Gap Sets for the Spectra of Regular Graphs with Minimum Spectral

arXiv:2106.13129v1 [math.CO] 24 Jun 2021 a -alAddresses E-mail eateto ahmtc,K .ToiUiest fTechno of University Toosi N. K. Mathematics, of Department interval aiis epciey epoeta (1 that prove We respectively. families, √ w eune fgah,∆ spect graphs, minimum of with sequences graphs two quartic and cubic connected large a nevl.A ypout eso htteeigenvalues the that show we by-product, a As intervals. gap t iiu ievle ovret 1 to converge eigenvalue) minimum its n h ievle fΓ of eigenvalues the and neet ihfwecpin,alteegnauso conne simple. of are Keywords: eigenvalues gap spectral the minimum all with exceptions, graphs few whic with property following interest: the on depend heavily results above M ahmtc ujc lsicto (2010): Classification Subject Mathematics AMS a esfrtesetao eua rpswith graphs regular of spectra the for sets Gap 17) udstae5 Goaiu) 04 abr,Germany Hamburg, 20146 (Geomatikum), 55 Bundesstraße olwn eetwr yKl´radSra,w td asin gaps study we Sarnak, Kollár and by work recent Following / 2 b , eateto ahmtc,Uiest fHamburg, of University Mathematics, of Department ]i a nevlfrΓ for interval gap a is 3] − 3 , − pcrlgp a e nsetu,Cbcgah uri grap Quartic graph, Cubic spectrum, in set Gap gap, Spectral : [email protected] √ aymAbdi Maryam 5 .O o 66-31 ern Iran Tehran, 16765-3381, Box O. P. i atclr t iiu ievle ovret (1 to converge eigenvalue) minimum its particular, (in iiu pcrlgap spectral minimum n yn nteitra [ interval the in lying n n Γ and n ue2,2021 25, June ecnetr htteetoaeide maximal indeed are two these that conjecture We . a n Abstract hc r oe‘ymti’aogteetwo these among ‘symmetric’ more are which − √ , bai Ghorbani Ebrahim M Abdi), (M. 1 √ 3as 13 ]i a nevlfr∆ for interval gap a is 5] − n 4 ed oifiiy h roso the of proofs The infinity. to tends , − e (1+ [email protected] √ 05C50 17) a eo independent of be can h tdcbcadquartic and cubic cted / a a.W ou on focus We gap. ral ](n nparticular, in (and 2] f∆ of a , b n h pcr of spectra the n n [( and , E Ghorbani) (E. yn nthe in lying − √ h 33) − + 1 logy, / 2 1 Introduction

The eigenvalues of a simple graph G are defined as the eigenvalues of its adjacency matrix. The spectrum of G is the list of its eigenvalues. The difference between the two largest eigenvalues of G is called the spectral gap of G. Aldous and Fill (see [5, p. 217]) posed a conjecture on the maximum relaxation time of the random walk in regular graphs. This conjecture can be rephrased in terms of the spectral gap as follows: the spectral gap 2rπ2 of a connected r-regular graph on n vertices is at least (1 + o(1)) 3n2 . Brand, Guiduli, and Imrich [7] (confirming a conjecture by L. Babai (see [12])) determined the structure of connected cubic (i.e. 3-regular) graphs with minimum spectral gap. For every even n 10, such a graph is proved to be unique. We denote this family of graphs by ≥ Fcub (see Section 2 for the precise description). Abdi, Ghorbani and Imrich [2] showed that the 2π2 spectral gap of the graphs of is (1+o(1)) 2 , confirming the Aldous–Fill conjecture for Fcub n r = 3. Abdi and Ghorbani [1] gave a ‘near’ complete characterization of the graphs which have the minimum spectral gap among the connected quartic (i.e. 4-regular) graphs with a fixed order. They showed that such graphs belong to a specific family, here denoted by

∗ (see Section 2 for the precise description). Based on this result, they established the Fqua Aldous–Fill conjecture for r = 4. They further conjectured that the graphs in a subfamily

of ∗ gives the unique minimal graph for every order n. Fqua Fqua Kollár and Sarnak [14] studied gaps in the spectra of large finite cubic graphs. It is known that the gap interval (2√2, 3) achieved in cubic Ramanujan graphs are maximal meaning that for any ǫ > 0, the interval (2√2 ǫ, 3) contains some eigenvalues of some − cubic Ramanujan graphs. Also the gap interval [ 3, 2) achieved in cubic line graphs are − − maximal. Kollár and Sarnak gave constraints on spectra in [ 3, 3] which are maximally − gapped and constructed examples which achieve these bounds. Among other results, they also showed that every point in [ 3, 3) can be gapped (meaning that every point is − contained in an open interval which is a gap set) by planar cubic graphs. They also posed the similar spectral gap questions for general regular graphs.

Problem 1.1 (Koll´ar and Sarnak [14]). Investigate the spectral gap questions more gen- erally for r-regular graphs (r> 3).

This is the main motivation in this work. Our goal is to determine what gaps can be achieved by large connected cubic and quartic graphs with minimum spectral gap. More precisely, our focus will be on the graphs in with orders 2 (mod 4) and the graphs Fcub ≡ in with orders 1 (mod 5) which are denoted by ∆ and Γ , respectively. These Fqua ≡ n n graphs are more ‘symmetric’ than the other graphs of their respective families. We prove that:

2 • (1, √5] is a gap interval for the graphs ∆n;

• [( 1+ √17)/2, 3] is a gap interval for the graphs Γ . − n We conjecture that the above two intervals are indeed maximal gap intervals. As a by- product, we obtain the following results. Here ρ(G) denotes the smallest eigenvalue of a graph G.

• The eigenvalues of ∆ lying in the interval 3, √5 converge to (1 √33)/2 as n − − − n tends to inﬁnity. In particular, lim ρ(∆n)=(1 √33)/2. n →∞ − • The eigenvalues of Γ lying in the interval [ 4, (1+ √17)/2] converge to 1 √13 n − − − as n tends to inﬁnity. In particular, lim ρ(Γn)=1 √13. n →∞ − The proofs of the above results heavily depend on the following properties on the simplicity of the eigenvalues which can be of independent interest.

