On the Number of Independent Sets in a Tree

We show in a simple way that for any k,m ∈ N, there exists a tree T such that the number of independent sets of T is congruent to k modulo m. This resolves a conjecture of Wagner (Almost all trees have an even number of independent sets, Electron. J. Combin. 16 (2009), # R93). 1 The number of independent sets in a tree A set of vertices in a graph G is called independent if the set induces no edges. We write i(G) for the number of independent sets in G; i(G) is often known as the Fibonacci number, or in mathematical chemistry as the Merrifield-Simmons index or the σ-index. The study was initiated by Prodinger and Tichy in [4]. In particular, they showed that among trees of the same order, the maximum and minimum Fibonacci numbers are attained by the star and the path respectively. The name stems from the fact that the Fibonacci numbers of paths are the usual Fibonacci numbers. Indeed, as the empty set is independent, i(P0) = 1, i(P1) = 2 and i(Pn) = i(Pn−1) + i(Pn−2) for n > 2. The inverse question asks for a positive integer k, whether there exists a graph G such that i(G) = k. Clearly there does as i(Kk−1) = k (note that the empty set is independent). The question becomes more interesting if we restrict ourselves to certain classes of graphs. For the class of bipartite graphs, Linek [3] answered the question affirmatively. Here we are interested in the class of trees. For k ∈ N, we say that k is constructible if there exists a tree T such that i(T ) = k. For example, 1, 2, 3 are constructible (from the paths P0, P1, P2 respectively) but 4 is not. In [3], Linek raised the following conjecture (see also [2]). Conjecture 1 ([3]). There are only finitely many positive integers that are not constructible. the electronic journal of combinatorics 17 (2010), #N18 1 An interesting paper of Wagner [5] looks at the number of independent sets modulo m. Wagner showed that the proportion of trees on n vertices with the number of independent sets divisible bym tends to 1 as n tends to infinity. In the same paper, Wagner [5] proposed a weaker version of Conjecture 1. Let C(m) = {i(T ) (mod m) : T a tree }. Conjecture 2 ([5]). For m ∈ N, C(m) = Zm. The aim of this paper is to prove Conjecture 2. In fact, we prove a stronger result. For a rooted tree (T, r), let i0(T, r) denote the number of independent sets not covering the root. Let D(m) = {(i0(T, r), i(T )) (mod m) : (T, r) a rooted tree}. Theorem 3. For m ∈ N, D(m) = Zm. First we note a recursion between the Fibonacci number of a rooted tree and its subtrees. Suppose r1, r2, · · · , rj are the neighbours of r, let (Tk, rk) be the subtree of T rooted at rk. Then we have [2, 5]