The degree sequence of Fibonacci and Lucas cubes

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1 The degree sequence of Fibonacci and Lucas cubes Sandi Klavžar Faculty of Mathematics and Physics University of Ljubljana, Slovenia and Faculty of Natural Sciences and Mathematics University of Maribor, Slovenia Michel Mollard CNRS Université Joseph Fourier Institut Fourier, BP rue des Maths, St Martin d Hères Cedex, France michelmollard@ujf-grenoblefr Marko Petkovšek Faculty of Mathematics and Physics University of Ljubljana, Slovenia markopetkovsek@fmfuni-ljsi Abstract The Fibonacci cube Γ n is the subgraph of the n-cube induced by the binary strings that contain no to consecutive s The Lucas cube Λ n is obtained from Γ n by removing vertices that start and end ith It is proved that the number of vertices of degree k in Γ n and Λ n is k n 2i i0 k i n k i and k i0 [2 i n 2i 2ik n k i i n 2i ] 2ik n k i, respectively Both results are obtained in to ays, since each of the approaches yields additional results on the degree sequences of these cubes In particular, the number of vertices of high resp lo degree in Γ n is expressed as a sum of fe terms, and the generating functions are given from hich the moments of the degree sequences of Γ n and Λ n are easily computed Key ords: Fibonacci cube; Lucas cube; degree sequence; generating function AMS subject classifications: 05C07, 05A5 Introduction A Fibonacci string is a binary string that contains no to consecutive s The Fibonacci cube Γ n, n 0, is defined as follos Its vertices are all Fibonacci strings

2 of length n, to vertices are adjacent if they differ in precisely one bit In particular, Γ 0 K, Γ K 2, and Γ 2 is the path on three vertices Alternatively, Γ n can be defined as the so-called simplex graph of the complement of the path on n vertices, cf [] Fibonacci cubes ere introduced as a model for interconnection netorks [7] and received a lot of attention afterards For different studies of their structure e refer to [2, 3, 6, 8, 2, 3] These cubes also found an application in theoretical chemistry There, perfect matchings in hexagonal graphs reflect the stability of the corresponding benzenoid molecules and the so-called resonance graphs capture the structure of the perfect matching It is appealing that Fibonacci cubes are precisely the resonance graphs of a special class of hexagonal graphs called fibonaccenes, the result proved in [0] We also mention that Fibonacci cubes led to the concept of the Fibonacci dimension of a graph [] and that they can be recognized in O EG log V G time [5] Lucas cubes form a class of graphs closely related to Fibonacci cubes The Lucas cube Λ n, n 0, is the subgraph of the n-cube induced by Fibonacci strings b b n such that not both b and b n are In particular, Λ 0 Λ K and Λ 2 Γ 2 is the path on three vertices For different aspects of Lucas cubes see [2, 6, 8, 9,, 6] In this paper e are interested in the degree sequence of Fibonacci and Lucas cubes One of our motivations is the fact that several partial results ere previously obtained in order to attack different problems on Fibonacci cubes In the seminal paper [7, Lemma 6] it as observed that the degrees of Γ n are at least n 2/3 and obviously not more than n More than ten years later, a recursive formula for computing the degree of any vertex of Γ n is given in [3] It depends on the recursive structure of Γ n and the value of the integer that represents the given vertex binary number This approach as further developed in [4] here the degrees are used to determine the domination number of the Fibonacci cubes In the main result on the degrees [4, Theorem 26] vertices of degrees n, n, n 2, and n 3 are explicitly described Hoever, the approach in general does not give the number of vertices of Γ n of a given degree, a fundamental property of a given family of graphs For k, n 0 let f denote the number of vertices of Γ n having degree k Then our first main result is: Theorem For all n k 0, f n 2i i k i n k i i0 In the next section e prove Theorem by deriving and solving a corresponding system of linear recurrences Then, in Section 3, several consequences of Theorem are presented A special emphasis is given on vertices of small and large degrees For instance, Corollary 34 in particular covers the degrees of the above-mentioned [4, Theorem 26] In Section 4 e give a direct approach to Theorem by considering degrees via the partition of V Γ n into strings of a given eight In this ay not only 2

