The Fibonacci Number of a Grid Graph and a New Class of Integer Sequences

Reinhardt Euler Faculté des Sciences B.P. 809 20 Avenue Le Gorgeu 29285 Brest Cedex France

                 Reinhardt.Euler@univ-brest.fr

Abstract Given a grid graph G of size mn, we study the number i(m, n) of independent sets in G, as well as b(m, n), the number of maximal such sets. It turns out that the initial cases b(1, n) and b(2, n) lead to a Padovan and a Fibonacci sequence. To determine b(m, n) for m > 2 we present an adaptation of the transfer matrix method, well known for calculating i(m, n). Finally, we apply our method to obtain explicit values of b(m, n) for m = 3, 4, 5 and provide the corresponding generating functions.

1 Introduction

Let G_{m,n} = (V, E) be the grid graph of size mn with vertex set

V = {(i, j) : 1 ≤ i ≤ m, 1 ≤ j ≤ n}

and edge set

E = {{(i, j), (i′, j′)} : |i − i′| + |j − j′| = 1}.

A set I ⊆ V is called an independent set if no two of its elements are joined by an edge. We let I_{m,n} denote the collection of independent sets in G_{m,n} and i(m, n) the total number of such sets. A maximal independent set B in G_{m,n} (maximal with respect to set-inclusion) is called a basis. Again, B_{m,n} represents the collection of bases in G_{m,n} and b(m, n) is the total number of such sets. We are interested in calculating i(m, n) and b(m, n) for any m, n ∈ ℕ together with the associated generating functions. The study of i(m, n) is closely related to the “hard-square model” as it is used in statistical physics. Of particular interest is the so-called “hard-square entropy constant” defined to be lim_{m,n→∞} i(m, n)^{1/mn} (see Baxter et al. [1]). Applications also include tiling and, more recently, efficient coding schemes in data storage (see Roth et al. [5]). A basic reference for work on i(m, n) is Calkin and Wilf [3]. More recently, Burstein et al. [2] have enumerated independent sets associated with several classes of (almost) regular graphs and calculated the corresponding generating functions.

Whereas this study of b(m, n) is new, distinguishing bases from independent sets is quite common: just think of matroid theory or weighted independent set problems over graphs with non-negative weights on the vertices. The question for the true value of lim_{m,n→∞} b(m, n)^{1/mn}, however, is open. We recently learned (see Weigt and Hartmann [9]) that maximal independent sets may be associated with meta-stable liquid states of certain hard-core lattice gas models under compaction.

The main question we will answer in this paper is that of calculating b(m, n) for m, n > 2 by a proper adaptation of the “transfer matrix method” as designed by Engel [4] for the calculation of i(m, n). The initial cases b(1, n) and b(2, n) turn out to produce a Padovan and a Fibonacci sequence, respectively. Padovan sequences have little history; their study goes back only to the beginning of the last century. In the 1990’s the architect Richard Padovan popularized these sequences and some closely related concepts such as the “plastic constant”, the Padovan-analogue to the golden number. We refer the interested reader to Stewart’s recreational article [8] for further details. Since i(m, n) is also called the Fibonacci number of G_{m,n} we think that, this analogy in mind, it would be convenient to call b(m, n) the Padovan number of G_{m,n}.

This paper is organised as follows: the next section is devoted to the Fibonacci number i(m, n). We review the transfer matrix method for the calculation of i(m, n) and the results obtained for m = 1, 2, 3, 4. Section 3 addresses the combinatorial structure of B_{1,n} and B_{2,n} as well as their respective cardinalities. In section 4 we present the new method to calculate b(m, n), and in section 5 we apply our method to explicitly describe b(m, n) for m = 3, 4, 5. Whenever convenient, we indicate the associated generating functions that we calculated using Maple. We conclude with a discussion of open problems.

