# Discrete Mathematics. The signless Laplacian spectral radius of bicyclic graphs with prescribed degree sequences

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2 Y Huang et al / Discrete Mathematics 311 (2011) Let B π denote the class of all bicyclic graphs with bicyclic degree sequence π In this paper, we determine the unique maximal graph in B π with respect to the signless Laplacian spectral radius of G B π Moreover, recall the notion of majorization Let π = (d 1,, d n ) and π = (d,, 1 d n ) be two non-increasing k sequences If i=1 d i k i=1 d i for k = 1, 2,, n 1 and n i=1 d i = n i=1 d i, then the sequence π is said to be majorized by the sequence π, which is denoted by π π Let π and π be two different tree (resp unicyclic) degree sequences with the same order Let G and G be the trees (resp unicyclic graphs) with the largest signless Laplacian spectral radius in G π and G π, respectively In [20] (resp [21]), the author proved that if π π, then µ(g) < µ(g ) Such a theorem is called the signless Laplacian majorization theorem for trees (resp unicyclic graphs) In Section 3, the signless Laplacian majorization theorem is proved to be true for bicyclic graphs As corollaries, all extremal bicyclic graphs having the largest signless Laplacian spectral radius are obtained in the set of all bicyclic graphs of order n (resp all bicyclic graphs with a specified number of pendant vertices, all bicyclic graphs with given maximum degree) 2 Preliminaries In this section, we introduce some definitions and lemmas which are useful in the presentations and proofs of our main results Let G be a graph with a root vertex v 1 V(G) Denote by dist(v, v 1 ) the distance between v V(G) and v 1 Besides, the distance dist(v, v 1 ) is called the height of vertex v in G, denoted by h(v) = dist(v, v 1 ) Definition 21 ([21]) Let G = (V, E) be a graph with a root v 1 V(G) A well-ordering of the vertices is called a breadthfirst-search ordering (BFS-ordering for short) if the following hold for all vertices u, v V : (1) u v implies h(u) h(v); (2) u v implies d(u) d(v); (3) suppose uv, xy E, uy, xv E with h(u) = h(x) = h(v) 1 = h(y) 1 If u x, then v y We call a graph that has a BFS-ordering of its vertices a BFS-graph Lemma 22 ([21]) Let G = (V, E) be a simple connected graph having the largest signless Laplacian spectral radius in G π and f be the Perron vector of Q (G) If V = {v 1,, v n } satisfies f (v i ) f (v j ) for i < j (ie, the vertices of V(G) are denoted with respect to f (v) in non-increasing order), then d(v i ) d(v j ) for i < j Moreover, if f (v i ) = f (v j ), then d(v i ) = d(v j ) Corollary 23 Let G = (V, E) be a simple connected graph having the largest signless Laplacian spectral radius in G π and f be the Perron vector of Q (G) If d(u) > d(v), then f (u) > f (v), where u, v V Proof Suppose f (v) f (u) By Lemma 22, d(v) d(u), a contradiction Lemma 24 ([21]) Let G = (V, E) be a simple connected graph having the largest signless Laplacian spectral radius in G π Then G has a BFS-ordering