• Let G be a graph in with order n 10. If n 2 (mod 4), then all the Fcub ≥ ≡ eigenvalues of G except 1 and 0 are simple. If n 0 (mod 4), then the only − ≡ non-simple eigenvalue of G is 1. −

• Non-simple eigenvalues of the graphs in ∗ of order n 11 belong to Fqua ≥ 2, 0, 1, 1 √2, ( 1 √5)/2 . n− ± − ± − ± o In passing, we remark that several results on gap sets for the spectra of speciﬁc families of graphs can be found in the literature. Jacobs et al. [13] showed that threshold graphs (that is graphs with no induced P , C nor 2K ) have no eigenvalues in ( 1, 0). 4 4 2 − This result was extended by Ghorbani [9] who showed that a graph G is a cograph (i.e. a P4-free graph) if and only if no induced subgraph of G has an eigenvalue in the interval ( 1, 0). Improving the aforementioned result of Jacobs et al., Ghorbani [10] proved − that besides 0, 1, threshold graphs have no eigenvalues in the maximal gap interval − ( 1 √2)/2, ( 1+ √2)/2 . This was ﬁrst conjectured by Aguilar et al. [3]. In [9], it − − − is also proved that bipartite P -free graphs have no eigenvalues in the intervals ( 1/2, 0) 5 − and (0, 1/2). The result of [10] is extended to the case that if the ‘generating sequence’ of a threshold graph is given by Andelic et al. [6] and to the spectrum of distance matrix by Alazemi et al. [4]. The rest of the paper is organized as follows. In Section 2, we present the precise description of the graphs in and ∗ , and in Section 3, we prove the results on Fcub Fqua simplicity of the eigenvalues of the graphs in these two families. This will be used in Section 4, to establish the results on the gap intervals and other properties of the spectra of the graphs ∆n and Γn.

3 2 Preliminaries

In this section we give the precise description of the families and ∗ and recall Fcub Fqua some basic facts about graph eigenvalues. L. Babai (see [12]) posed a conjecture on the structure of the connected cubic graph with minimum spectral gap. Brand, Guiduli, and Imrich [7] conﬁrmed the conjecture by proving that for any even n 10, the n-vertex graph given in Figure 1 is the unique graph ≥ with minimum spectral gap among connected cubic graphs of a given order n. Note that cubic graphs have even order. We denote this family of graphs by . Fcub

Figure 1: The graphs of of order n 2 (mod 4) and n 0 (mod 4), resp. Fcub ≡ ≡

For every n 2 (mod 4), we denote the n-vertex graph in by ∆ . This is the ≡ Fcub n upper graph of Figure 1.

Turning to quartic graphs, we deﬁne the family ∗ as follows. Similar to the graphs Fqua in , the graphs in ∗ have a path-like structure as illustrated in Figure 2. Further, Fcub Fqua their building blocks are those given in Figure 3; any middle block of G is M, and the left end block of G is one of D1,...,D5. The right end block is the mirror image of one of these ﬁve blocks.

Figure 2: The path-like structure

M D1 D2 D3 D4 D5

Figure 3: The building blocks of a minimal quartic garph

Theorem 2.1 (Abdi and Ghorbani [1]). For any n 11, a graph with minimum spectral ≥ gap among the connected quartic graphs of order n belongs to ∗ . Fqua

4 Figure 4: The graph Γn

In [1], we posed the following conjecture which uniquely speciﬁes the quartic graphs with minimum spectral gap of any given order. To this end, we deﬁne a subfmaily of Fqua ∗ which contains a unique graph of any order n 11. Let q and r< 5 be non-negative Fqua ≥ integers such that n 11=5q + r. Then the n-vertex graph in consists of q middle − Fqua blocks M and each end block is determined by r as follows. If r = 0, then both end blocks are D1. If r = 1, then the end blocks are D1 and D2. If r = 2, then both end blocks are

D2. If r = 3, then the end blocks are D2 and D3. Finally, if r = 4, then the end blocks are D1 and D5. Conjecture 2.2 ([1]). For every n 11, the n-vertex graph of is the unique graph ≥ Fqua with minimum spectral gap among connected quartic graphs of order n.

For every n 1 (mod 5), we denote the n-vertex graph in by Γ . The end blocks ≡ Fqua n of Γn are D1. This is the graph depicted in Figure 4. In the remainder of this section, we recall some basic facts on graph eigenvalues which will be used in our arguments. Let G be a graph and A(G) be its adjacency matrix. It is well known that the smallest eigenvalue ρ(G) satisﬁes the following: xA(G)x ρ(G) = min ⊤ . 0=x Rn 6 ∈ xx⊤

Also if x =(x1,...,xn), then

xA(G)x⊤ =2 xixj, ij XE(G) ∈ where E(G) is the edge set of G. If x is an eigenvector corresponding to an eigenvalue λ of G, then

λxi = xj, for i =1, . . . , n. (1) j: ijXE(G) ∈ We refer to (1) as the eigen-equation. The following lemma gives the well-known eigenvalue interlacing inequalities.

Lemma 2.3 ([8, p. 26]). If G is a graph with eigenvalues λ (G) λ (G), and H 1 ≥···≥ n is an induced subgraph of G with eigenvalues λ (H) λ (H), then 1 ≥···≥ m

λi(G) λi(H) λn m+i(G), for i =1, . . . , m. ≥ ≥ − 5 A partition Π = C ,...,C of the vertex set of G is called an equitable partition { 1 m} for G if for every pair of (not necessarily distinct) indices i, j 1,...,m , there is a ∈ { } non-negative integer qij such that each vertex v in the cell Ci has exactly qij neighbors in the cell Cj, regardless of the choice of v. The matrix Q = (qij) is called the quotient matrix of Π. The following lemma gives a characterization of the eigenvectors of G in terms of Π.

Lemma 2.4 ([8, p. 24]). Let G be a graph with an equitable partition Π. The spectrum of G consists of the spectrum of the quotient matrix of Π, with eigenvectors that are constant on the cells of Π, together with the eigenvalues belonging to eigenvectors whose components sum up to zero on each cell of Π.

3 Simplicity of eigenvalues of minimal regular graphs

In this section, we show that, except for two eigenvalues, namely 0, 1, any eigenvalue − of the graphs in is simple. We also prove that, except for at most eight eigenvalues, Fcub the eigenvalues of the graphs in ∗ are simple. Moreover, the number of exceptional Fqua eigenvalues can be reduced to four if Conjecture 2.2 is true.

3.1 Minimal cubic graphs

Each graph in possesses an equitable partition in which each cell has size 1 or 2, Fcub consisting of the vertices drawn vertically above each other as displayed in Figure 1. We denote this equitable partition by Π in this subsection.