2 The transfer matrix method for the calculation of i(m, n)

Let us start with the notion of “orthogonality”. The j-th column G^{j} = (V^{j}, E^{j}) of G_{m,n} is the subgraph induced by the vertex set {(1, j), (2, j), …, (m, j)} with I^{j} denoting the associated collection of independent sets. If now, for j_{1}, j_{2} ∈ {1, …, n}, I = {(i_{1}, j_{1}), …, (i_{p}, j_{1})} and K = {(k_{1}, j_{2}), …, (k_{q}, j_{2})} are members of I^{j_{1}} and I^{j_{2}}, respectively, we say that I is orthogonal to K, I ⊥ K, whenever {i_{1}, …, i_{p}}∩{k_{1}, …, k_{q}} = Ø. Moreover, for any I ∈ I^{j} we define the set l(I) = {(i_{1}, 1), …, (i_{p}, 1)}, which again is a member of I^{1}. For any k ∈ {1, …, n}, we will also have to count the number of independent sets in G_{m,k}, whose intersection with V^{k} equals a given set I, a number that we represent by i(m, k, I). The basic idea of the transfer matrix method (due to Engel [4], see also Stanley [7]) is to start with the values i(m, 1, I) = 1 for I ∈ I^{1}, and to calculate, for k = 1, 2, …, n − 1,

i(m, k + 1, J) = ∑_{I∈I^{k} and I⊥J} i(m, k, I) ∀ J ∈ I^{k+1},  (1)

in order to finally obtain

i(m, n) = ∑_{I∈I^{n}} i(m, n, I).  (2)

Equations (1) and (2) can be formulated as matrix-vector products by introducing the following matrix T ∈ {0, 1}^{F_{m+2}×F_{m+2}}:

∀ J, I ∈ I^{1} set T_{JI} = 1 iff J ⊥ I.

Clearly, for k = 1, …, n − 1,

i(m, k + 1, J) = ∑_{I∈I^{k}} T_{JI} · i(m, k, I) ∀ J ∈ I^{k+1}

and at the end i(m, n) = 1T^{n−1}1.

What really happens with this method is the following: at any stage k, I_{m,k} is partitioned into a fixed number of F_{m+2} classes, each of which is represented by a unique independent set in G^{k}. During the following stages only the size of the classes is increasing, and a simple summation at the end yields i(m, n). One may think of proceeding the same way to calculate b(m, n). However, as the example with m = n = 3 shows, not only do we lose the integrality of the transfer matrix. After a few steps we get fractional values supposed to represent the cardinalities of the classes. What we really need is to refine our partition, and this will be the subject of section 4.

What are the results we can obtain for small values of m? It is well known that i(1, n) equals F_{n+2}, the (n + 2)nd Fibonacci number. Other recurrence formulas are available for m = 2, 3, 4 (see Sloane [6]). In the following we just resume the related sequences together with their generating functions:

(i(1, n))_{n∈ℕ} = (2, 3, 5, 8, 13, 21, 34, 55, …)

with

∑_{n≥1} i(1, n)x^{n} = \frac{2x + x^{2}}{1 − x − x^{2}} ;

(i(2, n))_{n∈ℕ} = (3, 7, 17, 41, 99, 239, 577, 1393, …)

with

∑_{n≥1} i(2, n)x^{n} = \frac{3x + x^{2}}{1 − 2x − x^{2}} ;

(i(3, n))_{n∈ℕ} = (5, 17, 63, 227, 827, 2999, 10897, 39561, …)

with

∑_{n≥1} i(3, n)x^{n} = \frac{5x + 7x^{2} − x^{3} − x^{4}}{1 − 2x − 6x^{2} + x^{4}} ;

and finally

(i(4, n))_{n∈ℕ} = (8, 41, 227, 1234, 6743, 36787, 200798, 1095851, …)

with

∑_{n≥1} i(4, n)x^{n} = \frac{8x + 9x^{2} − 9x^{3} − 3x^{4} + x^{5}}{1 − 4x − 9x^{2} + 5x^{3} + 4x^{4} − x^{5}} .

If we work “backwards”, i.e., from column n to column 1 of G_{m,n}, we are able to express i(m, n) by a general formula, which for its “sum of products-form” might be interesting in its own. For this and any I ∈ I^{1}, we denote with s(m, I) the number of independent sets in I^{1}, that are orthogonal to I. Using the fact that i(m, 1) = F_{m+2} and assuming that I = {i_{1}, …, i_{p}} with i_{1} < i_{2} < ⋯ < i_{p}, s(m, I) can be written as follows:

s(m, I) = F_{i_{1}+1}F_{i_{2}−i_{1}+1} ⋯ F_{i_{p}−i_{p−1}+1}F_{m−i_{p}+2}.

We will not use this explicit form of s(m, I) further but, instead, show how to represent i(m, n) as a function of s(m, I)’s. We start with n = 2. Clearly,

i(m, 2) = ∑_{J∈I^{2}} i(m, 2, J).

