Generalized Fibonacci and Lucas cubes arising from powers of paths and cycles

# Generalized Fibonacci and Lucas cubes arising from powers of paths and cycles

## Abstract

The paper deals with some generalizations of Fibonacci and Lucas sequences, arising from powers of paths and cycles, respectively.

In the first part of the work we provide a formula for the number of edges of the Hasse diagram of the independent sets of the th power of a path ordered by inclusion. For such a diagram is called a Fibonacci cube, and for we obtain a generalization of the Fibonacci cube. Consequently, we derive a generalized notion of Fibonacci sequence, called -Fibonacci sequence. Then, we show that the number of edges of a generalized Fibonacci cube is obtained by convolution of an -Fibonacci sequence with itself.

In the second part we consider the case of cycles. We evaluate the number of edges of the Hasse diagram of the independent sets of the th power of a cycle ordered by inclusion. For such a diagram is called Lucas cube, and for we obtain a generalization of the Lucas cube. We derive then a generalized version of the Lucas sequence, called -Lucas sequence. Finally, we show that the number of edges of a generalized Lucas cube is obtained by an appropriate convolution of an -Fibonacci sequence with an -Lucas sequence.

###### keywords:
independent set, path, cycle, power of graph, Fibonacci cube, Lucas cube, Fibonacci number, Lucas number,
###### Msc:
[2010] 11B39, 05C38

## 1 Introduction

For a graph we denote by the set of its vertices, and by the set of its edges.

###### Definition 1.1.

For ,

• the -power of a path, denoted by , is a graph with vertices , , , such that, for , , if and only if ;

• the -power of a cycle, denoted by , is a graph with vertices , , , such that, for , , if and only if or .

Thus, for instance, and are the graphs made of isolated nodes, is the path with vertices, and is the cycle with vertices. Figure 1 shows some powers of paths and cycles.

###### Definition 1.2.

An independent set of a graph is a subset of not containing adjacent vertices.

Let , and be the Hasse diagrams of the posets of independent sets of , and , respectively, ordered by inclusion. Clearly, is a Boolean lattice with atoms (-cube, for short).

Before introducing the main results of the paper, we now provide some background on Fibonacci and Lucas cubes. Every independent set of can be represented by a binary string , where, for , if and only if . Specifically, each independent set of is associated with a binary string of length such that the distance between any two ’s of the string is greater than . Following fibocubes_enum () (see also survey_fibo ()), a Fibonacci string of order is a binary strings of length without consecutive ’s. Recalling that the Hamming distance between two binary strings and is the number of bits where and differ, we can define the Fibonacci cube of order , denoted , as the graph , where is the set of all Fibonacci strings of order and, for all , if and only if . One can observe that for the binary strings associated with independent sets of are Fibonacci strings of order , and the Hasse diagram of the set of all such strings ordered bitwise (i.e., for and , if and only if , for every ) is . Fibonacci cubes were introduced as an interconnection scheme for multicomputers in hsu (), and their combinatorial structure has been further investigated, e.g.  in klavzar (); fibocubes_enum (). Several generalizations of the notion of Fibonacci cubes has been proposed (see, e.g., gen_fibo (); survey_fibo ()).

###### Remark.

Consider the generalized Fibonacci cubes described in gen_fibo (), i.e., the graphs obtained from the -cube of all binary strings of length by removing all vertices that contain the binary string as a substring. In this notation the Fibonacci cube is . It is not difficult to see that cannot be expressed, in general, in terms of . Instead we have:

where is the subgraph of obtained by removing all strings that contain either or .

A similar argument can be carried out for cycles. Indeed, every independent set of can be represented by a circular binary string (i.e., a sequence of ’s and ’s with the first and last bits considered to be adjacent) , where, for , if and only if . Thus, each independent set of is associated with a circular binary string of length such that the distance between any two ’s of the string is greater than . A Lucas cube of order , denoted , is defined as the graph whose vertices are the binary strings of length without either two consecutive ’s or a in the first and in the last position, and in which the vertices are adjacent when their Hamming distance is exactly (see lucas ()). For the Hasse diagram of the set of all circular binary strings associated with independent sets of ordered bitwise is . A generalization of the notion of Lucas cubes has been proposed in gen_lucas ().

###### Remark.

Consider the generalized Lucas cubes described in gen_lucas (), that is, the graphs obtained from the -cube of all binary strings of length by removing all vertices that have a circular containing as a substring (i.e., such that is contained in the circular binary strings obtained by connecting first and last bits of the string). In this notation the Lucas cube is . It is not difficult to see that cannot be expressed, in general, in terms of . Instead we have:

To the best of our knowledge, our , and are new generalizations of Fibonacci and Lucas cubes, respectively.

