Arthur T. Benjamin

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.

How do I marry thee? Let me count the ways

Abstract A stable marriage problem of size 2n is constructed which contains 0(2”

Counting on Chebyshev Polynomials

) stable match- ings. This construction provides a new lower bound on the maximum number of stable matchings for problems of even size and is comparable to a known lower bound when the size is a power of 2. The method of construction makes use of special properties of the latin marriage problem, which we develop. 1. Introduction The stable marriage problem consists of n men and n women who are to be matched up into married couples. Each man ranks the women from most desirable to least desirable, and each woman does the same for the men. A matching is said to be unstable if there exists a man and woman who prefer each other to the partners they have been assigned. If no such people exist, the matching is said to be stable. In [2], Gale and Shapley proved that a stable matching always exists, but it need not be unique. The problem of determing the maximum number of possible stable matchings among all stable marriage problems of size n was posed by Knuth [S] and remains an open question. As reported in [3], Knuth established that this maximum number exceeds 2”12 for

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

where Tn is the Chebyshev polynomial of the first kind, defined by T0(x) = 1, Tx(x) = x, and for n > 2, Tn(x) = 2xTn.x{x) - Tn_2(x). (2) For example, T2(x) = 2x2 - 1, T3(x) = 4x3 - 3x, T4(x) = 8x4 - 8x2 + 1. This gen erates the familiar trigonometric identity eos(20) = 2 cos2 9 ? 1, and the less familiar cos(36>) = 4 cos3 9 - 3 cos 9 and cos(4#) = 8 cos4 9 - 8 cos2 9 + 1. If we change the initial conditions to be Uo(x) = 1 and Ux(x) = 2x, but keep the same recurrence Un(x) = 2xUn-i(x) - Un-2(x), we get the Chebyshev polynomials of the second kind. For instance, U2(x) = Ax1 ? 1, U3(x) = Sx3 - Ax, U4(x) = 16x4 - I2x2 + 1. The Chebyshev polynomials generate many fundamental sequences, including the constant sequence, the sequence of integers, and the Fibonacci numbers. Its easy to show that for all n > 0, Tn(l) = 1 and Un(l) = n + l, Tn(~l) = (-If, Un(-l) = (? l)n(n + 1). When we substitute complex numbers, such as x = i/2, the Fibonacci and Lucas numbers appear. Specifically,

A Combinatorial Proof of Vandermonde's Determinant

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.

Combinatorial Proofs of Fermat's, Lucas's, and Wilson's Theorems

that is as easy as playing cards. Let Vn denote the Vandermonde matrix with (i, j)th entry vij = x j i (0 ≤ i, j ≤ n). Since the determinant of Vn is a polynomial in x0, x1, . . . , xn, it suffices to prove the identity for positive integers x0, x1, . . . , xn with x0 ≤ x1 ≤ · · · ≤ xn. We define a Vandermonde card to possess a suit and a value, where a card of Suit i has a value from the set {1, . . . , xi}. (In our examples, we will let Suits 0, 1, 2, 3, and 4 be represented by suits ,♣,♦,♥, and ♠, respectively.) Hence there are x0 + x1 + · · · + xn different Vandermonde cards, but we have at our disposal an unlimited supply of each. First we do some card counting. Card Counting Question 1. How many ways can Vandermonde cards be arranged in n+1 rows, where row 0 is empty, row 1 has one card of Suit 1, row 2 has two cards of Suit 2, row 3 has three cards of Suit 3, . . . , and row n has n cards of Suit n? The order of the cards is important, and we are allowed to repeat values of cards within each row. We call such an arrangement a Vandermonde table associated with the identity permutation π = 012 . . . n, an example of which is given in Figure 1.

Counting on Continued Fractions

(L(p)[Q], Q)k = (PQ, PQ)k+m = +IIell2m |> IIPII\QII, (5) showing that L(p) is positive. Therefore, we see that all eigenvalues of L(p) are positive and, on the basis of (5), that II P II| furnishes a lower bound for them. Furthermore, equality holds in (4) if and only if either P 0 or II P II| is the smallest eigenvalue of L(p) and Q is an eigenvector corresponding to it (unless Q = 0). A particular case in which equality holds in (4) occurs when P = P(y) belongs to Nm and Q = Q(z) to Nk, where y E RP, z E Rq, and IR = RP x Rq. Added in proof. Professor Luo Xuebo, who was one of his coauthors Ph.D. supervisors, died in March 2004. Zhu-Jun Zheng expresses his deep respect for and everlasting memory of his deceased colleague and mentor.

MAGIC SQUARES INDEED

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