consistent with the Perron vector f of Q (G) in such a way that u v implies f (u) f (v) From the Perron Frobenius Theorem of nonnegative matrices [15], we have Lemma 25 ([15]) If H is a proper connected subgraph of a connected simple graph G, then µ(h) < µ(g) Suppose that uv E(G) (resp uv E(G)) Denote by G uv (resp G + uv) the graph obtained from G by deleting (resp adding) the edge uv Lemma 26 ([20]) Let G = (V, E) G π and f be a Perron vector of Q (G) (1) Suppose uw i E and vw i E for i = 1,, k Let G = G + vw vw k uw 1 uw k If G is connected and f (v) f (u), then µ(g ) > µ(g) (2) If d(u) > d(v) and f (u) f (v), then there exists a connected graph G G π such that µ(g ) > µ(g) Lemma 27 ([20]) Let G = (V, E) G π and f be a Perron vector of Q (G) Assume that v 1 u 1, v 2 u 2 E, and v 1 v 2, u 1 u 2 E Let G = G + v 1 v 2 + u 1 u 2 v 1 u 1 v 2 u 2 If G is connected and (f (v 1 ) f (u 2 ))(f (v 2 ) f (u 1 )) 0, then µ(g ) µ(g) Moreover, the last equality holds if and only if f (v 1 ) = f (u 2 ) and f (v 2 ) = f (u 1 ) Corollary 28 Let G = (V, E) be a simple connected graph that has the largest signless Laplacian spectral radius in G π and f be a Perron vector of Q (G) Assume that v 1 u 1, v 2 u 2 E, and v 1 v 2, u 1 u 2 E Let G = G + v 1 v 2 + u 1 u 2 v 1 u 1 v 2 u 2 Then the degree sequence of G is π Moreover, (1) if G is connected and f (v 1 ) > f (u 2 ), then f (v 2 ) < f (u 1 ); (2) if G is connected and f (v 1 ) = f (u 2 ), then f (v 2 ) = f (u 1 )

3 506 Y Huang et al / Discrete Mathematics 311 (2011) Proof Obviously, the degree sequence of G is π Now suppose that f (v 2 ) f (u 1 ) in (1), and f (v 2 ) f (u 1 ) in (2) Since (f (v 1 ) f (u 2 ))(f (v 2 ) f (u 1 )) 0, by Lemma 27, we both have µ(g ) > µ(g), a contradiction An internal path of G is a path (or cycle) P with vertices v 1,, v k (or v 1 d(v 2 ) = = d(v k 1 ) = 2, where k > 1 = v k ) such that d(v 1 ), d(v k ) 3 and Lemma 29 ([14]) Let G be a simple connected graph and uv be an edge on an internal path of G If G uv is obtained from G by the subdivision of the edge uv into the edges uw and wv, then µ(g uv ) < µ(g) Let G be a simple connected nontrivial graph and u V(G) Let G(k, l) be the graph obtained from G by attaching two paths uu 1 u 2 u k and uv 1 v 2 v l at vertex u respectively, where k, l 1 Lemma 210 ([13]) If k l 2, then µ(g(k, l)) > µ(g(k + 1, l 1)) Lemma 211 ([11,19]) Let π = (d 1,, d n ) and π = (d 1,, d n ) be two non-increasing graphic sequences If π π, then there exist a series of graphic degree sequences π 1,, π k such that (π =)π 0 π 1 π k π k+1 (=π ), and only two components of π i and π i+1 are different from 1, where i = 0, 1,, k Lemma 212 ([16]) Let m(v) = u N(v) d(u)/d(v) Then d(u)(d(u) + m(u)) + d(v)(d(v) + m(v)) µ(g) max uv E(G) d(u) + d(v) Lemma 213 ([17]) Let G be a connected graph with at least one edge Then µ(g) λ(g) + 1, where λ(g) is the Laplacian spectral radius of G, and is the maximum degree of G The first equality holds if and only if G is bipartite, and the second equality holds if and only