Jennifer J. Quinn

Recounting Fibonacci and Lucas Identities

Arthur Benjamin (benjamin@nriath.hnric.edu) earned his B.S. in Applied Mathematics from Carnegie Mellon and his Ph.D. in Mathematical Sciences from Johns Hopkins where he studied discrete optimization under Alan J. Goldman. Since 1989, he has taught at Harvey Mudd College, where he is currently an associate professor. He is editor of the Spectrum book series for MAA, and an Associate Editor of Mathematics Magazine. He was recently awarded the MAAs Haimo Award for distinguished college teaching.

The Fibonacci Numbers -- Exposed More Discretely

In the previous article, Kalman and Mena [5] propose that Fibonacci and Lucas sequences, despite the mathematical favoritism shown them for their abundant patterns, are nothing more than ordinary members of a class of super sequences. Their arguments are beautiful and convinced us to present the same material from a more discrete perspective. Indeed, we will present a simple combinatorial context encompassing nearly all of the properties discussed in [5]. As in the Kalman-Mena article, we generalize Fibonacci and Lucas numbers: Given nonnegative integers a and b, the generalized Fibonacci sequence is

Strong chromatic index of subset graphs

The strong chromatic index of a graph G, denoted sq(G), is the minimum number of parts needed to partition the edges of G into induced matchings. For 0 5 k 5 1 5 m, the subset graph S,(k, 1) is a bipartite graph whose vertices are the k- and 1-subsets of an m element ground set where two vertices are adjacent if and only if one subset is contained in the other. We show that sq(Sm(k, 1)) = (&) and that this number satisfies the strong chromatic index conjecture by Brualdi and Quinn for bipartite graphs. Further, we demonstrate that the conjecture is also valid for a more general family of bipartite graphs. @ 1997 John Wiley & Sons, Inc.

An Alternate Approach to Alternating Sums: A Method to DIE for.

would you be inclined to try to prove these by counting arguments? Probably not. We confess that there was a time that when we saw alternating sum identities like the ones above, we would attack it with noncombinatorial proof techniques, like induction or generating functions. After all, how can an object be added a negative number of times? Perhaps it can be tackled using the Principle of InclusionExclusion (abbreviated P.I.E.), but that brings another level of complexity to the combinatorial proof. But now we have an alternate opinion. Now when we see an alternating sum, we can usually prove it combinatorially with a method that is as easy as pie and even easier than P.I.E.! As we’ll demonstrate, this technique offers new insights to identities involving binomial coefficients, Fibonacci numbers, derangements, and other combinatorial structures.

Counting on Continued Fractions

have the same numerator. These fractions simplify to 1 a3993 and 103993 respectively. _T3_102 355 In this paper, we provide a combinatorial interpretation for the numerators and denominators of continued fractions which makes this reversal phenomenon easy to see. Through the use of counting arguments, we illustrate how this and other important identities involving continued fractions can be easily visualized, derived, and remembered. We begin by defining some basic terminology. Given an infinite sequence of integers ao ? 0, a, ? 1, a2 ? 1,. let [ao, a n.a,] denote the finite continued fraction

Composite Fermions and Integer Partitions

We utilize the KOH theorem to prove the unimodality of integer partitions with at most a parts, all parts less than or equal to b, that are required to contain either repeated or consecutive parts. We connect this result to an open question in quantum physics relating the number of distinct total angular momentum multiplets of a system of N fermions, each with angular momentum ?, to those of a system in which each Fermion has angular momentum ?*=??N+1.

Semimetal-semiconductor transition in InAsGaSb heterostructures

Abstract In a heterostructure consisting of neighboring layers of InAs and GaSb in a large band-gap host like AlSb, the bottom of the InAs conduction band lies ∼150meV below the top of the GaSb valence band. However, because the electron and hole subbands are weakly coupled for non-zero values of the wavevector k ∥ along the layer, an anticrossing of the levels occurs at k ∥ = k c , where k 2 c = (2μ/ħ 2 ) (Δ−e v 0 −e c 0 ). Here Δ is the band overlap, e c 0 and e v 0 are the lowest electron and hole subband energies, and μ is the reduced mass. The splitting is in the order of 5–10 meV. For an intrinsic material, the Fermi level lies in this gap, so that no semimetallic state occurs. If the InAs layers are sufficiently wide, or if an electric field can be applied across the heterostructure (by fabricating front and back gates) it is possible to have the first excited electron subband e c 1 realize an anticrossing with e v 0 as well. Then a true semimetallic state can occur with a circle of electrons in the e c 1 subband near k ∥ = 0, and a ring of holes near the anticrossing in the hybridized valence subband. A semimetal-to-semiconductor transition should occur, which is driven by shifting the single-particle subband energies with an applied electric or magnetic field.

Random Approaches to Fibonacci Identities

Many combinatorialists live by Mach’s words, and take it as a personal challenge. For example, nearly all of the Fibonacci identities in [5] and [6] have been explained by counting arguments [1, 2, 3]. Among the holdouts are those involving infinite sums and irrational quantities. However, by adopting a probabilistic viewpoint, many of the remaining identities can be explained combinatorially. As we shall demonstrate, even the “irrational-looking” Binet’s formula for the n-th Fibonacci number

Summing Cubes by Counting Rectangles

But that’s not the end of the story. As we’ll see, we can apply this “checkerboard logic” to derive the famous formulas for summing squares and cubes. Recall that the sum of the first n cubes is ∑n k=1 k 3 = (n+1 2 )2. This formula can be proved by induction, telescoping sums, combinatorially [1], or geometrically [2]. Our four-line proof suggests that we should be able to directly count rectangles on an n × n checkerboard and arrive at the same conclusion. But which rectangles are being counted by the k3 term? We claim that these rectangles have upper right corner (x, y) with largest coordinate k (i.e., max{x, y} = k). Such a rectangle would lie inside a k × k rectangle but not inside a (k − 1)× (k − 1) rectangle. See Figure 2. From our four-line proof, there are ( k + 1 2 )2 − ( k

Transformation of statistics in fractional quantum Hall systems

A Fermion to Boson transformation is accomplished by attaching to each Fermion a tube carrying a single quantum of flux oriented opposite to the applied magnetic field. When the mean field approximation is made in Haldanes spherical geometry, the Fermion angular momentum IF is replaced by IB = IF -