But i(m, 2, J) is just the number of those sets in I^{1}, that are orthogonal to l(J); therefore,

i(m, 2) = ∑_{I∈I^{1}} s(m, I).

For n = 3, we have again

i(m, 3) = ∑_{J∈I^{3}} i(m, 3, J).

But

i(m, 3, J) = ∑_{K∈I^{2} and K⊥J} i(m, 2, K);

thus,

i(m, 3) = ∑_{K∈I^{2}} s(m, l(K))i(m, 2, K),

which equals

∑_{I∈I^{1}}(s(m, I))^{2}.

For n = 4, we have

i(m, 4) = ∑_{J∈I^{4}} i(m, 4, J) = ∑_{J∈I^{4}} ∑_{K∈I^{3} and K⊥J} i(m, 3, K)

= ∑_{K∈I^{3}} s(m, l(K))i(m, 3, K) = ∑_{K∈I^{3}} s(m, l(K)) ( ∑_{L∈I^{2} and L⊥K} i(m, 2, L)) ,

which equals

∑_{I∈I^{1}} s(m, I) ( ∑_{J∈I^{1} and J⊥I} s(m, J)) .

Finally, for n = 5 we obtain

i(m, 5) = ∑_{J∈I^{5}} i(m, 5, J) = ∑_{J∈I^{5}} ∑_{K∈I^{4} and K⊥J} i(m, 4, K)

= ∑_{K∈I^{4}} s(m, l(K))i(m, 4, K) = ∑_{K∈I^{4}} s(m, l(K)) ( ∑_{L∈I^{3} and L⊥K} ( ∑_{M∈I^{2} and M⊥L} s(m, l(M)))) ,

which equals

∑_{I∈I^{1}} s(m, I) ( ∑_{K∈I^{1} and K⊥I} ( ∑_{J∈I^{1} and J⊥K} s(m, J))) .

Since in this last expression, any s(m, J) is counted s(m, I ∪ J) times,

i(m, 5) = ∑_{I,J∈I^{1}} s(m, I)s(m, I ∪ J)s(m, J).

We proceed the same way for any n > 5 to obtain the following result. Theorem 2.1 We have

(i) i(m, 1) = F_{m+2}, (ii) i(m, 2) = ∑_{I∈I^{1}} s(m, I), (iii) i(m, 3) = ∑_{I∈I^{1}}(s(m, I))^{2}, (iv) i(m, 4) = ∑_{I∈I^{1}} s(m, I) (∑_{J∈I^{1} and J⊥I} s(m, J)),

(v) i(m, 5) = ∑_{I,J∈I^{1}} s(m, I)s(m, I ∪ J)s(m, J).

More generally, for p ≥ 3,

(vi) i(m, 2p) = ∑_{I_{1},...,I_{p}∈I^{1}} s(m, I_{1})s(m, I_{1}∪I_{2}) ⋯ s(m, I_{p−2}∪I_{p−1}) (∑_{J∈I^{1} and J⊥I_{p−1}} s(m, J)),

(vii) i(m, 2p + 1) = ∑_{I_{1},...,I_{p}∈I^{1}} s(m, I_{1})s(m, I_{1} ∪ I_{2}) ⋯ s(m, I_{p−1} ∪ I_{p})s(m, I_{p}).

3 B_{1,n} and B_{2,n}, their structure and cardinality

Let m = 1 and G_{1,n} be the corresponding grid graph, a path of length n; also, let B be a maximal independent set in G_{1,n} (an example is depicted in Figure 1).

Figure 1: The basis B = {(1, 1), (1, 4), (1, 6), …, (1, n − 2), (1, n)} for m = 1.

Obviously, B cannot contain two consecutive vertices on this path, nor can the complement of such a set contain three such vertices. How can we obtain b(1, n + 1)? We take a basis B from B_{1,n+1} and consider the elements it can have in common with the set {n − 2, n − 1, n, n + 1}. There are four possibilities for this intersection: {n − 2, n}, {n − 1, n + 1}, {n}, and {n − 2, n + 1}. The first two possibilities lead to a total number of b(1, n − 1), the second to b(1, n − 2) bases over {1, …, n + 1}, i.e.,

b(1, n + 1) = b(1, n − 1) + b(1, n − 2) for n ≥ 3  (3)
(with b(1, 1) = 1, b(1, 2) = b(1, 3) = 2).