if = n 1 3 Main results To begin with, we give a characterization of the degree sequences of connected bicyclic graphs as follows Proposition 31 ([1,11]) Let π = (d 1, d 2,, d n ) be a non-increasing sequence Then π is graphic if and only if n i=1 d i is even and k d i k(k 1) + i=1 n min{k, d i }, where 1 k n (1) i=k+1 Proposition 32 Let π = (d 1, d 2,, d n ) be a positive non-increasing integer sequence with even sum and n 4 If π is connected bicyclic graphic (also called a bicyclic degree sequence), then n i=1 d i = 2n + 2 and (1) holds Let θ(p, q, r) denote the θ-graph, which is obtained from three vertex-disjoint paths (of orders p + 1, q + 1 and r + 1 respectively, where p, q, r > 0 and at most one of p, q, r equals 1) by identifying the three initial (resp terminal) vertices of them Let C(n 1, n 2 ) denote the graph obtained from two cycles C n1 and C n2 (of orders n 1 and n 2 respectively, where n 1, n 1 3) by identifying a vertex of C n1 with a vertex of C n2 For a prescribed non-increasing bicyclic degree sequence π = (d 1, d 2,, d n ) with n 4, by Proposition 32, π should be one of the following four cases Moreover, we construct a special bicyclic graph B π with degree sequence π as follows (1) d n = 2 and d 1 = 4 Then d i = 2 (2 i n 1) Let B π = C(3, n 2) (2) d n = 2 and d 1 = 3 Then d 2 = 3 and d i = 2 (3 i n 1) by Proposition 32 Let B π = θ(1, 2, n 2) (3) d n = 1 and d 2 = 2 Then we have d 1 > 4, d i = 2 (i = 3, 4, 5), and d j 2 (6 j n 1) Denote by B π the connected graph of order n obtained from C(3, 3) and d 1 4 paths of almost equal lengths (ie, their lengths are different at most by one) by identifying the maximum degree vertex of C(3, 3) with one end vertex of each path of the d 1 4 paths (4) d n = 1 and d 2 3 Then d 1 d 2 3, and we use the breadth-first-search method to define a special bicyclic graph B π with degree sequence π as follows Select a vertex v 01 as a root and begin with v 01 of the zeroth layer Put s 1 = d 1 and select s 1 vertices {v 11,, v 1,s1 } of the first layer such that they are adjacent to v 01, and v 11 is adjacent to v 12 and v 13 Thus d(v 01 ) = s 1 = d 1 Next we construct the second layer Put d(v 1i ) = d i+1 (i = 1,, s 1 ) and select s 2 = s 1 d(v i=1 1i) s 1 4 vertices {v 21,, v 2,s2 } of the second layer such that d(v 11 ) 3 vertices are adjacent to v 11, d(v 12 ) 2 (resp d(v 13 ) 2) vertices are adjacent to v 12 (resp d(v 13 )), and d(v 1i ) 1 vertices are adjacent to v 1i for i = 4,, s 1 In general, assume that all vertices of the tth (t 2) layer have been constructed and are denoted by {v t1,, v t,st } Now using the induction