In the first part of this paper (which is an extended version of endm () — see remark at the end of this section) we evaluate , i.e., the number of independent sets of , and , i.e., the number of edges of . We then introduce a generalization of the Fibonacci sequence, that we call -Fibonacci sequence and denote by . Such integer sequence is based on the values of . Our main result (Theorem 3.4) is that, for , the sequence is obtained by convolving the sequence with itself.

In the second part we deal with power of cycles, and derive similar results for this case. Specifically, we compute , i.e., the number of independent sets of , and , i.e., the number of edges of . Further, we introduce a generalization of the Lucas sequence, that we call -Lucas sequence and denote by . Such integer sequence is based on the values of . The analogous of Theorem 3.4 in the Lucas case (Theorem 5.4) states that, for , the sequence is obtained by an appropriate convolution between the sequences and .

###### Remark.

As mentioned before, this work is an extended version of endm (). (The extended abstract endm () has been presented at the conference Combinatorics 2012, Perugia, Italy, 2012, and consequently published in ENDM.) Specifically, Section 2 of the present paper is (endm, , Section 2) enriched with some details and full proofs, while Section 3 is (endm, , Section 3), with corrected notation and some remarks added. Lemma 3.1, Lemma 3.2, and Theorem 3.4 have remained as they were in endm (): we decided to keep full proofs of these results, to make the content clearer and in order for the paper to be self-contained.

## 2 The independent sets of powers of paths

For , we denote by the number of independent -subsets of .

###### Remark.

counts the number of binary strings with ’s.

###### Lemma 2.1.

For ,

 p(h)n,k=(n−hk+hk) .

This is Theorem 1 of hoggatt (). An alternative proof follows.

###### Proof.

By Definition 1.2, any two elements of an independent set of must satisfy . It is straightforward to check that whenever , . It is also immediate to see that when our lemma holds true.

Suppose now . We can complete the proof of our lemma by establishing a bijection between independent -subset of and -subsets of a set with elements. Let be the set of all -subsets of a set , and the set of all independent -subsets of . Consider the map such that, for any , with ,

 f({bi1,bi2,…,bij,…,bik})={vi1,vi2+h,…,vij+(j−1)h,…,vik+(k−1)h}.
###### Claim 0.

The map associates an independent -subset of with each -subset .

To see this we first remark that is a -subset of . Furthermore, for each pair , with , we have

 ij+t+(j+t−1)h−(ij+(j−1)h)=ij+t−ij+th>h.

Hence, by Definition 1.1, . Thus, is an independent set of .

###### Claim 0.

The map is bijective.

It is easy to see that is injective. Then, we consider the map such that, for any , with ,

 f−1({vi1,vi2,…,vij,…,vik})={bi1,bi2−h,…,bij−(j−1)h,…,bik−(k−1)h}.

Following the same steps as for , one checks that is injective. Thus, is surjective.

We have established a bijection between independent -subsets of and -subsets of a set with elements. The lemma is proved. ∎

Some values of are shown in Tables 13 (Section 6). The coefficients also enjoy the following property: .

For , the number of independent sets of is

 p(h)n=∑k≥0p(h)n,k=⌈n/(h+1)⌉∑k=0p(h)n,k=⌈n/(h+1)⌉∑k=0(n−hk+hk).
###### Remark.

Denote by the element of the Fibonacci sequence: , , and , for . Then, is the number of vertices of the Fibonacci cube of order .

The following, simple fact is crucial for our work.

###### Lemma 2.2.

For ,

 p(h)n={n+1if  n≤h+1,p(h)n−1+p(h)n−h−1if  n>h+1.

A proof of this Lemma can also be obtained using the first part of (hoggatt, , Proof of Theorem 1).

###### Proof.

For , by Definition 1.2, the independent sets of have no more than element. Thus, there are independent sets of .

Consider the case . Let be the set of all independent sets of , let be the set of the independent sets of that contain , and let . The elements of are in one-to-one correspondence with the independent sets of , and those of are in one-to-one correspondence with the independent sets of . ∎

Tables 4 displays a few values of .

## 3 Generalized Fibonacci numbers and generalized Fibonacci cubes

Figure 2 shows a few Hasse diagrams . Notice that, as stated in the introduction, for each , is the Fibonacci cube .

Let be the number of edges of . Noting that in each non-empty independent -subset covers exactly independent -subsets, we can write

 H(h)n=⌈n/(h+1)⌉∑k=1kp(h)n,k=⌈n/(h+1)⌉∑k=1k(n−hk+hk) . (1)
###### Remark.

counts the number of edges of .