But (3) is nothing but the definition of a Padovan sequence (see Stewart [8]), whose explicit form is

(b(1, n))_{n∈ℕ} = (1, 2, 2, 3, 4, 5, 7, 9, 12, 16, …) ,

and whose generating function is given by

∑_{n≥1} b(1, n)x^{n} = \frac{x + 2x^{2} + x^{3}}{1 − x^{2} − x^{3}} .

Now let us turn to m = 2 and consider a basis B ∈ B_{2,n} (see Figure 2 for an example).

Figure 2: A basis for m = 2.

Either (1, n) ∈ B or (2, n) ∈ B, in which case we can augment B by the vertex (2, n+1) or (1, n + 1), respectively, to obtain b(2, n) many bases within B_{2,n+1}. Moreover, if the vertices (1, n) and (2, n − 1) are in B, B \ {(1, n)} ∪ {(1, n + 1)} will be in B_{2,n+1}, and the same holds for B \ {(2, n)} ∪ {(2, n + 1)} in case that (2, n) and (1, n − 1) are elements of B. We hereby obtain the remaining b(2, n − 1) many bases of B_{2,n+1} so that altogether

b(2, n + 1) = b(2, n) + b(2, n − 1) for n ≥ 2  (4)
(with b(2, 1) = b(2, 2) = 2),

or more explicitly,

(b(2, n))_{n∈ℕ} = (2, 2, 4, 6, 10, 16, 26, 42, 68, 110, …) ,

which is nothing but a Fibonacci sequence with generating function

∑_{n≥1} b(2, n)x^{n} = \frac{2x}{1 − x − x^{2}} .

4 B_{3,n} and the general case

To calculate b(m, n) for m, n > 2 we maintain the basic ideas underlying the transfer matrix method for the determination of i(m, n):

1. Creation of a partition of B_{m,n} that is valid at any stage.

2. Construction of a transfer matrix T reflecting the contribution of a class at stage k to any other one at stage k + 1.

3. Determination of b(m, n) by means of T and the cardinality of the classes.

To create our partition it will be convenient to analyse the bases with respect to their structure within the last two columns. (Recall that in the independent set case I_{m,n} had been separated into F_{m+2} many classes whose elements had an identical structure within the last column). We first need to generate B_{m,3} from B_{m,2}, for which the following, general result will be useful:

Theorem 4.1 Given B_{m,n} and a set of vertices S = {(s_{1}, n), …, (s_{p}, n)} ⊆ V^{n} let r(S) denote the set {(s_{1}, n + 1), …, (s_{p}, n + 1)}. Then B_{m,n+1} is the collection of all maximal sets within {B′ : B′ = B \ I ∪ r(I), for B ∈ B_{m,n} and I ∈ I^{n}}.

Proof: Let B ∈ B_{m,n} and C = V \ B its complement with respect to V. Then C is a minimal cover of E, i.e., C contains at least one element from all the edges and C is minimal with respect to this property. We let C_{m,n} denote the collection of all these covers. Now consider the set of edges E^{+} that we add to G_{m,n} to obtain G_{m,n+1}. Obviously,

E^{+} = {{(i, n), (i, n + 1)} : i = 1, …, m} ∪ {{(i, n + 1), (i + 1, n + 1)} : i = 1, …, m − 1}.

It is easy to verify that the collection C^{+} of minimal covers of E^{+} is the following:

C^{+} = {C = I ∪ r(Ī) : I ∈ I^{n}, Ī = V^{n} \ I}.

By minimality, such a cover C cannot contain two consecutive vertices (i, n), (i + 1, n) ∈ V^{n} with i ∈ {1, …, m − 1}, and any edge {(i, n), (i, n + 1)} with i ∈ {1, …, m} not covered by (i, n) has to be covered by (i, n + 1), these latter vertices covering at the same time all edges {(i, n + 1), (i + 1, n + 1)} with i ∈ {1, …, m − 1}. Now to obtain the collection C_{m,n+1} of minimal covers of the edges of G_{m,n+1} we simply have to form the sets C ∪ C′ for any C ∈ C_{m,n}, C′ ∈ C^{+} and to retain the minimal ones. By complementation we get B_{m,n+1}.