Theorem 3.1. Let G be of order n 10. If n 2 (mod 4), then all the ∈ Fcub ≥ ≡ eigenvalues of G except 1 and 0 are simple. If n 0 (mod 4), then the only non-simple − ≡ eigenvalue of G is 1. − Proof. Let λ be an eigenvalue of G with eigenvector x. In view of Lemma 2.4, x is either constant on the cells of Π or it is orthogonal to each cell of Π. Accordingly, we call x of the ﬁrst or second type, respectively. First, let x be of the ﬁrst type. So x gives rise to an eigenvector for the quotient matrix Q of G (with respect to Π) with the same eigenvalue λ. We show that λ is a simple eigenvalue of Q. One can easily see that due to the path-like structure of the graphs of , Q is a tridiagonal matrix with non-zero subdiagonal entries. We also know Fcub that the quotient matrix of an equitable partition is diagonalizable (see [11, p. 197]). So Q is a diagonalizable tridiagonal matrix with non-zero subdiagonal entries. The submatrix

6 obtained from deleting the first row and the last column of Q λI is an upper triangonal − matrix whose diagonal entries are all non-zero and so it is non-singular. Therefore, the nullity of Q λI is 1. It follows then that the geometric multiplicity of λ is 1 and because − Q is diagonalizable, the algebraic multiplicity of λ is 1 as well. Therefore, λ is a simple eigenvalue of Q. Now, let x be of the second type. In this case, we show that λ 0, 1, ( 1 √5)/2 . ∈ − − ± The components of x sum up to zero on each cell of the partition Π. In particular, the components of x on the cut vertices are zero. Note that x = 0 and we can always find a 6 non-K2 block B such that the components of x are not all zero on it. Firstly, let B be a middle block. The components of x on B are as shown in Figure 5a. Since the vertices i and i + 1 belong to a cell and x is of the second type, we have xi + xi+1 = 0. Also by the eigen-equation, we see that λx = x . It follows that λ = 1. Next, let B be an end i i+1 − block. The components of x on B are as shown in Figure 5b or 5c. In the first case, if x = 0, then λx = 0, so λ = 0. Otherwise, x = 0 and from λx = x , we obtain that 3 6 3 1 6 1 − 1 λ = 1. In the second case, we have λx = x . So if x = 0, then x = 0. Hence from − 3 5 5 6 3 6 λx = x + x and λx = x , we obtain λ2 + λ 1 = 0, that is λ = ( 1 √5)/2. If 5 − 5 3 3 5 − − ± x = 0, then x = λx = 0. Therefore, x = 0 and then by λx = x , we have λ = 1. 5 3 5 1 6 1 − 1 −

xi x x 1 3 x1 x3 x5

0 0 0 0

xi+1 = xi x1 x3 x1 x3 x5 − − − − − − (a) (b) (c)

Figure 5: The blocks of a minimal cubic graph and the components of an eigenvector of the second type

Note that there is no eigenvalue of G with eigenvectors x and y of the first and second type, respectively. Otherwise, x + y is an eigenvector and so it is either of the first or second type. If x + y is of the first type, then all the components of y on the non-cut vertices should be zero. The components of y on cut vertices are already zero, as it is of the second type. This means y = 0, a contradiction. A similar argument works if x + y is of the second type. Therefore, if 0 and ( 1 √5)/2 are eigenvalues of G with − ± eigenvectors of the second type, then they cannot be an eigenvalue of Q. Now, if λ / 0, 1, ( 1 √5)/2 , then x is of the first type. So the multiplicity of ∈ − − ± λ for G is the same as its multiplicity for Q. Thus, λ is a simple eigenvalue of G. The above argument shows that the eigenvalues ( 1 √5)/2 have an eigenvector of second − ± type which come from the end block of Figure 5c. Since there is at most one such an end

7 block in G, the multiplicity of ( 1 √5)/2 is at most 1. Also by the above argument, the − ± eigenvalue 0 of the second type comes from the end block of Figure 5b, and thus when n 0 (mod 4), 0 is a simple eigenvalue. This completes the proof. ≡

3.2 Minimal quartic graphs

Similar to the minimal cubic graphs, members of ∗ have an equitable partition with Fqua cells of size 1 or 2 consisting of the vertices drawn vertically above each other in the building blocks of Figure 3 (with the exceptions that the ﬁrst cell in the blocks D1 and

D5 has three vertices and the ﬁrst cell in the blocks D2 and D4 has four vertices). We denote this equitable partition by Π in this subsection.

Theorem 3.2. Let G ∗ be a graph of order n 11. Then non-simple eigenvalues ∈ Fqua ≥ of G belong to Λ := 2, 0, 1, 1 √2, ( 1 √5)/2 . n− ± − ± − ± o Proof. Let λ be an eigenvalue of G with eigenvector x. In view of Lemma 2.4, x is either constant on the cells of Π or x is orthogonal to each cell of Π. Accordingly, we call x of the first or second type, respectively. First, let x be of the first type. So x gives rise to an eigenvector for the quotient matrix Q of G (with respect to Π) with the same eigenvalue λ. First, suppose that G does not contain D3 or its mirror image. In this case, Q is a tridiagonal matrix with non-zero subdiagonal entries. Further, Q is diagonalizable, and so similar to the proof of Theorem 3.1, all the eigenvalues of Q are simple. Next, assume that one or both end blocks of G are D3. In this case, Q is not tridiagonal anymore. However, we show that the nullity of Q λI is 1. − We first show that λ = 1. For a contradiction, let λ = 1. By the eigen-equation, we 6 can write the components of an eigenvector of λ (as shown in Figure 6a) in terms of x1. It can be seen that x = x = 2x and x = x = x . Hence we must have x = x , 6 21 − 1 8 23 1 9 24 x11 = x26, and so on. In this way, the components of x on the first three middle blocks of G are repeated periodically, see Figure 6b. In particular, the components of x on the cut vertices of G are obtained in terms of x1, as indicated in Figure 7a. Now suppose that the right end block of G is D1, see Figure 7b. Since xn 10 and xn 5 are two consecutive − − cut vertices, according to Figure 7a, one of the following occurs: 4 1 xn 5 = 5xn 10, xn 5 = xn 10, or xn 5 = xn 10. − − − − −5 − − 4 − Again, by the eigen-equation, on the vertices of Figure 7b, we can write these components 5 7 7 in terms of xn. It follows that xn 10 = xn and xn 5 = xn. So xn 5 = xn 10, a − 2 − − 2 − − 5 − 8 x3 x6 x9 x11 x14 x16 x19 x21

x5 x8 x13 x18 x23

(a)

x3 x1 x6 x9 x11 x14 x16 x19 x21 x9 x11 x14 x16 x19 x21

x5 x8 x13 x18 x8 x13 x18 x8

(b)

Figure 6: The graph G with at least one block D3 and the components of an eigenvector of λ =1

x1 x1 5x1 4x1 x1 5x1 4x1 − −

(a)

xn−9 xn−7 xn−4 xn−2 xn−5 xn−10

(b)

Figure 7: The graph G with at least one block D3 and the components of an eigenvector of λ = 1 on cut vertices contradiction. (Note that x , x = 0 since otherwise by the eigen-equation x = 0.) A 1 n 6 similar argument works if the right end block is one of D2,...,D5.