3 that Theorem is reproved, but i the vertices of a given degree and eight are enumerated thus giving an additional information on the Fibonacci semilattice [4] and the Lucas semilattice [6] and ii the ay to our second main theorem is paved Denoting by l, k, n 0, the number of vertices of Λ n having degree k, e prove in Section 5: Theorem 2 For all n k 0 ith n 2, l i0 [ i 2 2i k n n 2i k i i 2i k n ] n 2i 2 k i Finally, in Section 6, e reprove Theorem 2 by the method of generating functions This approach is somehat more involved than the one taken in Section 5, hoever it can be further used to obtain several additional properties of the sequence of degrees of the Fibonacci and Lucas cubes 2 Proof of Theorem The vertex set of Γ n naturally decomposes into the sets A n and B n consisting of those strings that start ith a, and those strings that do not start ith a, respectively Hence A 0, B 0 {λ} here λ is the empty string, and for n, A n {α α B n } and B n {0α α A n B n } Since every vertex in A n, n 2, necessarily starts ith 0, A n induces Γ n 2 in Γ n On the other hand, B n induces Γ n in Γ n Moreover, each vertex α of A n has exactly one neighbor in B n, namely 0α We no give the key definition that ill enable us to compute the degree sequence of Γ n For any n and any 0 k n, let a, respectively b, be the number of vertices of A n, respectively B n, of degree k Consider a vertex x A n of degree k Then it is of degree k in the subgraph Γ n 2 of Γ n induced by A n Since x lies in exactly one of the corresponding sets A n 2 and B n 2, e get a a k,n 2 b k,n 2 Similarly, a vertex y B n either has a neighbor in A n if it starts ith 00 or has no neighbor in A n In the first case, it is a vertex of the corresponding set B n, in the second case, a vertex of A n Therefore, b b k,n a Hence the degree sequences in the subgraphs induced by A n and B n satisfy the 3

4 system of linear recurrences and initial conditions a a k,n 2 b k,n 2 k, n 2, 3 b b k,n a k, n, 4 a k,0 a 0,n 0 k 0, n 0, a,, a k, 0 k 2, b 0,0, b k,0 b 0,n 0 k, n Their generating functions Ax, y 0 a x k y n and Bx, y 0 b x k y n therefore satisfy the system of linear algebraic equations hose solution is Ax, y xy xy 2 Ax, y xy 2 Bx, y, Bx, y xybx, y yax, y Ax, y Bx, y xy y xy xy xy 2 xy 3, 5 xy xy 2 xy 3 Write u xy, v xy 2 Then Bx, y uv xy 3 uv xy 3 uv x t y 3t uv t t 0 t t x t y 3t xy i xy 2 j i j t,i,j 0 t t ij x tij y 3ti2j i j t,i,j 0 i t j t x tij y 3ti2j i t t,i,j 0 Note that these geometric and binomial series expansions are valid in a neighborhood of the origin eg, if x, y < /2 then xy 3 /uv < 2/2, xy < /4, xy 2 < /8 No introduce ne summation variables k t i j, n 3t i 2j, t j Then i k, j k n 2, t n k, so Bx, y, 0 n 2 k x k y n, n k hence b n 2 6 k n k 4

5 From 3 and 4 or, alternatively, from 5 e obtain a b k,n b k 2,n 2 b k,n 2 k 2, n 2 7 Denote Then t, n 2 k n k t k,n, t k 2,n 2, t k,n 2, [ ] n 2 n 2 2 t k,n 2, k k 2 n k n 2 2 n 2 2 k n k k n k n 2 2 k n k by using Pascal s identity tice, hence it follos from 6 and 7 that a n 2 2 k n k 0 n 2 k 2, n 2 8 k n k Here e replaced by and noted that n 0 k n k 0 for n It is easy to check that 8 holds hen k {0, } or n {0, } as ell Finally, from 6 and 8 e obtain f a b by using Pascal s identity once more 3 Consequences of Theorem n 2 k n k Let F n be the nth Fibonacci number: F F 2, F n F n F n 2 for n 3 Since V Γ n F n2, Theorem immediately implies: Corollary 3 For any n 0, F n2 n k0 i0 i n k i n 2i k i 9 5

