A Family of Subgraphs of Fibonacci Cubes

August 30, 2020 12:9 112-IJFCS 2050031

International Journal of Foundations of Computer Science Vol. 31, No. 5 (2020) 639–661 c World Scientiﬁc Publishing Company

DOI: 10.1142/S0129054120500318

k-Fibonacci Cubes: A Family of Subgraphs of Fibonacci Cubes

Omer¨ E˘gecio˘glu∗ Department of Computer Science University of California Santa Barbara Santa Barbara, California 93106, USA [email protected]

Elif Saygı Department of Mathematics and Science Education Hacettepe University, Ankara 06800, Turkey [email protected]

Z¨ulf¨ukar Saygı Department of Mathematics TOBB University of Economics and Technology Ankara 06560, Turkey [email protected]

Received 29 November 2019 Accepted 19 February 2020 Published 26 August 2020 Communicated by Oscar Ibarra

Hypercubes and Fibonacci cubes are classical models for interconnection networks with interesting graph theoretic properties. We consider k-Fibonacci cubes, which we obtain as subgraphs of Fibonacci cubes by eliminating certain edges during the fundamental recursion phase of their construction. These graphs have the same number of vertices as Fibonacci cubes, but their edge sets are determined by a parameter k. We obtain properties of k-Fibonacci cubes including the number of edges, the average degree of a vertex, the degree sequence and the number of hypercubes they contain.

Keywords: Hypercube; Fibonacci cube; Fibonacci number.

1. Introduction An interconnection network can be represented by a graph G = (V,E) with vertex set V denoting the processors and edge set E denoting the communication links between the processors. One of the basic models for these networks is the n- dimensional hypercube graph Qn, whose vertices are indexed by all binary strings

∗Corresponding author.

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640 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

of length n and two vertices are adjacent if and only if their Hamming distance is 1. Hsu [4], defined the n-dimensional Fibonacci cubes Γn as an alternative model of computation for interconnection networks and showed that they have interesting properties. Γn is a subgraph of Qn, where the vertices are those without two consecutive 1’s in their binary string representation. Γ0 is defined as Q0, the graph with a single vertex and no edges. Numerous graph theoretic properties of Fibonacci cubes have been studied. In [8], a survey of the some of the properties including representations, recursive construction, hamiltonicity, degree sequences and other enumeration results are given. The induced d-dimensional hypercubes Qd in Γn are studied in [3, 9, 12, 13, 16, 17] and the boundary enumerator polynomial of hypercubes in Γn is considered in [18]. The number of vertex and edge orbits of Fibonacci cubes are determined in [1]. A Fibonacci string of length n which indexes the vertices of Γn is a binary string b b . . . b such that b b = 0 for 1 i n 1. The additional requirement 1 2 n i · i+1 ≤ ≤ − b b = 0 for n 2 defines a subgraph of Γ called Lucas cube Λ , which was also 1 · n ≥ n n proposed as a model for interconnection networks [14]. Other classes of graphs have also been defined along the lines of Fibonacci cubes. For instance subgraphs of Qn where k consecutive 1’s are forbidden are proposed in [5] and Fibonacci (p, r)-cubes are presented in [2]. The generalized Fibonacci cube Qn(f) is defined as the subgraph of Qn by removing all the vertices that contain some forbidden string f [6]. With this formulation one has Γn = Qn(11). In this paper, we consider a special subgraph of Γn which is obtained by eliminating certain edges. These are called k-Fibonacci cubes (or k-Fibonacci graphs) k and denoted by Γn as they depend on a parameter k. The edge elimination which defines k-Fibonacci cubes is carried out at the step analogous to where the fundamental recursion is used to construct Γn from the two previous cubes by link edges. k The eliminated edges in Γn then recursively propagate according to the resulting fundamental construction which now depend on the value of the parameter k. k We obtain properties of Γn including the number of edges, the average degree of a vertex, the degree sequence and the number of hypercubes they contain.

2. Preliminaries

Fibonacci numbers are deﬁned by the recursion fn = fn 1 + fn 2 for n 2, with − − ≥ f0 = 0 and f1 = 1. Similarly, the Lucas numbers Ln are deﬁned by the recursion Ln = Ln 1 + Ln 2 for n 2, with L0 = 2 and L1 = 1. Any positive integer − − ≥ can be uniquely represented as a sum of non-consecutive Fibonacci numbers. This representation is usually called the Zeckendorf or canonical representation. Here we note that this representation corresponds to the Fibonacci strings. For some positive n integer i assume that 0 i fn+2 1. Then i can be written as i = bj fn j+2, ≤ ≤ − j=1 · − where b 0, 1 . This gives that the Zeckendorf representation of i is (b , b , . . . , b ) j ∈ { } P 1 2 n and it corresponds to the Fibonacci string b1b2 . . . bn. Note that here we assume the integer 0 has Zeckendorf representation (0, 0,..., 0). As an example, for n = 4, August 30, 2020 12:9 112-IJFCS 2050031