4 Y Huang et al / Discrete Mathematics 311 (2011) Fig 1 B π hypothesis, we construct all the vertices of the (t + 1)st layer Put d(v ti ) = d i+1+ t 1 j=1 s (i = 1,, s t ) and select j s t+1 = s t d(v i=1 ti) s t vertices {v t+1,1,, v t+1,st+1 } of the (t + 1)st layer such that d(v ti ) 1 vertices are adjacent to v ti for i = 1,, s t In this way, we obtain only one connected bicyclic graph with degree sequence π For example, for a given bicyclic degree sequence π = (6, 5, 4, 3, 2, 1 (8) ), B π is the bicyclic graph of order 13 shown in Fig 1 It can be seen that for a prescribed bicyclic degree sequence π, the graph B π constructed above has a BFS-ordering Now we show that B π is the only bicyclic graph that has the largest signless Laplacian spectral radius in B π Lemma 33 Let π = (d 1, d 2,, d n ) be a non-increasing bicyclic degree sequence with d n = 2 Then B π is the only connected bicyclic graph that has the largest signless Laplacian spectral radius in B π Proof Let G = (V, E) be a connected graph that has the largest signless Laplacian spectral radius in B π, and f be the Perron vector of Q (G) By Lemma 24, there exists a BFS-ordering of G with root v 1 such that v 1 v 2 v n, f (v 1 ) f (v 2 ) f (v n ), h(v 1 ) h(v 2 ) h(v n ), and d(v 1 ) d(v 2 ) d(v n ) Let V i = {v V h(v) = i} for i = 0, 1,, h(v n ) (=p) Hence we can relabel the vertices of G in such a way that V i = {v i1,, v i,si } with f (v il ) f (v ij ) f (v rk ) and d(v il ) d(v ij ) d(v rk ) for 0 i < r p, 1 l < j s i, and 1 k s r Clearly, s 1 = d(v 1 ) = d(v 01 ) = d 1 On the other hand, note that d n = n 2 and i=1 d i = 2n + 2 by Proposition 32 Then we need to consider the following two bicyclic degree sequences Case 1 π = (4, 2,, 2), where the frequency of 2 in π is n 1 Hence d(v 01 ) = 4 and d(v iji ) = 2 for 1 i p and 1 j i s i If v 1i v 1j E for all 1 i < j 4, it follows from π = (4, 2,, 2) that we can suppose there exist two cycles (v 01, v 11, w 1,, w n1, v 12, v 01 ) and (v 01, v 13, u 1,, u n2, v 14, v 01 ) without loss of generality, where n 1, n 2 1 and n 1 + n = n Since d(v 01 ) > d(v 11 ), by Corollary 23, we have f (v 01 ) > f (v 11 ) It follows from Corollary 28 that f (v 13 ) > f (w 1 ) Let G 1 = G v 11 w 1 v 13 u 1 + v 11 v 13 + w 1 u 1 Note that G 1 B π, f (v 11 ) f (u 1 ) (u 1 V 2 ) and f (v 13 ) > f (w 1 ), then by Lemma 27, we have µ(g 1 ) > µ(g), a contradiction If there exists an edge v 1i v 1j E (1 i < j 4), according to the degree sequence π = (4, 2,, 2), then G is isomorphic to C(3, n 2), that is, B π Case 2 π = (3, 3, 2,, 2), where the frequency of 2 in π is n 2 Thus d(v 01 ) = d(v 11 ) = 3 and d(v 12 ) = d(v 13 ) = d(v iji ) = 2 for 2 i p and 1 j i s i If v 11 v 12 E (or v 11 v 13 E), then it follows from π = (3, 3, 2,, 2) that G is isomorphic to θ(1, 2, n 2), namely, G = B π If v 11 v 12, v 11 v 13 E, according to the degree sequence π = (3, 3, 2,, 2), we consider the following two subcases Subcase 21 There exist two cycles (v 01, v 12, w 1,, w n1, v 13, v 01 ) and (v 11, v 21, u 1,, u n2, v 22, v 11 ), where n 1, n 2 0 and n 1 + n = n Since d(v 11 ) > d(w 1 ), by Corollary 23, we have f (v 11 ) > f (w 1 ) Note that w 1 v 21, v 11 v 12 E Let G 1 = G v 11 v 21 v 12 w 1 + v 11 v 12 + v 21 w 1 Then G 1 B π Since f (v 11 ) > f (w 1 ) and f (v 12 ) f (v 21 ), by Lemma 27, we get µ(g 1 ) > µ(g) This contradicts G having the largest signless Laplacian spectral radius in B π Subcase 22 There exist two