In the following, we assume that both end blocks of G are D3. If G has one D3, it will be enough to use the same argument only on the submatrix corresponding to D3. We have

1 λ 2 1 0  −  2 1 λ 0 1 −    2 0 λ 2   −   0 1 11 λ   −  Q λI =   . −      1 λ 1 1 0   −   2 λ 0 2   −   1 01 λ 2     −   0 1 21 λ  − m m   × 9 Now we perform the following elementary operations on the rows Ri and the columns 2 1 λ 1 Ci as follows: add − R1 to R3, add − R3 to R4, add − Rm 2 to Rm 1, and ﬁnally add 1 λ 4 2 − − 1 − 1− λ Cm to Cm 2. The resulting matrix is − − 1 λ 2 1 0  −  2 1 λ 0 1 −  4 λ2 λ 2   0 1− λ 1− λ− 2   − −   λ2 λ+2 3 3λ   0 0 − −   4 2    Q′ =  .    1 λ 1 1 0   −   λ2 λ 2   2 − − 0 2   1 λ   −   λ2+λ 2   0 − 2 2λ− 1 λ 2   − −     0 0 21 λ   −  2 So rank(Q λI) = rank(Q′). Since λ λ +2 = 0 for real λ, the submatrix obtained by − − 6 deleting the ﬁrst row and the last column of Q′ is upper triangular with non-zero diagonal.

This implies that rank(Q′) m 1 and so the nullity of Q λI is at most 1. Therefore, ≥ − − λ is a simple eigenvalue of Q. Now, let x be of the second type. We show that then λ belongs to Λ. The components of x sum up to zero on each cell of Π. In particular, the components of x on the cut vertices are zero. Note that x = 0 and we can always ﬁnd a block B such that the 6 components of x are not all zero on it. Firstly, let B be a middle block. The components of x on B are as shown in Figure 8. Since the vertices i and i + 1 belong to the same cell in Π and x is of the second type, we have xi + xi+1 = 0. By using the eigen-equation, it can be seen that λx = x . If x = 0, then λ = 1, otherwise x = 0 and from i − i i 6 − i+2 6 λx = x we again obtain λ = 1. Now we examine the end blocks of G. The end i+2 − i+2 − xi xi+2 0 0

xi = xi+1 xi+3 = xi+2 − − Figure 8: A middle block of G and the components of an eigenvector of the second type blocks and the components of x on them are illustrated in Figure 9. Accordingly, one of the following cases (i)–(v) occurs.

(i) B = D . Since x is of the second type, we have x + x + x = 0. If x = 0, then 1 1 2 3 4 6 λx4 = x1+x2+x3 = 0, so λ =0. If x4 = 0, then at least one of x1, x2, x3 must be non-

10 zero. By the symmetry, we may suppose that x = 0. So from λx = x +x = x , 1 6 1 2 3 − 1 we have that λ = 1. − (ii) B = D . We have x + x + x + x = 0. First, suppose x = 0. Then from 2 1 2 3 4 5 6 λx = x + x and λx = x + x , we obtain either λ = 1 or 0 = x = x . If 1 − 1 5 3 − 3 5 − 6 1 3 the latter holds, then λx = x +2x . From λx = x x , then it follows that 5 − 5 1 1 5 − 1 λ2 +2λ 1 = 0, that is λ = 1 √2. Next, suppose x = 0. Then at least one of − − ± 5 x ,...,x must be non-zero. By the symmetry, we may assume that x = 0. Hence, 1 4 1 6 from λx = x + x + x = x , we have λ = 1. 1 2 3 4 − 1 − (iii) B = D . If x = 0, then λx = 0, so λ = 0. Suppose that x = 0. If x = 0, 3 5 6 5 5 1 6 then from λx = x , we have λ = 1. If x = 0, then from λx = x x and 1 − 1 − 1 3 6 − 3 λx = x x , it follows that either λ = 0or 0 = x = x and subsequently 6 3 − 6 6 3 − 6 λ = 2. − (iv) B = D . We have x + x + x + x = 0. If x = 0, then λx = x , so λ = 1. 4 1 2 3 4 7 6 7 − 7 − Now, suppose that x =0. If x = 0, then λx = x + x implies that x = 0 and 7 1 1 − 1 5 5 then x3 = λx5 = 0. Since all the components of x on the ﬁrst cell cannot be zero, we have 0 = x = x . Therefore, λx = x implies that λ = 1. Next, suppose 6 2 − 4 2 − 2 − that x =0. If x = 0, then from λx = x + x , we have λ = 1. If x = 0, then 1 6 5 1 − 1 5 − 5 6 from λx = x + x and λx = x + x , we conclude that λ = 1or0 = x = x . 1 − 1 5 3 − 3 5 − 6 1 3 If the latter holds, then from λx = x + x and λx = 2x , we obtain λ = 1 or 1 − 1 5 5 1 λ = 2. − (v) B = D . We have x + x + x = 0. If x = 0, then λx = x implying that 5 1 2 3 8 6 8 − 8 λ = 1. Thus, assume that x = 0. If all the components of x on the ﬁrst cell are − 8 zero, then from λx = x and λx = x + x , we see that x , x = 0. Now from 4 6 6 − 6 4 4 6 6 λx = x and λx = x + x , it follows that λ2 + λ 1=0, so λ =( 1 √5)/2. 4 6 6 − 6 4 − − ± Otherwise, by the symmetry, we may suppose that x = 0. Hence, from λx = x , 1 6 1 − 1 we obtain λ = 1. −

x3 x3 x x x3 x1 x5 x7 2 4 x1 x5 x1 x6 x2 x4 x6 x8 x x1 0 0 5 0 0 x1 0

x2 x5 x1 x6 x2 x5 x7 x3 x4 x4 − x3 − x4 x3 x4 x6 x8 − − − − − − − − D1 D2 D3 D4 D5

Figure 9: End blocks of G and the components of an eigenvector of the second type