6 We next give an alternative proof avoiding Fibonacci cubes of Corollary 3 Set F n n k i n 2i k0 i0 n k i k i If k > n and i then n k i < 0, thus i n k i 0 If k > n and i 0, or if i > k, then n 2i k i 0 Thus, after interchanging the order of summation and using Vandermonde s convolution, F n i0 k0 i n k i n 2i k i i0 n i n 2i Since for i > n /2 e have n i n 2i 0, e can restrict our summation range to, say, 0 i n, and obtain n n n i n i F n, n 2i i i0 here the last equality holds because n i 0 Using the ell-knon identity m m i i0 i Fm, see [5, p 289, equation 630], e conclude that F n F n2 An interesting problem, useful for applications such as domination or coloring, is to determine for some fixed integer m the number of vertices of degree Γ n m and δγ n m As usually, and δ denote the maximum and the minimum degree When m is small, the folloing to corollaries sho that in both cases almost all the terms in the sum of Theorem vanish Corollary 32 For 0 m n, f n m,n Proof By Theorem, f n m,n m i m/2 n m i0 i0 n 2i i 0 n m i m i n 2i i n m i m i If m i < 0 e have m i 0, thus e can assume that i m If 2i < m e have m i > i and again m i 0 Corollary 33 Let δ δγ n n2 3 For n > 0 and m n δ, f δm,n δm iδ m 2 n 2i i δ m i δ m 2i n 6

7 Proof We have i > 0, thus i n k i i k2i n Rerite the sum in Theorem for k δ m and observe that if i δ m 2 2 then δ m 2i n 3δ m 2 m 2 n 4 3δ n 3 Hence in this case i δm2i n 0 Our last to results in particular give the asymptotic behavior of the number of vertices of degrees Γ n m and δγ n m hen n Corollary 34 Let m 0 and let n 2m 2 Then ; m 0, 2; m, n ; m 2, f n m,n 3n 8; m 3, i m/2 n 2 /2 3n/2 2; m 4, 2n 2 6n 0; m 5 More generally, f n m,n is a polynomial in n of degree m/2 Its leading coefficient is m/2! hen m is even, and m/2 m/2! hen m is odd Proof When i m and n 2m 2 e have n 2i 0, thus n 2i n m i n 2i m i Hence, having in mind Corollary 32, the first values are thus obtained directly from m n 2i i m i m i Consider no this sum for some fixed m For all i, n 2i m i is a polynomial in n ith leading term nm i m i!, and m i is independent of n Thus fn m,n is a polynomial in n Its leading monomial is obtained from the term corresponding to the minimal i such that m i 0, hich is equivalent to 2i m and further to i m/2 Hence the minimal such i is m/2, and deg f n m,n m m/2 m/2 If m is even, then m i hen i m/2 m/2, thus the leading term is m/2! nm/2 If m is odd, then m i m/2 hen i m/2, thus in this case the leading term is m/2 m/2! n m/2 Corollary 35 Let δ δγ n n2 3 Then ; m 0, n 3p, 2 p p 4; m 0, n 3p, p 2; m 0, n 3p 2, f δm,n pp p 8; m, n 3p, 6 20 pp p 3 24p 2 8p 4 ; m, n 3p, 24 p p 2 p 2 5p 2 ; m, n 3p 2 More generally, for all m 0, f δm,n is : 7