It is clear that generating B_{m,n+1} this way will produce a large number of non-minimal covers. For n = 2 a thorough analysis of the bases in B_{m,2}, however, will allow us to keep this effort at a minimum level.

What is the structure of a basis B ∈ B_{m,2}? For an answer we consider two elements of B that are consecutive within the first column, i.e., which can be at distance 2, 3, 4, respectively, together with the position of the first or last element of B in that column (Figure 3 illustrates the four cases):

Case i) (s, 1) and (s + 2, 1) are in B for s ∈ {1, …, m − 2}: then, by maximality, (s + 1, 2) has to be in B, too.

Case ii) (s, 1) and (s + 3, 1) are in B: but then either (s + 1, 2) or (s + 2, 2) must be in B.

Case iii) (s, 1) and (s + 4, 1) are in B: then (s + 2, 2) has to be in B.

Case iv) If (s, 1) ∈ B for s = 2 or m − 1 (or if (s, 1) ∈ B for s = 3 or s = m − 2, and (1, 1) or (m, 1) ∉ B, respectively), then (1, 2) or (m, 2), respectively, has to be in B.

Case i) gives rise to the definition of an alternating path AP as induced by a maximal alternating sequence in column 1:

AP = {(s, 1), (s + 1, 1), …, (s + 2t − 1, 1), (s + 2t, 1)}

Figure 3: Substructures of a basis B ∈ B_{m,2}.

with

{(s, 1), (s + 2, 1), …, (s + 2t − 2, 1), (s + 2t, 1)} ⊆ B.

In such a case, the vertices (s + 1, 2), (s + 3, 2), …, (s + 2t − 1, 2) have to be in B, too.

We now come to the proper generation of B_{m,3}. Again, let B be a basis in B_{m,2}. It is our aim to describe all independent sets I ∈ I^{2} for which B′ = B \ I ∪ r(I) (see Theorem 2) is a member of B_{m,3}. To this end we start with I := Ø and check column 2 of B from top to bottom with respect to the substructures studied above within column 1: whenever we find an alternating path AP we modify I according to the following rule:

Rule i) Reduce AP ∩ B to those vertices (v, 2), for which v = 1 or v = m, or for which both (v − 1, 1) and (v + 1, 1) are contained in B; choose among the remaining ones an arbitrary set BL = {(b_{1}, 2), …, (b_{p}, 2)} and set I := I ∪ BL. Within AP \ B, choose the set

W = {(s + 1, 2), …, (b_{1} − 3, 2), (b_{1} + 3, 2), …, (b_{2} − 3, 2), (b_{2} + 3, 2), …, (s + 2t − 1, 2)}

and set I := I ∪ W. Also, whenever the vertex (s, 2) or (s + 2t, 2) (with s > 2, s + 2t < m − 1) has been chosen to be in BL, add vertex (s − 2, 2) or (s + 2t + 2, 2) to I, respectively.

We are now left with a second check of column 2 with respect to substructures ii), iii), iv): for those not yet treated in rule 1 we have to select 1 or 2 elements according to the following rules:

Rule ii) Choose vertex (s + 1, 2) or (s + 2, 2) and add it to I.

Rule iii) Choose vertex (s + 2, 2) (or vertices (s + 1, 2) and (s + 3, 2) if (s + 2, 1) is in B) and add it (them) to I.

Rule iv) Choose (1, 2) or (m, 2) if (2, 2) or (m − 1, 2) is in B \ I, respectively, and add it to I; if (3, 2) is the first, or (m−2, 2) the last element in B, then choose one of (1, 2), (2, 2) or (m − 1, 2), (m, 2), respectively, and add it to I.

Set I has been constructed not only to be a member of I_{2}: the set B′ will always be a maximal independent set, i.e., a member of B_{m,3}. On the other hand, any such maximizing I appears as a particular case of our construction so that altogether we generate the complete family B_{m,3}. Also observe that due to the different possibilities to choose a set BL and a white vertex in the other cases, every basis in B_{m,2} generates a number of bases in B_{m,3}, which equals a power of 2. Finally, our construction allows to partition B_{m,3} into p equivalence classes, two bases being considered equivalent if they have the same intersection with V^{2} ∪ V^{3}. Any such intersection can be represented by a representing graph P_{m,2} (see Figure 4 for an illustration of the case m = 3).

Figure 4: Generation of B_{m,3} from B_{m,2} and the 8 representing graphs P_{m,2}.