11 Conjecture 2.2, if true, yields an improvement on Theorem 3.2 as follows. From the proof of Theorem 3.2, we observe that, the multiplicity of the eigenvalue 1 is at least − twice the number of the middle blocks of a quartic graph G ∗ . So it can be seen ∈ Fqua that, the multiplicity of the eigenvalue 1 is at least (2n 38)/5. If Conjecture 2.2 is − − true, then the minimal quartic graph G would not contain D4. Also D3 and D5 can only appear at most once as an end block of G. From the proof of Theorem 3.2, and by the fact that G has no eigenvalue with eigenvectors of both types (by a similar argument as in the proof of Theorem 3.1), then it follows that the multiplicity of eigenvalues 1 and ( 1 √5)/2 are at most one. Thus the non-simple eigenvalues of G will be restricted to − ± 0, 1, 1 √2 . − − ± Similarly, for cubic graphs with minimum spectral gap, the multiplicity of the eigenvalue 1 is at least equal to the number of the middle blocks of G. So it is at least − (n 12)/4. Therefore 1 is always an eigenvalue of cubic and quartic graphs with mini- − − mum spectral gap which has the maximum multiplicity.

4 Gap intervals for minimal regular graphs

In this section, we determine the gap intervals in the spectrum of the graph sequences ∆n and Γn. We also specify the limit of the smallest eigenvalues of ∆n and Γn.

4.1 Gap intervals for the spectrum of ∆n

Remark 4.1. Let λ / 1, 0 be an eigenvalue of ∆ with eigenvector x =(x ,...,x ). ∈ {− } n 1 n By the symmetry of ∆n, x′ = (xn,...,x1) is also an eigenvector for λ. By Theorem 3.1,

λ is simple. It follows that x = x′ implying that ±

xi = xn i+1 , for i =1, . . . , n. (2) | | | − |

We start with a lower bound on the smallest eigenvalue of ∆n.

Theorem 4.2. If n 14, then ρ(∆ ) 1 √2. ≥ n ≥ − −

Proof. For convenience, let us set G := ∆n and ρ := ρ(G). If n = 14 or 18, then by direct computation we see that ρ 1 √2. So we suppose that n 22. Since G is not a ≥ − − ≥ union of complete graphs, we have ρ< 1. So, by the proof of Theorem 3.1, ρ has a unit − eigenvector x that is constant on each cell of the equitable partition Π of G and further by Remark 4.1, the components of x satisfy (2).

12 Consider a set of cliques (K2 and K3) of G that covers the edges of G, as shown in Figure 10.

x3 xn−3 x1 xn−1 x5 xn−4

x2 xn x4 xn−2

Figure 10: The edge clique cover of G: each normal line indicates a K2, each dashed triangle is a K3; thick dashed edges are covered by two K3.

Let H be the spanning subgraph of G induced by the thick dashed edges, see Figure 11.

Figure 11: The subgraph H of G

Let M be the vertex-clique incidence matrix of G. So, the rows and columns of M are indexed by the vertices and the non-trivial cliques of G, respectively, and its (i, j)-entry is 1 if i-th vertex belongs to the j-th cilque and 0 otherwise. Also let A and B be the adjacency matrices of G and H, respectively, and C be the matrix whose entries are all zero except for C5,5 = Cn 4,n 4 = 1. Note that the (i, j)-entry of MM ⊤ is equal to the − − inner product of the two rows of M corresponding with the vertices i and j. Each vertex of G belongs to two cliques of the above clique covering except for the vertices 5 and n 4 which appear in three cliques. This implies that the diagonal entries of MM ⊤ and − 2I + C are the same. Now, if ij / E(G), i and j do not appear in any clique together. If ∈ ij E(G) E(H), then i and j appear in exactly one clique together and if ij E(H), ∈ \ ∈ they appear in two cliques together. From this argument, it follows that the oﬀ-diagonal entries of MM ⊤ coincide with those of A + B. It turns out that

MM ⊤ =2I + A + B + C.

Therefore,

ρ = xAx⊤ =(xM)(xM)⊤ 2 xBx⊤ xCx⊤. − − − 2 We have xBx⊤ =2 ij E(H) xixj and since x5 = xn 4 , xCx⊤ =2x5. It follows that ∈ | | | − | P ρ 2 2 x x 2x2. (3) ≥ − − i j − 5 ij XE(H) ∈

13 xs

xr xt

Figure 12: A middle block of G and the components of x

Note that if ij E(H), then x x = x2. Suppose that x , x , and x are the components ∈ i j i r s t of x on the cells of a middle block of G, as depicted in Figure 12. By using the eigen-equation on the vertex s, we have x =(x + x )/(ρ 1). Then we s r t − see that x2 k(x2 + x2), in which k =2/(ρ 1)2. Let S be the set of cut vertices of G, s ≤ r t − so

x x k x2 2kx2 +2x2 i j ≤ i − 5 1 ij XE(H) Xi S ∈ ∈

2 2 2 = k 1 2 xixj 4x  2kx +2x , − − 3 − 5 1 ij XE(H)  ∈  in which the last equality follows from the facts that

2 2 x3 = x4 = xn 3 = xn 2 , 1= x =2 xixj + xi +4x3. | | | | | − | | − | k k ij XE(H) Xi S ∈ ∈ Therefore,

k(1 4x2 2x2)+2x2 x x − 3 − 5 1 . (4) i j ≤ 1+2k ij XE(H) ∈ From (3) and (4), we obtain

2k(1 4x2 2x2)+4x2 ρ 2 − 3 − 5 1 2x2. (5) ≥ − − 1+2k − 5 By the eigen-equation, we see that 1 1 x = (ρ 1)x and x = (ρ2 ρ 4)x , (6) 3 2 − 1 5 2 − − 1 from which it follows that (ρ 1)2(ρ2 ρ 4)2x2 4 ρ 2 − − − 1 . (7) ≥ − − 2(ρ2 2ρ + 5) − ρ2 2ρ +5 − − 2 2 The facts that x = 1, n 22, xi = xn i+1, x1 = x2, and x7 = x8 yield k k ≥ − 1 2x2 +2x2 + x2 + x2 +2x2 + x2 + x2 . (8) 1 3 5 6 7 9 10 ≤ 2 14 Similar to (6), we can also write x6, x7, x9, and x10 in terms of x1. By plugging in all these into (8), we obtain 8 x2 , (9) 1 ≤ f(ρ) where f(t)= t12 4t11 15t10 +64t9 +88t8 364t7 284t6 +840t5+527t4 584t3 245t2 16t+280. − − − − − − − Now from (7) and (9), we get the following inequality in terms of ρ:

g(ρ) > 0, (10) f(ρ)(ρ2 2ρ + 5) − where

g(t)= t15 4t14 14t13 + 74t12 + 17t11 510t10 + 700t9 + 1708t8 4853t7 − − − − 3716t6 + 12026t5 + 6770t4 8061t3 3474t2 40t + 3984. − − − − We observe that both f(t) and t2 2t + 5 have no real zeros and so they are always − positive. So (10) implies that g(ρ) > 0. Also g(t) has a unique real zero which is greater than 2.406. This means that if t 2.406, then g(t) < 0. It follows that − ≤ − ρ> 2.406 > 1 √2. − − − Now, we are ready to prove the main result of this subsection.