cycles (v 01, v 11, w 1,, w n1, v 12, v 01 ) and (v 01, v 11, u 1,, u n2, v 13, v 01 ), where n 1, n 2 1 and n 1 + n = n Since d(v 11 ) > d(u n2 ), it follows from Corollary 23 that f (v 11 ) > f (u n2 ) Note that w 1 u n2, v 11 v 13 E Let G 1 = G v 11 w 1 v 13 u n2 + v 11 v 13 + u n2 w 1 Then G 1 B π Since f (v 11 ) > f (u n2 ) and f (v 13 ) f (w 1 ) (w 1 V 2 ), then by Lemma 27, µ(g 1 ) > µ(g), which is a contradiction All in all, we conclude that G is isomorphic to B π Lemma 34 Let π = (d 1, d 2,, d n ) be a non-increasing bicyclic degree sequence with d n = 1 and d 2 = 2 Then B π is the only connected bicyclic graph that has the largest signless Laplacian spectral radius in B π Proof Let H be a connected bicyclic graph that has the largest signless Laplacian spectral radius in B π By Proposition 32, we have d 1 > 4 and d i 2 (2 i n) Then H must be the graph of order n obtained from C(n 1, n 2 ) and d 1 4 paths P i of order p i for i = 1, 2,, d 1 4 by identifying the maximum degree vertex of C(n 1, n 2 ) with one end vertex of each P i

6 Y Huang et al / Discrete Mathematics 311 (2011) From Claim 1, we conclude that G contains a θ-graph θ(p, q, r) as its induced subgraph Let x, y be the vertices such that d θ(p,q,r) (x) = d θ(p,q,r) (y) = 3 If v 01 {x, y} (or v 11 {x, y}), then there is a hanging tree on v 01 (or v 11 ), and an edge u 1 u 2 E(C i ) such that u 1, u 2 v 01 and u 1 v 01 E (or u 1 v 11 E) By Lemma 35, f (u 2 ) > f (v 01 ) (or f (u 2 ) > f (v 11 )), which is a contradiction Hence {v 01, v 11 } = {x, y} Claim 2 V(C 1 ) = V(C 2 ) = 3 By contradiction, suppose V(C i ) > 3 (i = 1 or 2) Note that there exists a hanging tree on v 01 (or on v 11, or a hanging tree on v 12 since d n = 1) Take an edge u 1 u 2 E(C i ) such that u 1, u 2 v 01, u 2 v 11, and u 1 v 01 E (or u 1 v 11 E, or u 1 v 12 E) By Lemma 35, f (u 2 ) > f (v 01 ) (or f (u 2 ) > f (v 11 ), or f (u 2 ) > f (v 12 )), a contradiction It follows from Claim 2 that G contains θ(1, 2, 2) as its induced subgraph Finally, we need to show that v 12, v 13 θ(1, 2, 2) By contradiction, suppose v 1h θ(1, 2, 2) (h = 2 or 3) Then there is a hanging tree on v 1h Take an edge u 1 u 2 θ(1, 2, 2) such that u 2 v 01, v 11 and u 1 v 1h E Then by Lemma 35, f (u 2 ) > f (v 1h ), which is a contradiction Therefore, v 11 v 12, v 11 v 13 E Combining this with the BFS-ordering of G given above, we obtain that G = B π The main result of this section follows from Lemmas 33, 34 and 37 Theorem 38 Let π be a bicyclic degree sequence Then B π Laplacian spectral radius in B π is the only connected bicyclic graph that has the largest signless Now the signless Laplacian majorization theorem is proved to be true for bicyclic graphs in the following theorem Theorem 39 Let π and π be two different non-increasing bicyclic degree sequences with the same order If π π, then µ(b π ) < µ(b π ) Proof By Lemma 211, without loss of generality, assume that π = (d 1,, d n ) and π = (d,, 1 d ) n with d i = d i (i p, q), and d p = d 1, p d q = d + q 1 (1 p < q n) Let