Let now be the number of independent -subsets of containing the vertex , and let, for , ,

###### Lemma 3.1.

For , and ,

 T(n,h)k,i=k−1∑r=0¯p(h)i−h−1,r ¯p(h)n−i−h,k−1−r.
###### Proof.

No independent set of containing contains any of the elements . Let and be non-negative integers whose sum is . Each independent -subset of containing can be obtained by adding to a -subset such that

(a) is an independent -subset of ;

(b) is an independent -subset of .

Viceversa, one can obtain each of this pairs of subsets by removing from an independent -subset of containing . Thus, is obtained by counting independently the subsets of type (a) and (b). The remark that the subsets of type (b) are in bijection with the independent -subsets of proves the lemma. ∎

###### Remark.

counts the number of strings such that: (i) , and (ii) .

In order to obtain our main result, we prepare a lemma.

###### Lemma 3.2.

For positive ,

 ⌈n/(h+1)⌉∑k=1n∑i=1T(n,h)k,i=H(h)n.
###### Proof.

The inner sum counts the number of -subsets exactly times, one for each element of the subset. That is, . Hence the lemma follows directly from Equation (1). ∎

Next we introduce a family of Fibonacci-like sequences.

###### Definition 3.3.

For , and , we define the -Fibonacci sequence whose elements are

 F(h)n={1if  n≤h+1,F(h)n−1+F(h)n−h−1if  n>h+1.

The first values of the -Fibonacci sequences, for , are shown in Table 5. From Lemma 2.2, and setting for , and , we have that,

 F(h)i=¯p(h)i−h−1,  for each  i≥1. (2)

Thus, our Fibonacci-like sequences are obtained by prepending ’s to the sequence . Therefore, we have:

• ;

• is the Fibonacci sequence;

• more generally, .

In the following, we define the discrete convolution operation , as follows.

 (F(h)∗F(h))(n)≐n∑i=1F(h)iF(h)n−i+1 (3)
###### Theorem 3.4.

For , the following holds

 H(h)n=(F(h)∗F(h))(n).
###### Proof.

The sum counts the number of independent sets of containing . We can also obtain such a value by counting the independent sets of both , and . Thus, we have:

 ⌈n/(h+1)⌉∑k=1T(n,h)k,i=¯p(h)i−h−1¯p(h)n−h−i.

Using Lemma 3.2 we can write

 H(h)n=⌈n/(h+1)⌉∑k=1n∑i=1T(n,h)k,i=n∑i=1⌈n/(h+1)⌉∑k=1T(n,h)k,i=n∑i=1¯p(h)i−h−1¯p(h)n−h−i.

By Equation (2) we have By (3), the theorem is proved. ∎

We display some values of in Table 6 (Section 6).

###### Remark.

For , we obtain the number of edges of by using Fibonacci numbers:

 H(1)n=n∑i=1FiFn−i+1.

The latter result is (mediannature, , Proposition 3).

## 4 The independent sets of powers of cycles

For , we denote by the number of independent -subsets of .

###### Remark.

For , counts the number of binary strings with ’s.

###### Lemma 4.1.

For , and ,

 c(h)n,k=nk(n−hk−1k−1) .

Moreover, , and , for each .

###### Proof.

Fix an element , and let . Any independent set of containing does not contain the elements preceding and the elements following . Thus, the number of independent -subsets of containing equals

 p(h)n−2h−1,k−1=(n−hk−1k−1) .

The total number of independent -subsets of is obtained by multiplying by , then dividing it by (each subset is counted times by the previous proceeding). The case , as well as the cases , can be easily verified. ∎

Some values of are displayed in Tables 79.

For , the number of all independent sets of is

 c(h)n=∑k≥0c(h)n,k=⌈n/(h+1)⌉∑k=0c(h)n,k , (4)
###### Remark.

Denote by the element of the Lucas sequence , , and , for . Then, for , is the number of elements of the Lucas cube of order .

Some values of are shown in Table 10. The coefficients satisfy a recursion that closely resembles that of Lemma 2.2.

###### Lemma 4.2.

For ,

 c(h)n={n+1if  n≤2h+1,c(h)n−1+c(h)n−h−1if  n>2h+1. (5)
###### Proof.

The case can be easily checked. The case is discussed at the end of this proof. Let , and let be the set of the independent sets of . Let be the subset of these sets that (i) do not contain , and that (ii) do not contain any of the following pairs: . Let then be the subset of the remaining independent sets of .