Theorem 4.3. (i) The eigenvalues of ∆ lying in the interval 3, √5 converge to n − − (1 √33)/2 as n tends to inﬁnity. In particular, lim ρ(∆n)=(1 √33)/2. n − →∞ −

(ii) ∆n has no eigenvalue in the interval (1, √5].

Proof. Let λ / 1, 0 be an eigenvalue of ∆ . So by the proof of Theorem 3.1, λ has ∈ {− } n an eigenvector x that is constant on each cell of Π. Furthermore, by Remark 4.1, the components of x satisfy (2). We can take x = λ (note that x = 0, otherwise by the 1 1 6 eigen-equation, x = 0). Consider three consecutive middle blocks of ∆n, as depicted in Figure 13, in which the labels of the vertices indicate the components of x. Using the

ai−2 ai−1 ai

bi−2 ci−2 bi−1 ci−1

Figure 13: Three middle blocks of ∆n and the components of x

15 eigen-equation, we obtain

λbi 2 ci 2 2ai 2 =0, λci 2 bi 2 2ai 1 =0, λai 1 ci 2 ai 1 bi 1 =0, − − − − − − − − − − − − − − − − − λbi 1 2ai 1 ci 1 =0, λci 1 bi 1 2ai =0. − − − − − − − − −

From these equations we can write ai 1 in terms of ai 2 and ai as: − −

2(ai 2 + ai) ai 1 = − . (11) − λ3 λ2 5λ +1 − − Let a1,...,am be the components of x on the non-cut vertices of ∆n, as shown in Figure 14. From (11) the following recurrence relation on the components of x follows:

ai = kai 1 ai 2, i =2,...,m +1, − − − (12) a = 1 (λ2 λ), a = 1 (λ5 λ4 7λ3 +3λ2 +4λ), 0 2 − 1 4 − − where k =(λ3 λ2 5λ+1)/2. Note that a and a are obtained using the eigen-equation − − 0 1 on the ﬁrst four cells of ∆n.

λ a0 a1 a2 am−1 am am+1

Figure 14: The cubic graph ∆n with m middle blocks

To solve the recurrence relation (12), we ﬁnd the zeros of its characteristic equation x2 kx +1=0, that is − R =(k √k2 4)/2, S =(k + √k2 4)/2. − − − Now we assume that λ I := ( 3, √5) (1, √5). This implies that k2 4 > 0 and so ∈ − − ∪ − R,S are reals. It turns out that the solution of (12) is λ(λ 1) a = − (P + 1)Rt (P 1)St , (13) t 4 − − in which λ3 + k +7λ +4 λ3 λ2 +9λ +9 P = − = − − . (14) √k2 4 (λ 1)(λ 3)(λ2 5)(λ + 1)2 − − − − m m p Since x satisﬁes (2), a = a +1 if m is even and a m−1 = a m+3 if m is odd. Suppose | 2 | | 2 | | 2 | | 2 | that m =2l is even (if m is odd, then the argument is the same). From (13), we have

(P + 1)Rl (P 1)Sl = ((P + 1)Rl+1 (P 1)Sl+1). − − ± − − Therefore, we have either

(P + 1)Rl(1 R)=(P 1)Sl(1 S), or (P + 1)Rl(1 + R)=(P 1)Sl(S + 1). − − − − 16 We observe that R, P 1, and 1 R do not vanish as λ I. Hence we obtain − ± ∈ P +1 S l 1 S P +1 S l 1+ S = − , or = . (15) P 1 R 1 R P 1 R 1+ R − − − Also since λ I, we see that ∈ S 1 S 1+ S 0 < < 1, 0 < − < 1, and 1 < < 0. (16) R 1 R − 1+ R − P +1 Thus by (15), we have P 1 < 1. On the other hand, by direct computation using (14) we −√ P +1 observe that for λ (1, 5), one has P 1 > 1. This implies that ∆n has no eigenvalue ∈ − √ √ S l 1 S in the interval (1, 5). Now let λ ( 3, 5). From (16), we have R 1±R 0, as l ∈ − − ± → tends to infinity. So, by (15), we must have P +1 0, as n tends to infinity. From (14), → it runs out that (λ 3)(λ2 λ 8)(λ + 1)2 0. As 3 <λ< √5, this occurs only if 1 √33 − − − → − − λ − as n tends to infinity. → 2 To establish that limn ρ(∆n)=(1 √33)/2, it suffices to show that ρ(∆n) < √5. →∞ − − To see this, let H be the induced subgraph on the first seven blocks of ∆n. It is seen that ρ(H) < √5. Thus we are done by interlacing. − It only remains to show that √5 are not eigenvalues of ∆ . For a contradiction, let ± n λ = √5 be an eigenvalue with eigenvector x. The components of x satisfy the recurrence 5 √5 5 3√5 relation (12) with k = 2, a = − , and a = − − . It follows that − 0 2 1 2

1 t at = 5 √5+4√5t ( 1) . 2 − − As before, we may assume that m = 2l is even and so a = a . If a = a , then | l| | l+1| l l+1 ( 1)l(5 + √5+4√5l) = 0 and if a = a , then 2√5( 1)l+1 = 0. Both of which lead − l − l+1 − to contradictions, and so √5 is not an eigenvalue of ∆n. A similar argument works for √5. The proof is now complete. − We conjecture that the gap interval above is indeed maximal.

Conjecture 4.4. (1, √5] is a maximal gap interval for the sequence ∆n.

4.2 Gap intervals for the spectrum of Γn

Remark 4.5. Let λ / Λ (see Theorem 3.2) be an eigenvalue of Γ with eigenvector ∈ n x = (x1,...,xn). By the symmetry of Γn, x′ = (xn,...,x1) is also an eigenvector for λ.