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7 = 5 + 2 = 1 f + 0 f + 1 f + 0 f gives that the Zeckendorf representation of · 5 · 4 · 3 · 2 7 is (1, 0, 1, 0) and this corresponds to the Fibonacci string 1010. The distance betweenk-Fibonacci two vertices cubes: Au specialand v familyin a connected of subgraphs graph of FibonacciG is defined cubes 3 as the length of a shortest path between u and v in G. For Qn and Γn this distance dimensionalcoincides with hypercube. the Hamming Then its distance vertex set (dH and) which edge is set the can number be written of different as bits in the string representation of the vertices. Let Qn = (V (Qn),E(Qn)) be the n- V (Q ) = 0, 1 n = b b . . . b b 0, 1 , 1 i n dimensional hypercube.n { Then} its{ vertex1 2 setn | andi ∈ edge { set} can≤ be≤ written} as E(Qn) = u, v n u, v V (Qn) and dH (u, v) = 1 . V (Qn) ={{0, 1 } |= b1∈b2 . . . bn bi 0, 1 , 1 } i n { } { | ∈ { } ≤ ≤ } n n 1 Of course V (Qn) = 2 and E(Qn) = n2 − . Similarly, we can write the vertex | E(Q| n) = u, v| u, v| V (Qn) and dH (u, v) = 1 . set and the edge set of Γ{{n = (V}(Γ | n),E∈(Γn)) as } n n 1 Of course V (Q ) = 2 and E(Q ) = n2 − . Similarly, we can write the vertex | n | | n | set and theV (Γ edgen) = setb1 ofb2 Γ. . .= bn (Vb(Γi ),E0, 1(Γ, ))1 asi n with bi bi+1 = 0 { n | ∈n { } n ≤ ≤ · } EV(Γ(Γn)) = = bu,b v . . .u, b v b V (Γ0n,)1 and, d1H (u,i v)n =with 1 . b b = 0 n {{{ 1 2} | n |∈i ∈ { } ≤ ≤ } i · i+1 } Note that theE(Γ number) = u, of v verticesu, v ofV the(Γ Fibonacci) and d ( cubeu, v) =Γn 1is. fn+2. n {{ } | ∈ n H } The fundamental decomposition [8] of Γn can be described as follows: Γn can Note that the number of vertices of the Fibonacci cube Γ is f . be decomposed into the subgraphs induced by the vertices thatn startn+2 with 0 and 10 The fundamental decomposition [8] of Γn can be described as follows: Γn can respectively. The vertices that start with 0 constitute a graph isomorphic to Γn 1 be decomposed into the subgraphs induced by the vertices that start with 0 and− 10 and the vertices that start with 10 constitute a graph isomorphic to Γn 2. This respectively. The vertices that start with 0 constitute a graph isomorphic− to Γn 1 decomposition can be written symbolically as − and the vertices that start with 10 constitute a graph isomorphic to Γn 2. This − decomposition can be writtenΓn symbolically= 0Γn 1 + 10Γ as n 2 . − − Γn = 0Γn 1 + 10Γn 2. In this representation the vertices are labeled− with− the Fibonacci strings. In Figure 1In the this first representation, six Fibonacci cubes the vertices are presented. are labeled The with labeling the Fibonacci of the vertices strings. follows In Fig. the 1 fundamentalthe first six recursion. Fibonacci For cubesn are2, presented. Γn is built The from labeling Γn 1 and of the Γn vertices2 by identifying follows the ≥ − − Γnfundamental2 with theΓ recursion.n 2 in Γn For1 byn what2, Γwen canis built call fromlink edges. Γn 1 Theand labels Γn 2 by of Γ identifyingn 1 stay − − − ≥ − − − unchangedΓn 2 with in the Γn Γ.n The2 in labels Γn 1 ofby the what vertices we can in call Γn link2 (whichedges. are The circled labels in of Figure Γn 1 stay 1) − − − − − areunchanged shifted up in by Γnf.n The+1 in labels Γn. of the vertices in Γn 2 (which are circled in Fig. 1) are − shifted up by fn+1 in Γn.

Fig. 1. Fibonacci cubes Γ0, Γ1,..., Γ5. Fig. 1. Fibonacci cubes Γ0, Γ1,..., Γ5.

Throughout the paper we use the Fibonacci string representation and integer representation of the vertices interchangeably. For example, the ﬁrst 8 vertices of Γ5 August 30, 2020 12:9 112-IJFCS 2050031

642 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

Throughout the paper we use the Fibonacci string representation and integer representation of the vertices interchangeably. For example, the ﬁrst 8 vertices of Γ5 are the ones with labels 0, 1,..., 7 whose string representations are 00000, 00001, 00010, 00100, 00101, 01000, 01001, 01010. In terms of the adjacency matrix An of Γn, the link edges are the edges corresponding to the identity matrices that appear in the matrix decomposition of An as shown on the left in Fig. 3.

3. k-Fibonacci Cubes In this section we introduce k-Fibonacci cubes, which are special subgraphs of Fibonacci cubes. We will indicate the dependence on k by a superscript and denote k these graphs by Γn. k For k 1, Γn is defined for n 0 by removing certain edges of Γn. We let k ≥ k ≥k k k Γ0 = Γ0 and Γ1 = Γ1. We define Γn for n 2 in terms of Γn 1 and Γn 2, in a ≥ − − manner that is similar to the fundamental decomposition of Γn. The difference is as follows:

In the construction of Γn from 00Γn 2 and its copy 10Γn 2 in Γn 1, there − − − are fn link edges uv with u = 00α 00Γn 2 and v = 10α 10Γn 2, which ∈ − ∈ − are added to the edge set for every Fibonacci string α of length n 2. The − various α here are the Fibonacci representation of the integers 0, 1, . . . , f n − 1, which are the vertices of Γn 2. − k k k k In our construction of Γn from 00Γn 2 and its copy 10Γn 2 in Γn, we − k − only include the link edges uv with u = 00α 00Γn 2 and v = 10α k ∈ − ∈ 10Γn 2 corresponing to the Fibonacci strings α that are the representations − k k of 0, 1, . . . , k 1 in Γn 2. In other words, only the ﬁrst k vertices in 00Γn 2 − k − k − and its copy 10Γn 2 in Γn contribute link edges, and the remaining link − edges that would have been included in the construction of Γn are discarded.

As long as f k, the construction for the k-Fibonacci cubes is identical to the n ≤ construction of the Fibonacci cubes with the same initial graphs, and therefore for f k we have Γk = Γ . Let n = n (k) be the smallest integer for which f > k. n ≤ n n 0 0 n0 For a given k, n (k) is the smallest integer n for which Γk = Γ . First few values 0 n 6 n are n0(1) = 3, n0(2) = 4, n0(3) = 5, n0(4) = 5. Using the Binet formula for the Fibonacci numbers, we calculate that n0 = n0(k) is given explicitly by 1 n = 1 + log √5k + √5 (1) 0 φ − 2 1+√5 where φ = 2 is the golden ratio. This sequence starts as 3, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10,...

in which the lengths of the runs are the Fibonacci numbers. February 18, 2020 22:15 WSPC/INSTRUCTION FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

k-Fibonacci cubes: A special family of subgraphs of Fibonacci cubes 5 August 30, 2020 12:9 112-IJFCS 2050031 k We remark that an equivalent deﬁnition of the graphs Γn could be made by starting the construction from the two Fibonacci cubes Γn 1 and Γn where n0 is 0− 0 the smallest integer for which fn0 > k, and then constructing only k link edges for k the construction of the Γn from the two previous ones after that. k k In Figure 2 we show the construction of the k-Fibonacci cubes Γ4 from Γ3 and Γk for k = 1, 2, 3. 2 k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 643

Fig. 2. The construction of Γk k from Γk k and Γk k for k = 1, 2, 3. Fig. 2. The construction of Γ4 4from Γ3 3and Γ2 2for k = 1, 2, 3.

