:0) N t n m (cid:1) ≤ k − and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k − X n =2 ( − n − n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ log 2 + 1to Proposition 4.8, where the second inequality follows from the fact number ofinteger solutions of the equation t + . . . + t n = k, t i ≥ (cid:0) k + n − k (cid:1) ≤ k − . Thisleads us to the following corollary. Corollary 4.9.
For n ≥ , we have: i) | R ,k | ≤ ak ! (cid:16) at (cid:17) k (4 k − . k ) , (ii) | R ,k | ≤ ak (cid:16) at (cid:17) k (1 . k ) (cid:12)(cid:12)(cid:12)(cid:12) ( − n + 1 n ! (cid:12)(cid:12)(cid:12)(cid:12) , and(iii) | R ,k | ≤ (log 2 + 1) 1 a (cid:16) at (cid:17) k (1 . k )4 k − . For 4 ≤ k ≤
10, the following table lists the values of the bounds on the R i,k (which we denote by S i,k ) from Corollary 4.9, assuming that a = 127 . t = 1000,and n ≥ | S i,k | →
0, as k → ∞ .) k S ,k S ,k S ,k . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × −
10 2 . × − . × − . × − A similar set of values can be computed for the case when ǫ = 3. In conclusion, wehave the following result, which follows directly from Corollaries 4.6 and 4.9, andthe discussion at the beginning of this subsection. Theorem 4.10.
The graph T m,n,k is not Ramanujan when n ≥ . We conclude this paper by enlisting a collection of triples ( m, n, k ) such that givenany ( m, n ) ∈ { ( x, y ) : 3 ≤ x ≤ ≤ y < } there exist at least one nontrivialunit k ∈ Z × n such that T m,n,k is Ramanujan. (3, 7, 2), (3, 9, 4), (3, 13, 3), (3, 14, 9), (3, 18, 7), (3, 19, 7), (3, 21, 4), (3, 26, 3), (3, 28,9), (3, 31, 5), (3, 35, 11), (3, 37, 10), (3, 38, 7), (3, 39, 16), (3, 42, 25), (3, 52, 9), (3, 56,9), (3, 62, 5), (3, 63, 25), (3, 74, 47), (3, 78, 55), (3, 117, 16), (4, 5, 2), (4, 8, 3), (4, 10,3), (4, 12, 5), (4, 13, 5), (4, 15, 2), (4, 16, 3), (4, 20, 3), (4, 24, 5), (4, 26, 5), (4, 30, 7),(4, 32, 7), (4, 39, 5), (4, 40, 3), (4, 48, 5), (4, 52, 5), (4, 60, 7), (4, 78, 5), (4, 80, 7), (4,104, 5), (4, 120, 7), (5, 11, 3), (5, 25, 6), (5, 31, 2), (5, 33, 4), (5, 41, 10), (5, 55, 16), (5,61, 9), (5, 71, 5), (5, 77, 15), (5, 93, 4), (5, 101, 36), (5, 121, 9), (5, 123, 10), (5, 131, 53),(5, 151, 8), (5, 155, 66), (5, 181, 59), (5, 183, 58), (5, 191, 39), (5, 217, 8), (5, 241, 87),(5, 251, 149), (5, 275, 141), (5, 311, 36), (5, 341, 70), (5, 363, 130), (6, 7, 2), (6, 8, 3), (6,9, 2), (6, 12, 5), (6, 13, 3), (6, 14, 3), (6, 16, 7), (6, 18, 5), (6, 19, 7), (6, 21, 2), (6, 24,5), (6, 26, 3), (6, 28, 3), (6, 31, 5), (6, 36, 5), (6, 37, 10), (6, 38, 7), (6, 39, 4), (6, 42, 5),(6,52, 3), (6, 56, 3), (6, 57, 11), (6, 62, 5), (6, 72, 5), (6, 74, 11), (6, 76, 7), (6, 78, 17), (6,84, 5), (6, 91, 4), (6, 93, 5), (6, 104, 17), (6, 111, 26), (6, 112, 9),(6, 114, 11), (6, 117, 4),(6, 124, 37), (6, 133, 26), (6, 148, 47), (6, 152, 7), (6, 156, 17), (6, 168, 5), (6, 171, 68),(6, 182, 23), (6, 186, 5), (6, 208, 55), (6, 222, 47), (6, 234, 29), (6, 248, 37), (6, 266, 45),(6, 279, 88), (6, 296, 85), (7, 29, 7), (7, 43, 4), (7, 71, 30), (7, 87,7), (7, 113, 16), (7, 145,16), (7, 197, 114), (7, 211, 58), (7, 213, 172), (7, 215, 176), (7, 239, 44), (7, 339, 16), (8,5, 2), (8,8, 3), (8, 10, 3), (8, 12, 5), (8, 13, 5), (8, 15, 2), (8, 16, 3), (8, 17, 2), (8, 20, 3),(8, 24, 5), (8, 26, 5), (8, 30, 7), (8, 32, 3), (8, 34, 9), (8, 39, 5), (8, 40, 3), (8, 48, 5), (8,51, 8), (8, 52, 5), (8, 60, 7), (8, 64, 7), (8, 68, 9), (8, 73, 10), (8, 78, 5), (8, 80, 7), (8, 85,8), (8, 89, 12), (8, 96, 5), (8, 102, 19), (8, 104, 5), (8, 113, 18), (8, 120, 7), (8, 136, 9), (8,146, 51), (8, 160, 13), (8, 170, 53), (8, 178, 37), (8, 194, 33), (8, 204, 19), (8, 219, 10), (8,226, 69), (8, 272, 9), (8, 292, 83), (8,340, 93), (8, 356, 101), (8, 388, 33) Acknowledgements.
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Department of Mathematics, Indian Institute of Science Education and ResearchBhopal, Bhopal Bypass Road, Bhauri, Bhopal 462 066, Madhya Pradesh, India
E-mail address : [email protected] URL : https://home.iiserb.ac.in/ e kashyap/ Department of Mathematics, Indian Institute of Science Education and ResearchBhopal, Bhopal Bypass Road, Bhauri, Bhopal 462 066, Madhya Pradesh, India
E-mail address : [email protected] URL : https://home.iiserb.ac.in/ e sidhu/ The University of Oklahoma, School of Computer Science, 110 W. Boyd St., DevonEnergy Hall, Rm. 230, Norman, OK 73019, USA
E-mail address : [email protected] URL : Department of Electrical Engineering and Computer Science, Indian Institute ofScience Education and Research Bhopal, Bhopal Bypass Road, Bhauri, Bhopal 462 066,Madhya Pradesh, India
E-mail address : [email protected]@iiserb.ac.in