Since λ is simple (by Theorem 3.2), it follows that x = x′, that is ±

xi = xn i+1 , for i =1, . . . , n. (17) | | | − | 17 We ﬁrst establish a lower bound on the smallest eigenvalues of Γn.

Theorem 4.6. If n 11, then ρ(Γ ) 1 √3. ≥ n ≥ − −

Proof. For convenience, let us set G := Γn and ρ := ρ(G). If n = 11, 16, or 21, then by direct computation we see that ρ 1 √3. So we suppose that n 26. Note that ≥ − − ≥ ρ < 2.601 < 1 √2. This follows from interlacing and the fact that the smallest − − − eigenvalue of the induced subgraph on the ﬁrst four blocks of G is less than 2.601. So, − by Theorem 3.2, ρ has a unit eigenvector x that is constant on each cell of Π and by Remrak 4.5, it satisﬁes (17).

Consider a set of cliques (K2,K3, and K4) of G that covers the edges of G, as shown in Figure 15.

x2 x4 xn−4 xn−2

x6 x1 xn

x3 x5 xn−3 xn−1

Figure 15: The edge clique cover of Γn: each normal line indicates a K2, each dashed triangle is a K3, each dotted tetrahedral is a K4; dash-dotted edges are covered by a K3 and a K4.

Let H be the spanning subgraph of G induced by the edges that are covered by two cliques, see Figure 16.

Figure 16: The subgraph H induced by the edges covered by two cliques

Let M be the vertex-clique incidence matrix of G. Also let A and B be the adjacency matrices of G and H, respectively, and C be the matrix whose entries are all zero except for C5,5 = Cn 4,n 4 = 2 and C6,6 = Cn 5,n 5 = 1. Note that the (i, j)-entry of MM ⊤ is − − − − equal to the inner product of the two rows of M corresponding with the vertices i and j. Each vertex of G belongs to two cliques of the clique covering except for the vertices 5 and n 4 that belong to four cliques and the vertices 6 and n 5 that belong to three − − cliques. This implies that the diagonal entries of MM ⊤ and 2I + C are the same. Now, if ij / E(G), then i and j together do not appear in any clique. If ij E(G) E(H), then i ∈ ∈ \ and j appear in exactly one clique together and if ij E(H), they appear in two cliques ∈ together. From this argument, it follows that the oﬀ-diagonal entries of MM ⊤ coincide

18 with those of A + B. So

MM ⊤ =2I + A + B + C. Therefore,

ρ = xAx⊤ =(xM)(xM)⊤ 2 xBx⊤ xCx⊤. − − −

We have xBx⊤ = 2 ij E H xixj and since x5 = xn 4 and x6 = xn 5 , xCx⊤ = ( ) − − 2 2 ∈ | | | | | | | | 4x5 +2x6. It followsP that

ρ 2 2 x x 4x2 2x2. (18) ≥ − − i j − 5 − 6 ij XE(H) ∈ Note that when ij E(H), then x x = x2. Suppose that x , x , x , and x are the ∈ i j i r s t u components of x on the cells of a middle block of G, as depicted in Figure 17.

xs xt xr xu

Figure 17: A middle block of G and the components of x

By using the eigen-equation on the vertices s and t, we have

(ρ 1)x +2x (ρ 1)x +2x x = − r u , and x = − u r . s ρ2 2ρ 3 t ρ2 2ρ 3 − − − − 2 2 2 2 2 2(ρ 1) +8 − It turns out that xs + xt k(xr + xu) in which k = (ρ2 2ρ 3)2 . Let S be the set of cut ≤ − − vertices of G. Then

x x 2k x2 2kx2 i j ≤ i − 6 ij XE(H) Xi S ∈ ∈

2 2 2 =2k 1 2 xixj 6x 4x  2kx , − − 1 − 4 − 6 ij XE(H)  ∈  where the last equality follows from the fact that

1= x =2 x x + x2 +6x2 +4x2. k k i j i 1 4 ij XE(H) Xi S ∈ ∈ Therefore,

2k (1 6x2 4x2 x2) x x − 1 − 4 − 6 . (19) i j ≤ 1+4k ij XE(H) ∈

19 Combining (18) and (19), we get

4k(1 6x2 4x2 x2) ρ 2 − 1 − 4 − 6 4x2 2x2. ≥ − − 1+4k − 4 − 6

Using the eigen-equation, we can write x4, x6 in terms of x1 as 1 1 x = (ρ 2)x , and x = (ρ2 2ρ 6)x . 4 2 − 1 6 2 − − 1 We substitute these and the value of k in terms of ρ in the last inequality to obtain f(ρ)x2 16ρ2 + 32ρ 80 ρ 2+ 1 − − , (20) ≥ − 2ρ4 8ρ3 + 12ρ2 8ρ + 98 − − where

f(t)= t8 +8t7 12t6 36t5 + 27t4 + 236t3 + 254t2 1056t 636. − − − − − The smallest zero of f is larger than 2.034. As ρ < 1 √2, f(ρ), i.e. the coeﬃcient 2 − − − of x1 in (20), is negative. On the other hand, using the facts that n 26, x = 1, 2 2 ≥ k k xi = xn i+1 for i =1,...,n, and x being constant on each cell of Π, we can obtain that − 1 3x2 +2x2 + x2 +2x2 +2x2 + x2 +2x2 . (21) 1 4 6 7 9 11 12 ≤ 2

We can also write x7, x9, x11, and x12 in terms of x1 by using the eigen-equation. Plugging in all these into (21) yields 64 x2 , (22) 1 ≤ g(ρ) where g(t)= t12 8t11 4t10+148t9 103t8 1020t7+714t6+3024t5 960t4 2464t3+1024t2+3712. − − − − − − Now from (20) and (22), we come up with h(ρ) > 0, (23) g(ρ)(ρ4 4ρ3 +6ρ2 4ρ + 49) − − where h(t)= t17 10t16 +10t15 +188t14 494t13 1236t12 +4124t11 +5136t10 6907t9 34850t8 − − − − − 66910t7 +162036t6 +356704t5 185984t4 329408t3 +192576t2 +126592t +532608. − − − We observe that both g(t) and t4 4t3 +6t2 4t + 49 have no real zeros and so they are − − positive for any real t. So (23) implies that h(ρ) > 0. Note that h(t) has a unique real zero which is greater than 2.71. This means that if t 2.71, then h(t) < 0. It follows − ≤ − that ρ> 2.71 > 1 √3. − − − 20 Now, we are ready to prove the main result of this subsection. Theorem 4.7. (i) The eigenvalues Γ lying in the interval [ 4, (1+√17)/2] converge n − − to 1 √13 as n tends to inﬁnity. In particular, lim ρ(Γn)=1 √13. n − →∞ − (ii) Γ has no eigenvalue in the interval [( 1+ √17)/2, 3]. n − Proof. Let λ / Λ (given in Theorem 3.2) be an eigenvalue of Γ . So, by the proof of ∈ n Theorem 3.2, λ has an eigenvector x that is constant on each cell and by Remark 4.5, it satisﬁes (17). We can take x = λ (note that x = 0 since otherwise by the eigen-equation 1 1 6 x = 0). Consider two consecutive middle blocks of Γn, as depicted in Figure 18, in which the labels of the vertices indicate the components of x.