k We remark that an equivalent definition of the graphs Γn could be made by Similar to the adjacency matrix An of Γn we can write the adjacency matrix startingk k the constructionk from the two Fibonacci cubes Γn0 1 and Γn0 where n0 is An of Γn. In Γn, we have k link edges instead of the fn link− edges of Γn. Hence in thek smallest integer for which fn0 > k, and then constructing only k link edges for An the link edges are thek edges corresponding to the k k identity matrices that the construction of the Γn from the two previousk ones after× that. appear in the matrix decomposition of An shown on the right in Figurek 3.k k In Fig. 2 we show the construction of the k-Fibonacci cubes Γ4 from Γ3 and Γ2 for k = 1, 2, 3. Examples for small k Similar to the adjacency matrix A of Γ we can write the adjacency matrix For k = 1, the graphs Γ1 are all trees.n Ifn we think of them as rooted at the all Ak of Γk . In Γk , we have knlink edges instead of the f link edges of Γ . Hence in zeron vertex,n thenn the next tree is obtained by making then root of the previousn tree Ak the link edges are the edges corresponding to the k k identity matrices that an child of the root of the present tree, i.e. by making the× previous tree a principal appear in the matrix decomposition of Ak shown on the right in Fig. 3. subtree of the current one. Note that thisn is different from the usual definition of FibonacciExamples fortrees, small wherek traditionally the two previous trees are made left and right principal subtrees of a new1 root vertex. For k = 1, the graphs Γn are all trees. If we think of them as rooted at the all zero vertex, then the next tree is obtained by making the root of the previous tree a child of the root of the present tree, i.e. by making the previous tree a principal subtree of the current one. Note that this is different from the usual definition of Fibonacci trees, where traditionally the two previous trees are made left and right principal subtrees of a new root vertex. The number of vertices in Γ1 is Γ = f . The height h of Γ1 satisfies h = 0, n | n| n+2 n n 0 h1 = 1 and hn = min hn 1, 1+hn 2 . Therefore the height is given by hn = n/2 . { − − } d e February 18, 2020 22:15 WSPC/INSTRUCTION FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

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6 Omer¨ E˘gecio˘glu,Elif Saygı, Z¨ulf¨ukarSaygı 644 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

Fig. 3. Left: the structure of the adjacency matrix An of the Fibonacci cube Γn in terms of An 1 − and An 2 in accordance with the fundamental decomposition. Here I is the fn fn identity − × k matrix, and the remaining elements are zeros. Right: the structure of the adjacency matrix An of Γk in terms of Ak and Ak where I is now the k k identity matrix, and the remaining n n 1 n 2 k × elements are zeros. − −

1 1 The number of vertices in Γn is Γn = fn+2. The height hn of Γn satisﬁes h0 = 0, Fig. 3. Left: the structure of the adjacency| | matrix An of the Fibonacci cube Γn in terms of hFig.1 = 3. 1 Left: and thehn structure= min h ofn the1, adjacency1+hn 2 matrix. ThereforeA of the the Fibonacci height is cube given Γ byin termshn = ofn/A 2 . An 1 and An 2 in accordance with the fundamentaln decomposition. Here I is then fn fn identityn 1 − − { − − } × d −e matrix,andFigureAn and2 in the4shows accordance remaining theﬁrst with elementssix thek fundamental are-Fibonacci zeros. Right: decomposition. cubes the (trees) structure Herefor ofk theI=is 1.adjacency the Inf then Appendix matrixfn identityAk − × nk matrix,k and the remainingk elementsk are zeros. Right: the structure of the adjacency matrix An ofwe Γn redrawin terms these of An trees1 and inA anothern 2 where formIk is (Figure now the 7).k Notek identity that matrix, our k-Fibonacci and the remaining cubes of Γk in terms of Ak− and Ak− where I is now the k× k identity matrix, and the remaining elementsn are zeros. n 1 n 2 k × elements are zeros. − −

The number of vertices in Γ1 is Γ = f . The height h of Γ1 satisﬁes h = 0, n | n| n+2 n n 0 h1 = 1 and hn = min hn 1, 1+hn 2 . Therefore the height is given by hn = n/2 . { − − } d e Figure 4 shows the ﬁrst six k-Fibonacci cubes (trees) for k = 1. In the Appendix we redraw these trees in another form (Figure 7). Note that our k-Fibonacci cubes

Fig. 4. The ﬁrst six k-Fibonacci cubes for k = 1. Fig. 4. The ﬁrst six k-Fibonacci cubes for k = 1.

Fig. 4 shows the first six k-Fibonacci cubes (trees) for k = 1. In the Appendix wefor redrawk = 1 are these the trees “Fibonacci in another Trees” form as (Fig.given 7). in Hsu’s Note thatoriginal our paperk-Fibonacci on Fibonacci cubes forcubesk = [4]. 1 are the “Fibonacci Trees” as given in Hsu’s original paper on Fibonacci 2 cubesFor [4].n 3, Γn consists of fn 1 squares (Q2’s, or 4-cycles) glued by their edges ≥ 2 − andForfn n1 pendant3, Γn consists vertices of infn the1 manner squares shown (Q2’s, in or Figure 4-cycles) 8 in glued the Appendix. by their edges − ≥ 3 Fig. 4. The first− six k-Fibonacci cubes for k = 1. 4 andForfn 1npendant4, Γn verticesis constructed in the manner from f shownn+1 in2 Fig. squares 8 in and the Appendix. for n 4, Γn is − ≥ 3 − ≥ 4 constructedFor n from4, Γ 2fisn constructed3 squares in from thef manner2 shown squares in Figure and for 9n and Figure4, Γ 10is ≥ n − n+1 − ≥ n constructedin the Appendix, from respectively.2fn 3 squares in the manner shown in Figs. 9 and 10 in the for k = 1 are the “Fibonacci− Trees” as given in Hsu’s originalk paper on Fibonacci Appendix,We consider respectively. the number of induced hypercubes in Γn in Section 4.4 in more detail.cubesWe [4]. consider the number of induced hypercubes in Γk in Sec. 4.4 in more detail. 2 n For n 3, Γn consists of fn 1 squareskk (Q2’s, or 4-cycles) glued by their edges ToTo demonstrate demonstrate≥ the the structure structure− of of Γ Γnn forfor other other values values of of nn andand kk,, the the graphs graphs of of and44 fn 1 1313pendant vertices in the manner shown in Figure 8 in the Appendix. ΓΓ1212 andand− Γ Γ1212 areare depicted depicted in in Fig. Figure 11 11in the in the Appendix. Appendix. For n 4, Γ3 is constructed from f 2 squares and for n 4, Γ4 is ≥ n n+1 − ≥ n constructed from 2fn 3 squares in the manner shown in Figure 9 and Figure 10 − k 4.in theBasic Appendix, Properties respectively. of k-Fibonacci Cubes Γn k In thisWe section, consider we the present number the of basic induced properties hypercubes of k-Fibonacci in Γn in Section cubes. The 4.4 in results more includedetail. the number of edges, the average degree of a vertex, degree sequence and k the numberTo demonstrate of hypercubes the structure contained. of Γn for other values of n and k, the graphs of 4 13 Γ12Byand definition Γ12 are ofdepicted Γk we in know Figure that 11V in(Γ thek ) = Appendix.V (Γ ) = f . n | n | | n | n+2 August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 645

4.1. The number of edges of k-Fibonacci cubes

Let E(n) denote the number of edges of Γn and Ek(n) denote the number of edges k of Γn. E(n) is given by (see, [4, Lemma 5]) 1 E(n) = (2(n + 1)f + nf ). (2) 5 n n+1

Clearly, Ek(n) = E(n) for n < n0 and Ek(n) satisﬁes the recursion

E (n) = E (n 1) + E (n 2) + min k, f . (3) k k − k − { n} For n n the recursion in (3) reduces to ≥ 0 E (n) = E (n 1) + E (n 2) + k. (4) k k − k − Then by induction on n and using the recursion (4), we directly obtain the following result.

k Proposition 1. The number of edges of Γn is given by E (n) = (k + E(n 2))f + (E(n 1) E(n 2))f k, k 0 − t+3 0 − − 0 − t+2 − where t = n n . − 0

By using the classical identity Lt = ft 1 + ft+1 of the Lucas numbers, along with − Proposition 1 and (2) we obtain the following result.