f be the Perron vector of Q (B π ) By Lemma 24, there exists a BFS-ordering of B π with root v 1 such that v 1 v 2 v n, f (v 1 ) f (v 2 ) f (v n ), h(v 1 ) h(v 2 ) h(v n ), and d(v i ) = d i for i = 1, 2,, n If π = (4, 2,, 2), then π = (5, 2,, 2, 1) or (4, 3, 2,, 2, 1); if π = (3, 3, 2,, 2), then π = (3, 3, 3, 2,, 2, 1) or (4, 3, 2,, 2, 1) or (4, 2,, 2) By Theorem 38, it can be directly checked that there exists a vertex v k (k p, q) such that v k v q E(B π ) and v kv p E(B π ) Let B 1 = B π + v kv p v k v q Then by Lemma 26, µ(b π ) < µ(b 1) Since B 1 B π, it follows from Theorem 38 that µ(b π ) < µ(b 1) µ(b π ) In the following, we may assume that π (4, 2,, 2) and (3, 3, 2,, 2) Note that d q 1, if 5 q n, 2, if 2 q 4 Then d q = d q + 1 2, if 5 q n, Case 1 d q 3, if 2 q 4 2, if 6 q n, 3, if 3 q 5, 4, if q = 2 By Theorem 38, there exists a vertex v k with k > q such that v k v q E(B π ) and v kv p E(B π ) Let B 1 = B π + v kv p v k v q Then by Lemma 26, µ(b π ) < µ(b 1) Since B 1 B π, by Theorem 38, we have µ(b π ) < µ(b 1) µ(b π ) Case 2 q = 2 and d 2 = 3 Then d = 2 2 Subcase 21 d = 1 4 Then d 1 = 3 and n 5 Hence π = (4, 2,, 2), and then π = (3, 3, 2,, 2) Such a subcase has been proved Subcase 22 d > 1 4 Suppose that B π has k pendant vertices It follows that π = (k + 4, 2,, 2, 1,, 1), and then π = (k + 3, 3, 2,, 2, 1,, 1) By Theorem 38 and Lemmas 212 and 213, we obtain that d(u)(d(u) + m(u)) + d(v)(d(v) + m(v)) µ(b π ) max uv E(B π ) d(u) + d(v) k + 5 = (B π ) + 1 < µ(b π ) Case 3 q = 5 and d 5 = 2 Then d = 5 1 Subcase 31 d n = 1 and d 2 = 2 If p = 1, then d > 1 d 1 > 4 and d i = d i = 2 (2 i 4), but there does not exist such graph with degree sequence π by Proposition 32 Hence p = 2, and then d = 2 d = 3 By Theorem 38, v 4 v 5 E(B π ) but v 2 v 4 E(B π ) Let B 1 = B π + v 2v 4 v 4 v 5 Then by Lemma 26, µ(b π ) < µ(b 1) Since B 1 B π, by Theorem 38, µ(b π ) < µ(b 1) µ(b π ) Subcase 32 d n = 1 and d 2 3 By Theorem 38, there exists a vertex u such that uv 5 E(B π ) but uv p E(B π ) Let B 1 = B π + uv p uv 5 Then by Lemma 26, µ(b π ) < µ(b 1) Since B 1 B π, by Theorem 38, µ(b π ) < µ(b 1) µ(b π ) Let B n denote the set of all connected bicyclic graphs of order n Denote by B n (s) the set of all connected bicyclic graphs of order n with s pendant vertices, and B n, the set of all connected bicyclic graphs of order n with maximum degree, respectively

7 510 Y Huang et al / Discrete Mathematics 311 (2011) Corollary 310 Let G B n, where n 4 Then µ(g) µ(b π 1 ) with equality if and only if G is isomorphic to B π 1 with π 1 = (n 1, 3, 2, 2, 1,, 1), where the frequency of 1 in π 1 is n 4 (In other words, B π 1 is obtained from θ(1, 2, 2) and (n 4) K 2 by identifying one of the maximum degree vertices of θ(1, 2, 2) with one vertex of each K 2 ) Proof Let G B n with the non-increasing degree sequence π If π = π 1, then by Theorem 38, µ(g) µ(b π 1 ) with equality if and only if G is isomorphic to B π 1 If π π 1, it is easy to see that π, π 1 are two different bicyclic degree sequences, and π π 