8 a polynomial in p of degree 3m and leading coefficient 3m! for n 3p; a polynomial in p of degree 3m2 and leading coefficient 3m2! for n 3p; a polynomial in p of degree 3m and leading coefficient 3m! for n 3p2 Proof The first values are obtained by direct use of Corollary 33 Let m be fixed and consider the general case hen n 3p for some fixed m Then δ p, and by introducing a ne summation variable j i p e can rerite the sum of Corollary 33 as f pm,3p m j m 2 p 2j p j m j 2j m Notice that p 2j m j is a polynomial in p of degree m j, and pj 2jm is a polynomial in p of degree 2j m, therefore their product is of degree 2m j The maximum degree ill be obtained hen j is maximum, ie, j m Then p 2j m j and pj pm, thus the leading term is p 3m 2jm 3m 3m! The cases n 3p and n 3p 2 are treated similarly The maximum degree is obtained hen j is maximum, hich in these to cases is j m When n 3p e have p 2j m j and pj 2jm pm2 3m2, thus the leading term is p 3m2 3m2! When n 3p 2 then p 2j2 m j and pj 2jm pm2 3m, thus the leading term is p3m 3m! 4 A direct approach to Theorem The eight of a binary string is the number of s The vertex set of Γ n or Λ n is naturally partitioned into sets of strings of eight 0,,, n The purpose of this section is to determine the number of vertices in Γ n ith a given eight and degree As a consequence, e are able to give an alternative proof of Theorem as ell as a combinatorial interpretation of the summation expression From this approach e can also describe easily the set of vertices of a given eight and degree, and deduce quickly the degree sequence of Lucas cubes We leave the latter task for the next section and continue here ith the study of the structure of Fibonacci strings For n 0 denote F n the set of all Fibonacci strings of length n, L n the set of all Lucas strings of length n, Sn i,j {α F n ; α starts ith i and ends ith j}, i, j {0, } 8

9 Note that in the notation of Section 2, Sn, Sn,0 A n and Sn 0, Sn 0,0 B n In addition, for any integer m 0 e introduce the folloing Fibonacci strings: α m 0 m 0 β m 0 m γ m 0 m δ m 0 m We call the strings δ m degenerate Fibonacci strings Lemma 4 Every nondegenerate Fibonacci string can be uniquely decomposed as β m0 0 l 0 α m2 0 l 2 α mp 0 lp γ mp, here p 0, l 0,, l p 0, m,, m p, and m 0, m p 0 Moreover, m 0 and m p determine to hich of the sets Sn,, Sn,0, Sn 0,, or Sn 0,0 the string belongs Proof The proof of the existence of such a decomposition is by induction on the length of the string This is clearly true for strings of length n 2 Consider no a string s of length n > 2 Suppose first that s 0s, here s F n By induction, e have the folloing possibilities for s : s β m0 0 l 0 α mp 0 lp γ mp, hence s α m0 0 l 0 α mp 0 lp γ mp ; s 0 l 0 α mp 0 lp γ mp, hence s 0 l 0 α mp 0 lp γ mp ; s α mp 0 lp γ mp, hence s 0 α mp 0 lp γ mp ; s γ m, thus no s 0γ m ; s δ m, and hence s γ m Similarly, if s s, s F n, e have the folloing cases: s 0 l 0 α mp 0 l p γ mp, hence s β 0 l 0 α mp 0 l p γ mp ; s α m2 0 l 2 α mp 0 lp γ mp, hence s β m 0 l α m2 0 l 2 α mp 0 lp γ mp ; s γ m, but then s δ m is degenerate Hence in each of the cases e have obtained a decomposition of s in the expected form It is immediate to verify that strings from Sn, satisfy m 0 > 0 and m p > 0; strings from Sn,0 satisfy m 0 > 0 and m p 0; strings from Sn 0, satisfy m 0 0 and m p > 0; and strings from Sn 0,0 satisfy m 0 0 and m p 0 9