Corollary 2. For n n ,E (n) is given in closed form by ≥ 0 k 1 1 E (n) = (L + 3f )E(n 1) + (L + f )E(n 2) + (L + 2f 1)k k 2 t t 0 − 2 t t 0 − t t − 1 = (n0fn 1Lt+1 + (n0 1)fn Lt+2) + (ft+3 1)k 5 0− − 0 − where t = n n . − 0

k 4.2. Average degree of a vertex in Γn In [10] it is calculated that

2 2E(n) (2(n + 1)fn + nfn+1) 1 lim = lim 5 = 1 . n n →∞ nfn+2 →∞ nfn+2 − √5

It follows that the average degree of a vertex in Γn is asymptotically given by 1 1 n 0.553n. (5) − √5 ≈ k Now we consider the analogous problem for the k-Fibonacci cubes Γn for a ﬁxed k k and prove that the average degree of a vertex in Γn is independent of n. August 30, 2020 12:9 112-IJFCS 2050031

646 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

k Proposition 3. For a ﬁxed k the average degree of a vertex in Γn is asymptotically given by 1 1 1 (3 + √5) + 1 log √5k + √5 5 − √5 φ − 2 1 1.047 + 0.553 log √5k + √5 , ≈ φ − 2 1+√5 where φ = 2 is the golden ratio.

Proof. By the properties of the Fibonacci numbers we have

fn n0+3 1 n fn n0+2 n lim − = φ − 0 , lim − = φ− 0 . (6) n n →∞ fn+2 →∞ fn+2 k For k ﬁxed, using Proposition 1 and (6), the average degree of a vertex in Γn is given by

2Ek(n) n0 lim = 2φ− ((k + E(n0 2))φ + E(n0 1) E(n0 2)) n →∞ fn+2 − − − −

n0 n0 n0 1 = 2φ− φk + (5 √5)fn 1 + − fn (7) 10 − 0− √5 0 where we used the expression for E(n) in (2) for n = n 1 and n = n 2. To 0 − 0 − obtain the rate of growth of the exact formula for the asymptotic average degree given in (7), we further use the approximations φn0 1 fn , n0 1 + log √5k + √5 . 0 ≈ √5 ≈ φ − 2 Using these in the formula (7) gives the approximation for the average degree of a k vertex in Γn as desired.

k In Proposition 3, we show that the average degree of a vertex in Γn is independent of n, to illustrate this we compute asymptotic values of the average degree of k a vertex in Γn for some values of k in Fig. 5, where 1 k 300. 4 ≤ ≤ It is shown in [4] that Γn contains about 5 the number of edges of the hypercube for the same number of vertices. The analogous ratio goes to zero with increasing k n for Γn since the average degree is independent of n. We can calculate the rate of convergence by using the asymptotic expressions that appear in the proof of k Proposition 3. For ﬁxed k, the number of edges of Γn is about 1 (3 + √5) + 1 (5 √5) log (√5k + √5 1 ) 5 5 − φ − 2 log2 fn+2 1 1 1.508 + 0.796 log √5k + √5 ≈ n φ − 2 times the number of edges of Γn. August 30, 2020 12:9 112-IJFCS 2050031 February 18, 2020 22:15 WSPC/INSTRUCTION FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

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Fig. 5. Computational asymptotic values of the average degree of a vertex in Γk as a function Fig. 5. Computational asymptotic values of the average degree of a vertex in Γk asn a function of of k, 1 k 300. n k, 1 k≤ 300.≤ ≤ ≤ 4.3. Degree polynomials of Γk nk 4.3. Degree polynomials of Γn We start with the following known result for Γn that will be useful in our calcula- tions.We start For with the degree the following sequence known of Γk resultwe need for aΓn refinementthat will be of this useful theorem. in our calcula- nk tions. For the degree sequence of Γn we need a refinement of this theorem. Theorem 4 ([11]). For all n m 0 the number of vertices of Γn having degree Theorem 4 ([11]) For all n ≥m ≥ 0 the number of vertices of Γ having degree m is given by ≥ ≥ n m is given by s s n 2i i + 1 fn,m = n − 2i i + 1 , f = m− i n m i + 1 , n,m i= (n m)/2 m − in − m − i + 1 i=d(nX−m)/2e − − − where s = min(m, n m). d X− e where s = min(m, n − m). − We first define the bivariate degree enumerator polynomial Dk (x, y) for Γk in We first define the bivariate degree enumerator polynomial Dnk (x, y) for Γnk in which the degrees of the k vertices with labels 0, 1, . . . , k 1 aren kept track ofn by which the degrees of the k vertices with labels 0, 1, . . . , k − 1 are kept track of by the variable y, while the others are kept track of by the variable− x. the variable y, while the others are kept track of by the variable x. More precisely More precisely k 1 fn+2 1 k− 1 fn+2− 1 k − di − di Dnk(x, y) = y di+ x di, (8) Dn(x, y) = y + x , (8) i=0 i=k i=0 XX XXi=k where di is the degree of the vertex labeled i. This polynomial can be seen as a where di is the degree of the vertex labeled i. This polynomial can be seen as a refinement of the boundary enumerator polynomial Dn,0 given in [18,[18] Section. 3]. 2 2 2 Dn,0 As an example consider consider Γ Γ24 == Γ Γ23++ Γ Γ22asas shown shown in in Figure Fig. 6 6 in in which which the the first 4 3 2 2 2 2 k = 2 vertices are circled. From (8), we have D22(x, y) = y 2 + y + x and D32(x, y) = 3 2 2 2 2 2 23 y 3 + y 2 + 2x2 + x. Using D22(x, y) and D32(x, y) we can now develop D42(x, y) as y + y + 2x + x. Using D2(x, y) and D3(x, y) we can now develop D4(x, y) as follows:follows: By the definition of Γ2 there are k = 2 link edges between Γ2 and Γ2 . The first By the definition of Γ4 there are k = 2 link edges between Γ3 and Γ2. The first 2 2 one is the edge between the vertices labeled 0 Γ3 and 0 Γ2,, whereas whereas the the second second ∈ 3 ∈ 2 ∈ 2 ∈ 2 one is the edge between the vertices labeled 1 Γ3 and 1 Γ2.. Therefore, Therefore, only only the the ∈ 3 ∈ 2 degrees of these first 2 vertices in Γ2 and first∈ 2 vertices in∈ Γ2 increase by 1 in Γ2 . degrees of these first 2 vertices in Γ3 and first 2 vertices in Γ2 increase by 1 in Γ4. August 30, 2020 12:9 112-IJFCS 2050031 February 18, 2020 22:15 WSPC/INSTRUCTION FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

64810 OmerO.¨¨ E˘gecio˘glu,E. E˘gecio˘glu,Elif Saygı Saygı, & Z. Z¨ulf¨ukarSaygı Saygı

Fig. 6. Construction of Γ22 from Γ22 and Γ22. Fig. 6. Construction of Γ44 from Γ22 and Γ33.