If we now proceed to the generation of B_{m,4} from B_{m,3}, we can be sure to terminate with the same collection of representing graphs P_{m,2} (and this is true for any stage): the collection of independent sets represented by the P_{m,2}’s contains B_{m,2}, for any represented J there is a basis B from B_{m,2} such that J ⊆ B and B \ J contains only vertices of the first column.

What is still missing is an answer to the following question: what are the classes of stage k that produce (via Theorem 2) a given class at stage k + 1? For an answer let us say that class i contributes to class j if for some (this is sufficient!) B in class i there is an I in I^{3} such that B \ I ∪ r(I) is an element of (B_{m,4} and) class j.

We are now ready to introduce the notion of a “transfer matrix”: let p be the number of classes obtained from partitioning B_{m,3}, and let T ∈ {0, 1}^{p×p} be defined as follows:

T_{ij} = 1 iff class j contributes to class i .

Then we get the following Proposition 4.1 If for k=3,4,5,…, c_{i}^{k} denotes the cardinality of class i at stage k, the c_{i}^{3} being given for i = 1, …, p, then

c_{i}^{k+1} = ∑_{j=1}^{p} T_{ij}c_{j}^{k} for i=1,…,p, and b(m, k + 1) = ∑_{i=1}^{p} c_{i}^{k+1}.

To conclude this section we just mention that the initial size of matrix T can be reduced as follows: merge two classes into one whenever their representing graphs are (upside down) symmetric (note: T may then contain entries ≥ 2), or if they contribute to the same classes in the same way. These aspects will be further clarified in the next section, where a full solution of the cases m = 3, 4, 5 will be presented.

5 The cases m = 3, 4, 5

Let us start with m = 3. We have already seen how to generate B_{m,3} from B_{m,2}, as depicted in Figure 4. If we merge “symmetric” classes we come up with 6 classes, whose representing graphs are resumed in Figure 5:

Figure 5: Representing graphs for m = 3.

Now observe that classes 5 and 6 contribute to the same classes in the same way, namely class 1, so that we can merge them into one new class 5. The corresponding transfer matrix is

T (3) = [ \begin{matrix}0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \end{matrix} ]

which allows us, by application of Proposition 1, to calculate the class sizes c_{i}^{n} for i = 1, …, 5:

classes n=3 4 5 6 7 8 9 10
1  1 4 7 13 30 59 117 246
2  1 1 4 7 13 30 59 117
3  4 6 14 28 54 114 232 466
4  1 2 5 11 20 43 89 176
5  3 5 8 19 39 74 157 321

as well as the final values of b(3, n) for n > 3, hereby producing the sequence

(b(3, n))_{n∈ℕ} = (2, 4, 10, 18, 38, 78, 156, 320, 654, 1326, …) ,

whose generating function is

∑_{n≥1} b(3, n)x^{n} = \frac{2x + 2x^{2} + 4x^{3} − 2x^{4} − 2x^{6}}{1 − x − x^{2} − 3x^{3} + x^{4} + x^{5}} .

For m = 4 the partition of B_{4,n} is represented by the following graphs:

Figure 6: Representing graphs for m = 4.

As above, we can merge classes 6 and 7 to obtain the transfer matrix:

T (4) = [ \begin{matrix}1 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 0 & 0 \end{matrix} ]

together with the associated class cardinalities:

classes n=3 4 5 6 7 8  9 10
1  2  8 20 50 128 324 820 2078
2  2  6 12 32 82 204 520 1316
3  2  4 10 26 64 164 414 1048
4  4 10 26 64 164 414 1048 2656
5 2 2 8 20 50 128 324 820
6  6 12 32 82 204 520 1316 3330

We obtain the sequence:

(b(4, n))_{n∈ℕ} = (3, 6, 18, 42, 108, 274, 692, 1754, 4442, 11248, …) ,

whose generating function is

∑_{n≥1} b(4, n)x^{n} = \frac{3x + 3x^{2} + 3x^{3} − 3x^{4} − 3x^{5} − 2x^{6} + x^{7}}{1 − x − 3x^{2} − 3x^{3} + x^{4} + 2x^{5} + x^{6}} .

For m = 5, finally, we come up with 17 representing graphs: Figure 7: Representing graphs for m = 5.