1 By Theorems 38 and 39, we have µ(g) µ(b π ) µ(b π 1 ) with equality if and only if G is isomorphic to B π 1 Corollary 311 Let G B n (s), where n 4 (1) If n = 4, then G = θ(1, 2, 2) (2) If s = 0 and n 5, then µ(g) µ(c(3, n 3)) with equality if and only if G = C(3, n 2) (3) If n = 5 and s 1, then µ(g) µ(g ) with equality if and only if G is isomorphic to G, where G is obtained from θ(1, 2, 2) and K 2 by identifying one of the maximum degree vertices of θ(1, 2, 2) with one vertex of K 2 (4) If 1 s n 5, then µ(g) µ(b π 2 ) with equality if and only if G is isomorphic to B π 2 with π 2 = (s+4, 2,, 2, 1,, 1), where the frequency of 2 in π 2 is n s 1, and the frequency of 1 in π 2 is s (In other words, suppose n 5 = sq+t, 0 t < s, and B π 2 is obtained from C(3, 3), t paths of order q + 2, and s t paths of order q + 1 by identifying the maximum degree vertex of C(3, 3) with one end of each path of the s paths) Proof The result of (1) is obvious Now let G B n (s) with the non-increasing degree sequence π If s = 0 and n 5, then π = (4, 2,, 2) or (3, 3, 2,, 2) If n = 5 and s 1, then π = (4, 3, 2, 2, 1) or (3, 3, 3, 2, 1) Note that (3, 3, 2,, 2) (4, 2,, 2), and (3, 3, 3, 2, 1) (4, 3, 2, 2, 1) By Theorems 38 and 39, we obtain the results of (2) and (3) as desired (4) If 1 s n 5, let π = (d 1,, d n ) be the non-increasing degree sequence of G B n (s) Thus d n s > 1 and d n s+1 = = d n = 1 If π = π 2, then by Theorem 38, µ(g) µ(b π 2 ) with equality if and only if G is isomorphic to B π 2 If π π 2, it is easy to see that π, π 2 are two different bicyclic degree sequences, and π π 2 By Theorems 38 and 39, we have µ(g) µ(b π ) µ(b π 2 ) with equality if and only if G = B π 2 Hence the assertion holds Corollary 312 Let G B n,, where n 1 3 Then µ(g) µ(b π 3 ) with equality if and only if G = B π 3, where π 3 is defined as follows: n+2 (1) If 2 n 1, then π3 = (, n + 2, 2, 2, 1,, 1), where the frequency of 1 in π 3 is n 4 n+4 (2) If 3 < n+2 2, then π3 = (,, n 2 + 4, 2, 1,, 1), where the frequency of 1 in π 3 is n 4 (3) If < n+4 3, then let p = log 1 n( 2) 2 ( 1) 4 n ( 1)p 1 [ ( 1) 4] + 2 = Denote by ( 1)r 4 + q, if p = 1, ( 1)r + q, if p 2, where 0 q < 1 Hence π 3 = (,,, q + 1, 1,, 1), where the frequency of in π 3 is m = 1 + r, if p = 1, ( 1) p 2 [ ( 1) 4] r, if p 2 Proof Let π be the non-increasing degree sequence of G B n, If π = π 3, then by Theorem 38, µ(g) µ(b π 3 ) with equality if and only if G = B π 3 If π π 3, it is easy to see that π, π 3 are two different bicyclic degree sequences, and n+2 (1) if 2 n 1, then π π3 = (, n + 2, 2, 2, 1,, 1); n+4 (2) if 3 < n+2 2, then π π3 = (,, n 2 + 4, 2, 1,, 1) Then for (1) and (2), by Theorems 38 and 39, we have µ(g) µ(b π ) µ(b π 3 ) with equality if and only if G = B π 3 (3) If < n+4 3, by Theorem 38, B π3 can be constructed as follows Assume that B π 3 has (p + 2) layers Then there is one vertex in the layer 0 (ie, the root vertex); there are vertices in layer 1; there are ( 1) 4 vertices in layer 2;

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