The degrees of the other vertices remain the same. The vertices labeled 0, 1 Γ2 The degrees of the other vertices remain the same. The vertices labeled 0, 1∈ Γ42 are the vertices 0, 1 Γ2 whose degree information in D2(x, y) should be kept track∈ 4 are the vertices 0, 1∈ Γ32 whose degree information in D42(x, y) should be kept track of by the variable y.∈ Therefore3 we can write, 4 of by the variable y. Therefore we can write, 2 2 2 2 2 D42(x, y) = yD32(0, y) + D32(x, 0) + xD22(0, x) + D22(x, 0) D4(x, y) = yD3(0, y) + D3(x, 0) + xD2(0, x) + D2(x, 0) 4 3 3 2 ==yy4++yy3++xx3++ 3 3xx2++ 2 2x,x, andand this this can can also also be be seen seen from from Fig. Figure 6 using 6 using (8). (8). ByBy generalizing generalizing the the above above idea idea for for fixed fixedkk,, we we have have the the following following result result on on degree degree k k enumerator polynomial Dnk(x, y) for Γnk. enumerator polynomial Dn(x, y) for Γn. k k k Theorem 5. Assume that Dn0 1(x, y) = pn0 1(y) + qn0 1(x) and Dnk0 2(x, y) = Theorem 5. Assume that Dn0− 1(x, y) = pn0− 1(y) + qn0− 1(x) and Dn0− 2(x, y) = − − − k − k pn0 2(y) + qn0 2(x). Then the degree enumerator polynomial Dnk(x, y) for Γnk sat- pn0− 2(y) + qn0− 2(x). Then the degree enumerator polynomial Dn(x, y) for Γn sat- − − isfiesisfies the the recursion recursion k t+1 Dnk(x, y) = y t+1pn0 1(y) + ft+2qn0 1(x) + ft+1qn0 2(x) Dn(x, y) = y pn0− 1(y) + ft+2qn0− 1(x) + ft+1qn0− 2(x) − − − tt ii ++fftt+1+1xpxpnn0 22((xx)) + +xpxpnn0 11((xx)) fftt iixx 0−− 0−− −− ii=0=0 XX forfornn nn0 withwithtt==nn nn0.. ≥≥ 0 −− 0 Proof. We know that Γkk = 0Γkk + 10Γkk and there are k link edges between Proof. We know that Γnn = 0Γnn 11 + 10Γnn 22 and there are k link edges between k k −− −− ΓΓnk 1 andand Γ Γnk 2 forfornn nn0.. This This means means that that the the degrees degrees of of the the first firstkkverticesvertices of of n− 1 n− 2 ≥ 0 Γkk− and Γkk− increase≥ by 1 in Γkk. But the degrees of the other vertices remain Γnn 11 and Γnn 22 increase by 1 in Γnn. But the degrees of the other vertices remain −− −− k k thethe same. same. So So the the first firstkkverticesvertices of of Γ Γnk areare the the first firstkkverticesvertices of of Γ Γnk 1 whosewhose degree degree n n− 1 information in Dkk(x, y) should be kept track of by the variable y. Therefore,− for a information in Dnn(x, y) should be kept track of by the variable y. Therefore, for a fixed k, Dkk(x, y) satisfies the recursion fixed k, Dnn(x, y) satisfies the recursion k k k k k Dn(kx, y) = yDn k1(0, y) + Dn k1(x, 0) + xDn k2(0, x) + Dn k2(x, 0) (9) Dn(x, y) = yD−n 1(0, y) + D−n 1(x, 0) + xD−n 2(0, x) + D−n 2(x, 0) (9) − − − − for n n0. Now, writing for n≥ n0. Now, writing ≥ k Dnk(x, y) = pn(y) + qn(x), Dn(x, y) = pn(y) + qn(x), and using (9) we see that the polynomials p (y) and q (x) must individually satisfy and using (9) we see that the polynomials pn (y) and qn (x) must individually satisfy the recursions n n the recursions pn(y) = ypn 1(y) (10) pn(y) = ypn− 1(y) (10) − q (x) = q (x) + q (x) + xp (x) (11) qnn(x) = qnn 11(x) + qnn 22(x) + xpnn 22(x) (11) −− −− −− August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 649

for n n with initial values ≥ 0 pn 1(y), pn 2(y) and qn 1(x), qn 2(x). 0− 0− 0− 0− From (10) we immediately obtain the formula for p (y) for n n as n ≥ 0 n n0+1 pn(y) = y − pn 1(y). 0−

To solve the recursion for qn(x) from (11), we form the generating function

n Q(z) = qn(x)z . n n X≥ 0 Note ﬁrst that

n n0 n n0 1 n xpn 2(x)z = xpn 2(x)z + x x − − pn 1(x)z − 0− 0− n n n>n0 X≥ 0 X n0+1 n0 xpn0 1(x)z = xpn 2(x)z + − . 0− 1 xz − Multiplying (11) with zn and summing both sides for n n , Q(z) is found to be ≥ 0 n0 z xpn0 1(x)z Q(z) = (1 + z)qn 1(x) + qn 2(x) + xpn 2(x) + − 1 z z2 0− 0− 0− 1 xz − − − Using the expansions 1 z = f zn, = f zn 1 z z2 n+1 1 z z2 n n 0 n 0 − − X≥ − − X≥ and n z i n = fn ix z (1 z z2)(1 xz) − n 0 i=0 ! − − − X≥ X and setting t = n n , we ﬁnally obtain − 0 t i qn(x) = ft+2qn 1(x) + ft+1qn 2(x) + ft+1xpn 2(x) + xpn 1(x) ft ix . 0− 0− 0− 0− − i=0 X Using Theorem 5 with the initial conditions from Theorem 4 we obtain the following special cases.