10 To prove uniqueness, consider first a string β m0 0 l 0 α mp 0 lp γ mp from Sn,, thus ith m 0 > 0 and m p > 0 In the three possible cases l 0 > 0, l 0 0 and m > 0, p 0 such a string contains at least to consecutive 0 s, so the string is not degenerate On the other hand, it is clear that a nondegenerate string cannot be decomposed in to ays as β m0 0 l 0 α mp 0 l p γ mp Note also that the degenerate Fibonacci string δ m is of length n 2m, eight m, and the corresponding vertex of Γ n is of degree k m For all the other strings e have: Proposition 42 A Fibonacci string β m0 0 l 0 α m2 0 l 2 α mp 0 l p γ mp is of length n p i0 l i 2 p i0 m i p and eight p i0 m i, and the corresponding vertex of Γ n is of degree k p i0 l i p i0 m i Proof The assertion for the length and the eight follos immediately from definitions As for the degree, use the fact that changing a to 0 in a vertex from F n gives a vertex in F n, hile a 0 can be changed to only if it is not adjacent to, and thus not inside a block of the form α m, β m, and γ m We ill use the folloing classical results about composition of integers Lemma 43 Let a, b 0 Then the number of solutions of x x 2 x a b, ith x, x 2,, x a nonnegative integers, is ba b Lemma 44 Let a, b 0 Then the number of solutions of x x 2 x a b, ith x, x 2,, x a positive integers, is b b a In the rest e ill use some more notation Let s,, s,0, s0,, and s0,0 be the number of vertices of degree k in S, n, S,0 n, S 0, n, and S 0,0 n, respectively Let in addition Sn,, Sn,0, Sn 0,, and Sn 0,0, be the corresponding sets here each vertex is of eight, and let s,, s,0, s0,, and s0,0 be the number of vertices of degree k in these sets, respectively 0

11 Lemma 45 For all integers k, n, s 0,0 2 k n s 0, s,0 s, 2 k n 2 n 2, k n 2 2 k n k n 2 k Proof Assume first that k n A string from Sn 0,0 is decomposable as 0 l 0 α mp 0 l p here p 0, l 0, l, l p 0, and m, m p > 0 By Proposition 42 there is a - mapping beteen Sn 0,0 and the solutions of p n k 0, l 0 l p k ith l 0,, l p 0, m m p ith m, m p A string from Sn,0 is decomposable as β m0 0 l 0 α m2 0 l 2 α mp 0 l p ith p 0, l 0, l, l p 0, m 0, m, m p > 0 Thus there is a - mapping beteen S,0 n and the solutions of p n k 0, l 0 l p k ith l 0,, l p 0, m 0 m p ith m 0,, m p A nondegenerate string from S, n can be decomposed as β m0 0 l 0 α m2 0 l 2 α mp 0 lp γ mp ith p 0, l 0,, l p 0, m 0,, m p > 0 Thus there is a - mapping beteen these strings and the solutions of p n k 0, l 0 l p k ith l 0,, l p 0, m 0 m p ith m 0,, m p Assume that p n k 0, then by Lemmas 43 and 44 the number of solutions of, 2 and 3 are n 2 2k n k, n 2 2k n k, and n 2 2k n 2 k, respectively Assume no that n k < 0 Then there are no solutions of, 2 and 3, thus there are no nondegenerate strings of degree k in Sn 0,0, Sn,0 and Sn, Notice that e have because 0 implies n k < 0, a contradiction Suppose n k 2 Then e can rite 2 k n > 2 k n > 2 k n 2 k n 2 > 0, thus 2k n 2k n 2k n 2 0, 2 3

12 Assume that n k Then 2 k n > 2 k n > 0, therefore 2k n 2k n 0 Consider no n 2 2k n 2 k n 2 k n 2 k k k This number is zero if k > Otherise if k and n 2k it is, hich corresponds to the degenerate string δ k By symmetry e have s 0, s,0 A vertex of eight has degree at least k, thus there are no vertices of degree k in the sets Sn,, Sn,0, Sn 0,, Sn 0,0 if k n is not satisfied It is immediate to verify that the four formulas hold also Let f be the number of vertices of Γ n having degree k and eight Then e have: Theorem 46 For all integers k, n, ith k, n, n 2 f n k k Proof Clearly, f s, s,0 s0, s0,0 Applying Lemma 45 and three times the identity a b a b a b, e arrive at n 2 f 2 k n k Because > 0, e have 2k n n k Theorem immediately follos from Theorem 46 5 Proof of Theorem 2 Let l be the number of vertices of Λ n of degree k and eight, and let l p,q, for p, q {0, }, be the number of such strings in the set Sn p,q Lemma 5 For all n, k, such that n 2, k n and 0 n, l 0,0 s0,0 k,n s,0, l 0, l,0 s0, s,, l, 0 Proof A Lucas string that starts and ends ith 0 can be ritten as 0s, here either s S 0,0 n is of degree k, or s S,0 n is of degree k This gives the first equality Similarly e obtain the second equality, hile the last one is obvious 2