It is easy to see that the elements of are exactly the independent sets of . Indeed, is not a vertex of and the vertices of pairs , , , are adjacent in . On the other hand, to show that

 |Iout|=c(h)n−h−1

we argue as follows. First we recall (see the proof of Lemma 4.1) that the number of independent -subsets of that contain is . Secondly we claim that the number of independent -subsets of containing one of the pairs , , , is . To see this, consider the pair . The independent sets containing such a pair do not contain the vertices from to , do not contain the vertices from to , and do not contain the vertices from to . Thus, the removal of such vertices and of the vertices and turns into . Hence we can obtain all the independent -subsets of that contain the pair by simply adding these two vertices to one of the independent -subsets of . Same reasoning can be carried out for any other one of the pairs: , , .

Using Lemmas 2.1 and 4.1 one can easily derive that

 p(h)n−2h−1,k−1+hp(h)n−3h−2,k−2=c(h)n−h−1,k−1.

Hence, we derive the size of :

 |Iout|=c(h)n−h−1=∑k≥1p(h)n−2h−1,k−1 +h∑k≥2p(h)n−3h−2,k−2.

Summing up we have shown that , that is

 c(h)n=c(h)n−1+c(h)n−h−1.

The proof of the case is obtained in a similar way, observing that , and that . ∎

## 5 Generalized Lucas cubes and Lucas numbers

Figure 3 shows a few Hasse diagrams . Notice that, as stated in the introduction, for each , is the Lucas cube .

Let be the number of edges of . As done in Section 3 for the case of paths, we immediately provide a formula for :

 M(h)n=⌈n/h+1⌉∑k=0kc(h)n,k=n⌈n/h+1⌉∑k=0(n−hk−1k−1) . (6)
###### Remark.

For , counts the number of edges of . As shown in (lucas, , Proposition 4(ii)), .

As shown in the proof of Lemma 4.1, the value

 p(h)n−2h−1,k−1=(n−hk−1k−1)

is the analogue of the coefficient : in the case of cycles we have no dependencies on , because each choice of vertex is equivalent. We can obtain in terms of a Fibonacci-like sequence, as follows.

###### Proposition 5.1.

For , the following holds

 M(h)n=nF(h)n−h.
###### Proof.

Using Equation (2) we obtain:

 M(h)n=n⌈n/(h+1)⌉∑k=1¯p(h)n−2h−1,k−1=n¯p(h)n−2h−1=nF(h)n−h.

In analogy with Section 3, we introduce a family of Lucas-like sequences.

###### Definition 5.2.

For , and , we define the -Lucas sequence whose elements are

 L(h)n=⎧⎪⎨⎪⎩h+1if  n=1,1if  2≤n≤h+1,L(h)n−1+L(h)n−h−1if  n>h+1.

The first values of the -Lucas sequences, for , are displayed in Table 11. We have that,

 L(h)i=c(h)i−1,  for each  i>h+1. (7)

To prove the main result of this section, the following lemma is needed.

###### Lemma 5.3.

For , the following holds

 L(h)n+1=F(h)n+(h+1)F(h)n−h
###### Proof.

The result is proved by induction on . Indeed, Applying the inductive hypothesis, we have

 L(h)n+1=F(h)n−1+(h+1)F(h)n−h−1+F(h)n−h−1+(h+1)F(h)n−2h−1=F(h)n+(h+1)F(h)n−h.

Finally, our analogous of Theorem 3.4, for cycles, is the following.

###### Theorem 5.4.

For , the following holds

 M(h)n=(F(h)∗L(h))(n−h).
###### Proof.

By Proposition 5.1, the statement of the Theorem is equivalent to

 n−h∑i=1F(h)iL(h)n−h+1−i=nF(h)n−h. (8)

Let . We have

 n∑i=1F(0)iL(0)n+1−i=n∑i=12i−12n−i=n∑i=12n−1=n2n−1=nF(0)n.

Let . In this case the statement of the theorem reduces to the well known identity involving (classical) Fibonacci and Lucas sequences:

 n∑i=1FiLn−i+1=(n+1)Fn.

Let . We prove (8) by induction on . If , then

 n−h∑i=1F(h)iL(h)n−h+1−i=F(h)1L(h)1=h+1=nF(h)1.

Let , and suppose (inductive hypothesis) that (8) holds for every . Let (and note that ). We need to prove that

 m+1∑i=1F(h)iL(h)m+2−i=(m+h+1)F(h)m+1. (9)

We find it convenient to define, for , the following integer sequences, which extend and to a range of negative integers.

 ¯F(h)n=⎧⎪⎨⎪⎩1if  n=−h,0if  −h0. (10) ¯L(h)n=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩h+1if  n=−h,−hif  n=−h+1,0if  −h+1