Corollary 6. For k 1, 2, 3 the degree enumerator polynomials Dk (x, y) for Γk ∈ { } n n are given by

n 1 1 n − i Dn(x, y) = y + fn+1x + fn ix − i=2 X n 1 2 n n 1 − i Dn(x, y) = y + y − + fn 1x + fn i+1x − − i=2 X August 30, 2020 12:9 112-IJFCS 2050031

650 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

3 n n 1 n 2 n 1 2 Dn(x, y) = y + y − + y − + f1x − + Ln 1x − n 2 − i + 2 fn ix − i=3 X where Ln is the nth Lucas number.

k 4.4. Number of induced hypercubes in Γn k In this section, we count the number of d-dimensional hypercubes induced in Γn. We know that for d = 0 and d = 1 these numbers are equal to the number of vertices k and number of edges in Γn respectively. k We ﬁrst consider the number of squares (4-cycles) in Γn. This is the case where d = 2.

Definition 7. Let Z(i) denote the number of 1’s in the Zeckendorf representation of i for i 0 and define for m 0, the partial sums ≥ ≥ m P (m) = Z(i). (12) i=0 X These sequences start as 0, 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 3, 1, 2, 2, 2, 3, 2, 3, 3, 1,... for Z(i) and 0, 1, 2, 3, 5, 6, 8, 10, 11, 13, 15, 17, 20, 21, 23, 25, 27, 30, 32, 35, 38, 39,... for P (m). Let S(n) denote the number of squares of Γn and Sk(n) denote the number of squares of Γk for k 2. From [7] we know that n ≥ 1 S(n) = ((5n + 1)(n 2)fn + 6nfn 2). 50 − − k For n < n0 we have Sk(n) = S(n). By the fundamental decomposition of Γn, the k number of squares in Γn is the sum of three quantities: the number of squares in k k Γn 1, the number of squares in Γn 2 and the number of squares that are created − − k k by the addition of k link edges between Γn 1 and Γn 2 involving the vertices with labels 0, 1, . . . , k 1. − − − The number of squares of the last type above is equal to the number of edges in the subgraph of Γk induced by the first k vertices 0, 1, . . . , k 1. n − k Lemma 8. The number of edges in the subgraph of Γn induced by the first k vertices 0, 1, . . . , k 1 is P (k 1). − − Proof. For a vertex i in Γk with i 0, 1, . . . , k 1 , switching a 1 in the Zeckendorf n ∈ { − } representation to a 0 gives an adjacent vertex to i in 0, 1, . . . , k 1 . So i has Z(i) { − } August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 651

neighbors in the subgraph induced by the vertices 0 through k 1. Summing the − contributions over i gives the lemma.

It follows that Sk(n) satisﬁes the recursion

S (n) = S (n 1) + S (n 2) + P (k 1). (13) k k − k − − Using (13) and induction on n we directly obtain that for n n ≥ 0 S (n) = (P (k 1) + S(n 2))f k − 0 − t+3 + (S(n 1) S(n 2))f P (k 1), (14) 0 − − 0 − t+2 − − where t = n n . − 0 Numerical values of the number of squares for small values of k and n can be found in Table 1, where the entries given in boldface are the number of squares in the Fibonacci cube Γ itself. Note that this sequence for n 0 is n ≥ 0, 0, 0, 1, 3, 9, 22, 51, 111, 233, 474, 942,... and these numbers are the triple convo- x3 lution of the Fibonacci numbers. Their generating function is (1 x x2)3 and the 1 − − closed formula is given explicitly in [7] as 50 ((5n + 1)(n 2)fn + 6nfn 2). k − k − Let Qd(n) denote the number of d-dimensional hypercubes in Γn. Thus using this k k notation, Q1 (n) = Ek(n) and Q2 (n) = Sk(n) as they appear in (3) and (13), respectively. Furthermore, let P (k 1, n) denote the number of d-dimensional hypercubes d − contained in the subgraph of Γk induced by the vertices with labels 0, 1, . . . , k 1. n − Note that with this notation, P (k 1, n) = P (k 1) as it appears in recursion (14) 1 − − and P (k 1, n) = k, as it appears as the nonhomogeneous part of recursion (4). 0 −

k Table 1. Counting squares in Γn.

k n 3 4 5 6 7 8 9 10 Closed form n0(k) \ 2 1 2 4 7 12 20 33 54 fn 1 + fn 2 1 4 − − − 3 1 3 6 11 19 32 53 87 3fn 2 + 4fn 3 2 5 − − − 4 1 3 7 13 23 39 65 107 4fn 2 + 4fn 3 3 5 − − − 5 1 3 9 17 31 53 89 147 8fn 3 + 12fn 4 5 6 − − − 6 1 3 9 18 33 57 96 159 9fn 3 + 12fn 4 6 6 − − − 7 1 3 9 20 37 65 110 183 11fn 3 + 12fn 4 8 6 − − − 8 1 3 9 22 41 73 124 207 19fn 4 + 31fn 5 10 7 − − − 9 1 3 9 22 42 75 128 214 20fn 4 + 31fn 5 11 7 − − − 10 1 3 9 22 44 79 136 228 22fn 4 + 31fn 5 13 7 − − − 11 1 3 9 22 46 83 144 242 24fn 4 + 31fn 5 15 7 − − − 12 1 3 9 22 48 87 152 256 26fn 4 + 31fn 5 17 7 − − − 13 1 3 9 22 51 93 164 277 42fn 5 + 73fn 6 20 8 − − − 14 1 3 9 22 51 94 166 281 43fn 5 + 73fn 6 21 8 − − − August 30, 2020 12:9 112-IJFCS 2050031

652 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

k k Proposition 9. Let Qd(n) be the number of d-dimensional hypercubes in Γn, then k Qd(n) satisﬁes the recurrence relation k k k Qd(n) = Qd(n 1) + Qd(n 2) + Pd 1(k 1) − − − − where k 1 − Z(i) Pd 1(k 1) = . − − d 1 i=0 X − Proof. There are three types of d-dimensional hypercubes that contribute to k k k Qd(n): those coming from Γn 1, those coming from Γn 2, and the ones formed by − k − the k link vertices used in the construction of Γn. The d-dimensional hypercubes of the last type are counted by the number of (d 1)-dimensional hypercubes contained k − in the subgraph of Γn 1 induced by the vertices with labels 0, 1, . . . , k 1. For any − − of these vertices i we need to select d 1 ones among the Z(i) ones in i. Then by d 1− varying these d 1 ones we obtain 2 − vertices with labels in 0, 1, . . . , k 1 each − k { − } giving a (d 1)-dimensional hypercube in Γn 1. Therefore, a simple generalization − − of Lemma 8 gives the number of such hypercubes as k 1 − Z(i) Pd 1(k 1, n 1) = . − − − d 1 i=0 X − We can denote the above quantity as Pd 1(k 1) as the sum on the right does not − − depend on n. This completes the proof.

Let Qd(n) denote the number of d-dimensional hypercubes in Γn. This number and its q-analogue are determined in [9, Corollary 3.3] and [16, Proposition 3], respectively. Using these result we have the following.

