By René Schoof

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For n ≥ 1, let qn count the number of ways we can cover (or tile) a 2 × n chessboard using the 1 × 2 and 2 × 1 dominoes. Here q1 = 1, since a 2 × 1 chessboard requires one 2 × 1 (vertical) domino. 1 Fibonacci and Catalan Numbers: An Introduction, First Edition. Ralph P. Grimaldi. © 2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc. 33 34 TILINGS: DIVISIBILITY PROPERTIES OF THE FIBONACCI NUMBERS in Fig. 1 (c). So q2 = 2. For n ≥ 3, we consider the last (nth) column of a 2 × n chessboard.

Sn , where Si ⊆ A, for each 1 ≤ i ≤ n, and where S1 ⊆ S2 , S2 ⊇ S3 , S3 ⊆ S4 , S4 ⊇ S5 , . . , with (i) Sn−1 ⊆ Sn for n even, while (ii) Sn−1 ⊇ Sn for n odd and greater than 1. Determine an . (b) Answer the question in part (a) if A = {1, . . , m}, where m ≥ 2. 22 SOME INTRODUCTORY EXAMPLES 8. To raise money for the campus drive at their university, the sisters of Gamma Kappa Phi sorority are sponsoring a casino night. As a result, two of their pledges—namely, Piret and Columba—have been assigned to stack green poker chips and gold poker chips so that each stack contains ten chips, where no two adjacent chips are allowed to be green.

For a given chessboard C, this polynomial will be denoted by r(C, x) and, for any nonnegative integer k, the coefﬁcient of xk in r(C, x) will be the number of ways one can place k nontaking rooks on the chessboard C. This coefﬁcient will be denoted by rk (C, x). For any chessboard C, we have r0 (C, x) = 1, the number of ways to place no nontaking rooks on C. Also, r1 (C, x), the number of ways to place one nontaking rook on C, is simply the number of squares on the chessboard C. The rook polynomial for the chessboard in Fig.