13 Theorem 52 For all n, k, such that n k, 0 and n 2, n 2 n 2 l k n k 2 k n k Proof Assume first that k Since l l 0,0 2l0,, Lemmas 45 and 5 imply that n 2 n 2 l 2 k n k 2 k n k n 2 n k n k 2 k n k Using Pascal s identity e can group the first term ith one half of the third term, the second term ith one half of the fourth term, and the remaining half of the third term ith the remaining half of the fourth term to obtain 4 The only Lucas strings of degree k 0 are λ and 0, hence l 0,n 0 hen n 2 But in this case the right-hand side of 4 evaluates to 0 as ell Theorem 2 no follos immediately from Theorem 52 Corollary 53 Let n The number of vertices of eight n in L n is n l k0 n n n 2 Proof Note first that the result is true hen 0 or n Assume no that and n 2 Then 2k n n k and 2k n n k Hence e obtain from Theorem 52 by Vandermonde s convolution n l k0 n [ ] n 2 n 2 2 n k k n k k k0 n n 2 n 2 n 2 n e have the final expression Using Pascal s identity and n n 2 6 The method of generating functions In this section e approach Theorem 2 using generating functions It is relatively more complicated than the approach from the previous to sections On the other hand, it enables us to obtain many additional results as demonstrated at the end of the section by several examples 3

14 Clearly, F n Sn, Sn,0 Sn 0, Sn 0,0 for n 0, 5 L n Sn,0 Sn 0, Sn 0,0 for n 0, 6 Sn, 0F n 4 0 for n 4, 7 Sn,0 0F n 3 0 for n 3, 8 Sn 0, 0F n 3 0 for n 3, 9 Sn 0,0 0F n 2 0 for n 2 20 Equation 5 shos that V Γ n F n can be partitioned into four blocks hich, by 7 20, induce in Γ n ith n 4 a Γ n 4, a Γ n 3, a Γ n 3, and a Γ n 2, respectively By 5 again, each of these blocks can be further partitioned into four subblocks Sn, 0S, n 40 0S,0 n 40 0S0, n 40 0S0,0 n 40, 2 Sn,0 0S, n 3 0 0S,0 n 3 0 0S0, n 3 0 0S0,0 n 30, 22 Sn 0, 0S, n 30 0S,0 n 30 0S0, n 30 0S0,0 n 30, 23 Sn 0,0 0S, n 2 0 0S,0 n 2 0 0S0, n 2 0 0S0,0 n Proposition 6 The set of those edges of Γ n not contained ithin one of the four blocks in 5 equals 8 i M i here each M i is a perfect matching beteen a subblock and the union of a pair of subblocks of different blocks, as follos see Fig : M is a perfect matching beteen 0S 0,0 n 20 and 0S0,0 n 30 0S0, n 3 0, 2 M 2 is a perfect matching beteen 0S,0 n 20 and 0S,0 n 30 0S, n 3 0, 3 M 3 is a perfect matching beteen 0S 0,0 n 30 and 0S0,0 n 40 0S,0 n 4 0, 4 M 4 is a perfect matching beteen 0S 0, n 30 and 0S0, n 40 0S, n 4 0, 5 M 5 is a perfect matching beteen 0S 0,0 n 20 and 0S0,0 n 3 0 0S,0 n 3 0, 6 M 6 is a perfect matching beteen 0S 0, n 20 and 0S0, n 3 0 0S, n 3 0, 7 M 7 is a perfect matching beteen 0S 0,0 n 30 and 0S0,0 n 40 0S0, n 4 0, 8 M 8 is a perfect matching beteen 0S,0 n 30 and 0S,0 n 40 0S, n 4 0 Proof We need to analyze the external connections of each of the 6 subblocks of Γ n By ay of example e do this for the subblock 0S,0 n 30, in all the other cases the analysis is similar Each string σ 0S,0 n 30 is of the form σ 0τ00 here τ F n 5 So σ is adjacent to precisely one vertex 0τ0 S, n if τ ends ith then 0τ0 0S, n 4 0, otherise 0τ0 0S,0 n 4 0; 4