Corollary 10. Let Qd(n) denote the number of d-dimensional hypercubes in Γn. Then Qd(n) is given explicitly by

n+1 b 2 c n i + 1 i Q (n) = − . d i d Xi=d In particular, we have the formulas for the ﬁrst few dimensions d as follows: 1 Q (n) = E(n) = (2(n + 1)f + nf ), 1 5 n n+1 1 Q2(n) = S(n) = ((5n + 1)(n 2)fn + 6nfn 2), 50 − − n(n 2) Q3(n) = − (4(n 4)fn 3 + 3(n 3)fn 4). 150 − − − − k Using Corollary 10 and by solving the recurrence relation satisﬁed by Qd(n) given in Proposition 9 by induction, similar to the case of Proposition 1 and (14), we obtain August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 653

k k Theorem 11. Let Qd(n) denote the number of d-dimensional hypercubes in Γn. Then k Qd(n) = (Pd 1(k 1) + Qd(n0 2))ft+3 − − − + (Qd(n0 1) Qd(n0 2))ft+2 Pd 1(k 1), − − − − − − for n n with t = n n , where Q (n) denotes the number of d-dimensional ≥ 0 − 0 d hypercubes in Γn.

5. Remarks We end this work with three remarks. The first one is on a number of additional k graph theoretic properties of Γn. After that we present an observation on the computation of the partial sum P (k) defined in (12) for a given k. Finally we consider k a class of special subgraphs of Fibonacci cubes which are inspired by the Γn. k Remark 12. The diameter and the radius of Γn are directly related to the same properties of the Fibonacci cubes themselves. We start by noting the following nested structure of k-Fibonacci cubes. k For non-negative integers n and k we know that Γn can be obtained directly from Γ . It is either equal to Γ , or for n n , it is obtained from Γ by removing n n ≥ 0 n certain edges. Furthermore, for n n ,Γk can also be obtained from Γk+1 by ≥ 0 n n removing edges. Therefore Γ1 Γk Γ . (15) n ⊆ · · · ⊆ n ⊆ · · · ⊆ n 1 n We know that Γn is a tree with root 0 (the vertex with integer label 0). It follows that for u, v V (Γ1 ) ∈ n d(u, v) d(u, 0n) + d(v, 0n) = w (u) + w (v), (16) ≤ H H where wH denotes the Hamming weight. We have diam(Γn) = n as given in [4]. k k For n < n0 diam(Γn) = diam(Γn) = n. For n n0, we know that Γn is a subgraph k ≥ k of Γn, the vertices of Γn and Γn are the same and Γn has fewer edges. Therefore, k diam(Γn) diam(Γn) = n. On the other hand, using (15) and (16) for any u, v k ≥ ∈ V (Γn) we have d(u, v) d(u, 0n) + d(v, 0n) = w (u) + w (v) n. ≤ H H ≤ k Therefore, we have diam(Γn) = n. k By a similar argument one can show that the radius of Γn is equal to the radius of Γ , which is obtained in [15] as n . n d 2 e Remark 13. To our knowledge there is no direct formula for the partial sum P (k) defined in (12) for a given k. The trivial recurrence P (k) = P (k 1) + Z(k) − seems to require the computation of the Zeckendorf representation of every number up to k. Interestingly, the structure of the edges in k-Fibonacci cubes allows for a way to compute P (k). August 30, 2020 12:9 112-IJFCS 2050031

654 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

k Assume that k = fm. Then we know that the vertices of Γm 2 = Γm 2 are − labeled with the Zeckendorf representation of the integers 0, 1, . . . , k 1.− From − Lemma 8 we observe that P (k 1) equals to the number of edges of the Γm 2, that − − is, 1 P (k 1) = P (fm 1) = E(Γm 2) = (2(m 1)fm 2 + (m 2)fm 1). − − | − | 5 − − − −

Now we can easily generalize this idea. Assume that fm < k < fm+1. Then write k = f + r for some positive integer r. This means that P (k 1) is equal to the m − number of edges of the subgraph of Γm 1 induced by the first k vertices labeled − by 0, 1, . . . , k 1 from Lemma 8. Using the decomposition of Γm 1 = 0Γm 2 + − − − 10Γm 3, this number is equal to the sum of the number of edges in the subgraph of − Γm 1 induced by the first fm vertices labeled by 0, 1, . . . , fm 1 (which is Γm 2), − − − the number of edges in the subgraph of Γm 1 induced by the vertices labeled by − f , f + 1, . . . , k 1 (which is equal to the number of edges in the subgraph of m m − Γm 3 induced by the vertices labeled by 0, 1, . . . , r 1) and r which is the number − − of link edges between vertices labeled 0, 1, . . . , r 1 and the corresponding vertices − in Γm 2. − ` Let k 1 = s=1 fis denote the Zeckendorf representation of k 1 with i1 < − ` 1 − i < < i = m. We can write r = 1 + − f since k = f + r. Then we have 2 ··· ` P s=1 is m P (k 1) = P (f 1 + r) P − m − = P (f 1) + P (r 1) + r m − − ` ` 1 ` j − − = P (f 1) + 1 + f is − is s=1 j=1 s=1 ! X X X ` ` 1 − = E(Γi 2) + j fi − + r | s− | · ` j s=1 j=1 X X ` ` 1 1 − = (2(is 1)fi 2 + (is 2)fi 1) + j fi − + r. 5 − s− − s− · ` j s=1 j=1 X X Using this expansion, one can easily compute P (k 1) using the Zeckendorf repre- − sentation of k 1. − Example: Let k 1 = 15. The Zeckendorf representation of 15 is (100010). Next we need − to find the number of edges of the subgraph of Γ6 induced by the first k vertices labeled 0, 1,..., 15. By the fundamental decomposition Γ6 = 0Γ5 + 10Γ4. Now we can partition the vertex set of this graph as 0, 1,..., 12 and 13, 14, 15 . The { } { } subgraph with vertices 0, 1,..., 12 is isomorphic to the Γ Γ and the subgraph { } 5 ⊆ 6 with vertices 13, 14, 15 is isomorphic to the Γ Γ Γ . Note that there are 3 { } 2 ⊆ 4 ⊆ 6 August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 655

link edges between these subgraphs. Therefore, the total number of edges that we are looking for is equal to E(Γ ) + E(Γ ) + 3 = 20 + 2 + 3 = 25, | 5 | | 2 | which directly gives that P (15) = 25.