bi−2 ci−2 bi−1 ci−1

ai−2 ai−1 ai

Figure 18: Three middle blocks of Γn and the components of x

Using the eigen-equation, we obtain

(λ 1)bi 2 ai 2 2ci 2 =0, (λ 1)ci 2 2bi 2 ai 1 =0, λai 1 2ci 2 2bi 1 =0, − − − − − − − − − − − − − − − − − (λ 1)bi 1 ai 1 2ci 1 =0, (λ 1)ci 1 2bi 1 ai =0. − − − − − − − − − − − From these equations we can write ai 1 in terms of ai 2 and ai as: − − 4(ai 2 + ai) ai 1 = − . (24) − λ3 2λ2 7λ +4 − − Let a0,...,am be the components of x on the cut vertices of Γ, as shown in Figure 19.

λ a0 a1 am λ

Figure 19: The quartic graph Γn with m middle blocks

From (24), we obtain the following recurrence relation on the components of x:

ai = kai 1 ai 2, i =2, . . . , m, − − − (25) a = 1 (λ3 2λ2 6λ), a = 1 (λ6 4λ5 9λ4 + 32λ3 + 24λ2 24λ), 0 2 − − 1 8 − − − where k =(λ3 2λ2 7λ+4)/4. Note that a and a are obtained using the eigen-equation − − 0 1 on the first five cells of Γn. To solve the recurrence relation (25), we find the zeros of its characteristic equation x2 kx +1=0, that is − R =(k √k2 4)/2, S =(k + √k2 4)/2. − − − 21 Now we assume that λ I := 4, (1 + √17)/2 ( 1+ √17)/2, 3 . So k2 4 > 0 ∈ − − ∪ − − and so R,S are reals. It turns out that the solution of (25) is λ a = (P + Q)Rt (P Q)St , (26) t 4 − − in which λ5 +4λ4 +9λ3 34λ2 14λ + 24 P = − − − , Q = λ2 2λ 6. (27) 4√k2 4 − − − m m Since x satisfies (17), a 1 = a +1 if m is even and a m−1 = a m+1 if m is odd. We | 2 − | | 2 | | 2 | | 2 | suppose that m =2l is even (if m is odd, then the argument is the same). So from (26), we have

l 1 l 1 l+1 l+1 (P + Q)R − (P Q)S − = (P + Q)R (P Q)S . − − ± − − Therefore, we have either

l 1 2 l 1 2 l 1 2 l 1 2 (P +Q)R − (1 R )=(P Q)S − (1 S ), or (P +Q)R − (1+R )=(P Q)S − (1+S ). − − − − Note that R, S, 1 R2, and 1 S2 do not vanish since λ I. ± ± ∈ We further assume that λ ( 1+ √17)/2, 3 . So P + Q = 0 and we have either ∈ − 6 l 1 l 1 P Q R − 1 R2 P Q R − 1+ R2 − = − , or − = . (28) P + Q S 1 S2 P + Q S 1+ S2 − Also R 1 R2 1+ R2 > 1, − < 1, and > 1. (29) S 1 S2 − 1+ S2 − P Q P Q So by (28), we have that P +−Q > 1. However, from (27) it is seen that 1 < P −+Q < 1, a − contradiction. Therefore, Γ has no eigenvalue in the interval ( 1+ √17)/2, 3 . n − Next, assume that λ 4, ( 1 √17)/2 . So P Q = 0 and we have ∈ − − − − 6 l 1 l 1 P + Q S − 1 S2 P + Q S − 1+ S2 = − , or = . (30) P Q R 1 R2 P Q R 1+ R2 − − − Also S 1 S2 1+ S2 0 < < 1, 1 < − < 0, and 0 < < 1. R − 1 R2 1+ R2 − l 1 2 S − 1 S Thus R 1±R2 0, as l tends to infinity. From (30), it then follows that P + Q 0, ± → → where λ5 4λ4 9λ3 + 34λ2 + 14λ 24 P + Q = λ2 2λ 6 − − − . − − − 4√k2 4 − Therefore, (λ 2)(λ 3)(λ 4)(λ2 2 λ 12)(λ +1)2 0. As 4 <λ< (1+ √17)/2, − − − − − → − − we must have λ 1 √13 as n tends to infinity. → − 22 To establish that limn ρ(Γn)=1 √13, it suffices to show that ρ(Γn) < (1 + →∞ − − √17)/2. This is already done in the proof of Theorem 4.6, where we showed that ρ(Γn) < 2.601 < (1 + √17)/2. − − It only remains to show that ( 1 √17)/2 and 3 are not eigenvalues of Γ . For a − ± n contradiction, let ( 1+√17)/2 be an eigenvalue of Γn with eigenvector x. The components − √17 25 3√17+21 of x satisfy the recurrence relation (25) with k = 2, a = − , and a = . It − 0 4 1 4 turns out that ( 1)t at = − √17 25+4t(1 √17) . 4 − − As before, we may assume that m = 2l is even and al 1 = al+1 . If al 1 = al+1, then − − l | | | | l 2(1 √17)( 1) = 0 and if al 1 = al+1, then 25 √17 + 4(√17 1)l ( 1) = 0. − − − − − − − Both of which lead to contradictions, and so ( 1+ √17)/2 is not an eigenvalue of Γ . − n Similar arguments work for ( 1 √17)/2 and 3. So Γ has no eigenvalue in the interval − − n [( 1+ √17)/2, 3]. The proof is now complete. − We conjecture that the gap interval above is indeed maximal.

1+√17 Conjecture 4.8. [ − 2 , 3] is a maximal gap interval for the sequence Γn.

Acknowledgements

The authors thank Peter Sarnak for insightful comments. The second author carried this work during a Humboldt Research Fellowship at the University of Hamburg. He thanks the Alexander von Humboldt-Stiftung for ﬁnancial support.

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