15 0S, n 4 0 M 8 0S, n 3 0 0S,0 n 4 0 0S,0 n 3 0 0S 0, n 4 0 M 7 0S 0, n 3 0 0S 0,0 n 4 0 0S 0,0 n 3 0 M 4 Sn, Sn,0 M 6 M 3 0S, n 3 0 M 2 0S, n 2 0 M 5 0S,0 n 3 0 0S,0 n 2 0 0S 0, n 3 0 M 0S 0, n 2 0 0S 0,0 n 3 0 0S 0,0 n 2 0 Sn 0, Sn 0,0 Figure : Perfect matchings beteen subblocks and unions of subblocks of Γ n no vertices in Sn 0,, since each vertex of Sn,0 is at distance 2 or more from each vertex of S 0, n ; precisely one vertex in S 0,0 n, namely 00τ00 0S 0,0 n 2 0 When analyzing other subblocks, e find out in a similar ay that each vertex in 0S, n 40 0S,0 n 40 is adjacent to precisely one vertex in 0S,0 n 3 0; each vertex in 0S 0,0 n 30 is adjacent to precisely one vertex in 0S0,0 n 2 0; each vertex in 0S 0,0 n 20 is adjacent to precisely one vertex in 0S,0 n 3 0 0S0,0 n 3 0 Taken together, these facts imply that the external connections of the subblock 0S,0 n 3 0 are precisely the edges of M 5 M 8 ith one endpoint in 0S,0 n 3 0 5

16 It follos from 2 24 and from Proposition 6 that s, s, k 2,n 4 s,0 k 2,n 4 s0, k 2,n 4 s0,0 k 2,n 4 k 2, n 4, s,0 s, k,n 3 s,0 k 2,n 3 s0, k,n 3 s0,0 k 2,n 3 k 2, n 3, s 0, s, k,n 3 s,0 k,n 3 s0, k 2,n 3 s0,0 k 2,n 3 k 2, n 3, s 0,0 s, 2 s,0 k,n 2 s0, k,n 2 s0,0 k 2,n 2 k 2, n 2 Together ith the corresponding initial conditions, this system of recurrences implies that the generating functions S, x, y s, xk y n, 0 S,0 x, y 0 S 0, x, y 0 S 0,0 x, y 0 satisfy the system of linear algebraic equations s,0 xk y n, s 0, xk y n, s 0,0 xk y n S, x, y xy x 2 y 3 x 2 y 4 S, x, y S,0 x, y S 0, x, y S 0,0 x, y, S,0 x, y xy 2 xy 3 S, x, y S 0, x, y x 2 y 3 S,0 x, y S 0,0 x, y, S 0, x, y xy 2 xy 3 S, x, y S,0 x, y x 2 y 3 S 0, x, y S 0,0 x, y, S 0,0 x, y xy y 2 S, x, y xy 2 S,0 x, y S 0, x, y x 2 y 2 S 0,0 x, y hose solution is S, x, y S,0 x, y S 0, x, y S 0,0 x, y xy xy xy xy 2 xy 3, 25 xy 2 xy xy 2 xy 3, 26 xy 2 xy xy 2 xy 3 27 Expanding these rational functions into poer series e obtain s, n 2 2 k n 2 k s,0 n 2 s0, 2 k n k s 0,0 n 2 2 k n k 6,, 28