We can merge classes 6 and 11, 9 and 15, 14 and 16 to obtain a 14 × 14 transfer matrix:

T (5) = [ \begin{matrix}0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 1 & 1 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\ 2 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{matrix} ]

and the corresponding class cardinalities:

classes n=3 4 5 6 7  8  9  10
1  1  8 21 61 218 677 2097 6814
2  1  1 8 21 61 218 677 2097
3  2  6 22 66 206 676 2124 6692
4  1  4 15 51 148 485 1571 4898
5  4 14 40 134 414 1340 4200 13426
6  6 22 66 206 676 2124 6692 21476
7  4 10 40 126 386 1244 4012 12548
8 1 4 7 29 82 267 841 2708
9  4 10 28 102 324 988 3226 10242
10  2  4 22 64 188 642 2030 6318
11  4  8 28 82 278 832 2722 8588
12  4  7 29 82 267 841 2708 8491
13  2  3 11 36 123 357 1176 3765
14  2  7 21 72 209 691 2194 6929

which produce the sequence

(b(5, n))_{n∈ℕ} = (4, 10, 38, 108, 358, 1132, 3580, 11382, 36270, 114992, …) ,

with generating function

∑_{n≥1} b(5, n)x^{n} =

\frac{4x + 6x^{2} + 12x^{3} − 10x^{4} − 18x^{5} + 2x^{6} − 20x^{7} + 20x^{8} − 50x^{9} + 38x^{10} − 18x^{11} + 2x^{12} + 6x^{13} + 2x^{14}}{1 − x − 4x^{2} − 10x^{3} − 4x^{4} + 20x^{5} − x^{6} + 2x^{7} − 2x^{8} + 16x^{9} − 4x^{10} − x^{13}} .

We may want to compare this sequence with the corresponding one for the independent set case (taken from Sloane [6]):

(i(5, n))_{n∈ℕ} = (13, 99, 827, 6743, 55447, 454385, 3729091, 30584687, 250916131, 2058249165, …).

The conclusion with respect to their growth is immediate.

6 Conclusion

In this paper we have introduced a new class of integer sequences over grid graphs generalizing both Padovan and Fibonacci sequences, but still maintaining the essential property of independence. Counting the bases of a grid graph (or other graph families along the same line) and analysing their structure should therefore still be attractive for the applications mentioned in the introduction: tiling (our bases may have a very regular structure), efficient coding schemes (when closely related to the hard-square model) and statistical physics (in particular, the question of the true value of lim_{m,n→∞} b(m, n)^{1/mn}).

7 Acknowledgments

We are grateful to the referee for many valuable suggestions.

References

[1] R. J. Baxter, I. G. Enting and S. K. Tsang, Hard square lattice gas, J. Statist. Phys. 22 (1980), 465–489.

[2] A. Burstein, S. Kitaev and T. Mansour, Independent sets in certain classes of (almost) regular graphs, preprint, University of Kentucky, Lexington, 2004.

[3] N. Calkin and H. S. Wilf, The number of independent sets in a grid graph, SIAM J. Discrete Math. 11 (1997), 54–60.

[4] K. Engel, On the Fibonacci number of an m × n lattice, Fibonacci Quart. 28 (1990), 72–78.

[5] R. M. Roth, P. H. Siegel and J. K. Wolf, Efficient coding schemes for the hard-square model, IEEE Trans. Inform. Theory 47 (2001), 1166–1176.

[6] N. J. A. Sloane, The on-line encyclopedia of integer sequences, published electronically at http://www.research.att.com/∼njas/sequences/

[7] R. P. Stanley, Enumerative Combinatorics, Vol.1, Cambridge University Press, 1997.

[8] I. Stewart, Tales of a neglected number, Sci. Amer. 274 (1996), 102–103.

[9] M. Weigt and A. K. Hartmann, Glassy behavior induced by geometrical frustration in a hard-core lattice gas model, Europhys. Lett. 62 (2003), 533.

2000 Mathematics Subject Classification: Primary 11B39; Secondary 11B83, 05A15, 05C69. Keywords: Fibonacci number, Padovan number, transfer matrix method, independent set, grid graph.

(Concerned with sequences \underline{A000045} \underline{A000931} \underline{A001333} \underline{A051736} \underline{A051737} and \underline{A089936}.)

Received October 7 2004; revised versions received February 7 2005; April 12 2005. Published in Journal of Integer Sequences, May 17 2005.

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