Remark 14. In the construction of Γn there are fn link edges between Γn 1 and k − Γn 2. In this paper, we considered k-Fibonacci cubes Γn, in which the number of − k k link edges between Γn 1 and Γn 2 is a fixed number k for all n n0(k), where − − ≥ n0(k) is as given in (1). A simple variant of these graphs can also be defined by taking the number of link edges between Γn 1 and Γn 2 as a variable instead of a fixed number k. Given − − λ 0 λ 1, we loosely define here a class of graphs Γn for n 0, whose construction ≤ ≤ ≥ λ closely resembles that of the Fibonacci cubes themselves. We start with Γ0 = Γ0, λ λ λ λ Γ1 = Γ1. For n 2, Γn is defined in terms of Γn 1 and Γn 2, similar to the ≥ − − fundamental decomposition of Γn but by taking only the first λfn link edges that λ λ λ exist between 00Γn 2 and its copy 10Γn 2 in Γn 1, instead of all fn of them like − − k − we do for Γn or first k of them like we do for Γn. We have an explicit expression for the number of edges E(n) of Γn as given in λ (2). Let Eλ(n) denote the number of edges of Γn. Then Eλ(n) satisfies the recursion E (n) = E (n 1) + E (n 2) + λf λ λ − λ − n for n 2 and therefore ≥ E (n) = (1 λ)f + λE(n). λ − n Asymptotically, the average degree of a vertex in Γn is the expression in (5). By λ taking limits and using (5), the average degree of a vertex in Γn is found to be 1 1 λn, − √5 which reduces to the average degree in Γn for λ = 1, as expected.

Acknowledgments We would like to thank the reviewers for their useful comments which greatly improved the readability of this paper. The work of the second author was supported by TUB¨ ITAK˙ under Grant No. 117R032.

References [1] A. R. Ashraﬁ et al., Vertex and edge orbits of Fibonacci and Lucas cubes, Ann. Combin. 20 (2016) 209–229. [2] K. Egiazarian and J. Astola, On generalized Fibonacci cubes and unitary transforms, Appl. Algebra Eng. Commun. Comput. 8 (1997) 371–377. August 30, 2020 12:9 112-IJFCS 2050031

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[3] S. Gravier et al., On disjoint hypercubes in Fibonacci cubes, Discrete Appl. Math. 190–191 (2015) 50–55. [4] W.-J. Hsu, Fibonacci cubes — A new interconnection technology, IEEE Trans. Par- allel Distrib. Syst. 4 (1993) 3–12. [5] W.-J. Hsu and M.-J. Chung, Generalized Fibonacci cubes, in IEEE Proc. Int. Conf. Parallel Processing (1993), pp. 299–302. [6] A. Ili´c,S. Klavˇzarand Y. Rho, Generalized Fibonacci cubes, Discrete Math. 312 (2012) 2–11. [7] S. Klavˇzar,On median nature and enumerative properties of Fibonacci-like cubes, Discrete Math. 299 (2005) 145–153. [8] S. Klavˇzar,Structure of Fibonacci cubes: A survey, J. Comb. Optim. 25 (2013) 505– 522. [9] S. Klavˇzarand M. Mollard, Cube polynomial of Fibonacci and Lucas cube, Acta Appl. Math. 117 (2012) 93–105. [10] S. Klavˇzarand M. Mollard, Asymptotic properties of Fibonacci cubes and Lucas cubes, Ann. Comb. 18 (2014) 447–457. [11] S. Klavˇzar,M. Mollard and M. Petkovˇsek,The degree sequence of Fibonacci and Lucas cubes, Discrete Math. 311 (2011) 1310–1322. [12] M. Mollard, Maximal hypercubes in Fibonacci and Lucas cubes, Discrete Appl. Math. 160 (2012) 2479–2483. [13] M. Mollard, Non covered vertices in Fibonacci cubes by a maximum set of disjoint hypercubes, Discrete Appl. Math. 219 (2017) 219–221. [14] E. Munarini, C. P. Cippo and N. Zagaglia Salvi, On the Lucas cubes, Fibonacci Quart. 39 (2001) 12–21. [15] E. Munarini and N. Zagaglia Salvi, Structural and enumerative properties of the Fibonacci cubes, Discrete Math. 255 (2002) 317–324. [16] E. Saygı and O.¨ E˘gecio˘glu,Counting disjoint hypercubes in Fibonacci cubes, Discrete Appl. Math. 215 (2016) 231–237. [17] E. Saygı and O.¨ E˘gecio˘glu, q-cube enumerator polynomial of Fibonacci cubes, Discrete Appl. Math. 226 (2017) 127–137. [18] E. Saygı and O.¨ E˘gecio˘glu, Boundary enumerator polynomial of hypercubes in Fibonacci cubes, Discrete Appl. Math. 266 (2019) 191–199. February 18, 2020 22:15 WSPC/INSTRUCTION FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised August 30, 2020 12:9 112-IJFCS 2050031

k-Fibonacci cubes: A special family of subgraphs of Fibonacci cubes 19

Appendix A. Figuresk-Fibonacci of Cubes: some Ak Special-Fibonacci Family of cubes Subgraphs of Fibonacci Cubes 657

Appendix A. Figures of Some k-Fibonacci Cubes

Fig. 7. The ﬁrst six trees in Fig. 4 redrawn. Fig. 7. The ﬁrst six trees in Figure 4 redrawn. AugustFebruary 30, 2020 12:9 18, 2020 112-IJFCS 22:15 WSPC/INSTRUCTION 2050031 FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

20 Omer¨ E˘gecio˘glu,Elif Saygı, Z¨ulf¨ukarSaygı 658 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

Fig.Fig. 8. 8. The The ﬁrst ﬁrst eight eightkk-Fibonacci-Fibonacci cubes cubes for forkk== 2. 2. AugustFebruary 30, 2020 12:9 18, 2020 112-IJFCS 22:15 WSPC/INSTRUCTION 2050031 FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

k-Fibonacci cubes: A special family of subgraphs of Fibonacci cubes 21 k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 659

Fig. 9. The ﬁrst eight k-Fibonacci cubes for k = 3. Fig. 9. The ﬁrst eight k-Fibonacci cubes for k = 3. AugustFebruary 30, 2020 12:9 18, 2020 112-IJFCS 22:15 WSPC/INSTRUCTION 2050031 FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

22 Omer¨ E˘gecio˘glu,Elif Saygı, Z¨ulf¨ukarSaygı 660 O.¨ E˘gecio˘glu,E. Saygı & Z. Saygı

Fig. 10. The ﬁrst eight k-Fibonacci cubes for k = 4. Fig. 10. The ﬁrst eight k-Fibonacci cubes for k = 4. FebruaryAugust 30, 2020 18, 12:9 2020 112-IJFCS 22:15 WSPC/INSTRUCTION 2050031 FILE k-Fibonacci˙Cubes˙Egecioglu˙Saygi˙Feb19˙2020˙revised˙revised

k-Fibonacci cubes: A special family of subgraphs of Fibonacci cubes 23 k-Fibonacci Cubes: A Special Family of Subgraphs of Fibonacci Cubes 661

4 13 Fig. 11. The structure of Γ412 (top) and Γ1312 (bottom). Fig. 11. The structure of Γ12 (top) and Γ12 (bottom).