17 By noting that f s, s,0 s0, s0,0 and by using Pascal s identity repeatedly, e obtain again To recompute l, note that for n 3, L n 0F n 3 0 0F n 0F n 3 0 0S, n 0S,0 n 0S0, n 0S0,0 n Each σ 0F n 3 0 is of the form σ 0τ0 ith τ F n 3 Hence σ is adjacent to precisely one vertex in 0F n, namely 00τ0 0S 0,0 n Conversely, each vertex 00τ0 0S 0,0 n is adjacent to 0τ0 0F n 30 So for k, n 3, l f k,n 3 s, s,0 s0, s0,0 k,n f k,n 3 f s 0,0 k,n s0,0 29 Using and 28, this formula can be shon equivalent to 2 see Appendix A From and the values l 0,0 l 0,, l, 0 it is straightforard to compute the generating functions F x, y 0 Lx, y 0 f x k y n xy xxy 2 xy xy 2 xy 3, l x k y n xy x2 y 2 xxy 3 x 2 xy 4 xy xy 2 xy 3 from hich additional interesting information concerning the degree sequences f n k0 and l n k0 can be obtained easily For instance: Since the generating functions F x, y, Lx, y, S, x, y, S,0 x, y, S 0, x, y, S 0,0 x, y all have xy xy 2 xy 3 xy xy 2 xy 3 x 2 y 3 as their denominator, each of the sequences s {f, l, s,, s,0, s0,, s0,0 } satisfies the same recurrence s s k,n s k,n 2 s k,n 3 s k 2,n 3 for all large enough k and n 2 From y n n 0 k0 n f F x, y x y y y 2 n 0 F n2 y n it follos that V F n F n2, and from y n n 0 k0 n l Lx, y x y 2 y y 2 n 0 L n y n it follos that V L 0 L 0, V L n L n for n 7

18 3 From n 0 y n n k0 kf x F x, y x 2y y y n 0 it follos that EF n nf n 2n F n /5, and from n 0 y n n k0 kl x Lx, y x it follos that EL n nf n 22 yy2 nf n 2n F n y n 5 y y 2 2 n 0 2nF n y n 4 More generally, for each p 0 one can easily compute the generating functions of the sequences of the p-th moments n k0 kp f n0 resp n k0 kp l n0 of the degree sequences f n k0 resp l n k0 from the higher derivatives of F x, y resp Lx, y Since p x p F x, y x n 0 y n n k p f k0 here k p p j0 k j is the p-th falling poer of k, e have n n p y n y n S p,j k j f n 0 k p f k0 n 0 k0 p S p,j j0 n 0 j0 y n n k j f k0 p j0 S p,j j x j F x, y x here S p,j denotes Stirling numbers of the second kind Similarly, n 0 y n n k p l k0 p j0 S p,j j x j Lx, y x A Details of derivation of 2 from 29 on p 7 to be removed after revision From 29 e obtain, using and 28, l n k n 2 k n 2 2 k n k 8 2 k n 2 k n n 2 k n 2 k

19 This formula can be simplified as follos Shifting by in the first sum, and using Pascal s identity 2k n 2k n 2k n in the second one, e have l k n 2 2 k n k [ 2 k n 2 k n hence, by using Pascal s identity 2k n n 2 2 k n k 2 k n 2k n n 2 l 2 2 k n k n k n k [ n 2 n 2 2 k n k k 2 k n 2k n, 2 k n ] ] n 2 k n 2 k n 2 k Finally, by Pascal s identity n 2 k n 2 k n 2 k, e obtain 2 It can be verified directly that this formula holds also hen k 0 or n 2, so it holds for all k 0, n 2 Acknoledgement This ork as supported in part by the Proteus project BI-FR/08-09-PROTEUS- 002 and by the Ministry of Science of Slovenia under the grants P-0297 and P-0294 References [] S Cabello, D Eppstein, and S Klavžar The Fibonacci dimension of a graph submitted arxiv: v [mathco] [2] E Dedó, D Torri, and N Zagaglia Salvi The observability of the Fibonacci and the Lucas cubes Discrete Math, 255-3:55 63, 2002 [3] J A Ellis-Monaghan, D A Pike, and Y Zou Decycling of Fibonacci cubes Australas J Combin, 35:3 40, 2006 [4] S Fontanesi, E Munarini, and N Zagaglia Salvi On the Fibonacci semilattices In Algebras and combinatorics Hong Kong, 997, pages Springer, Singapore, 999, 9

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