By Bjorner A., Stanley R.P.

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For instance, N (2, 3) = 3, as shown by: We have in fact that N (2, n) = Fn+1 , 42 (23) where Fn+1 denotes a Fibonacci number, defined by the recurrence F1 = 1, F2 = 1, Fn+1 = Fn + Fn−1 . To prove equation (23), we need to show that N (2, 1) = 1, N (2, 2) = 2, and N (2, n + 2) = N (2, n + 1) + N (2, n). Of course it is trivial to check that N (2, 1) = 1 and N (2, 2) = 2. In any domino tiling of a 2 × (n + 2) rectangle, either the first column consists of a vertical domino, or else the first two columns consist of two horizontal dominos.

34 Hence if n > pq, then either (λ) ≥ p + 1 or λ1 ≥ q + 1. If we apply the Schensted correspondence to a permutation w of 1, 2, . . , n then we get a pair of reverse SYT of some shape λ, where λ is a partition of n. We have just shown that (λ) ≥ p + 1 or λ1 ≥ q + 1, so by Schensted’s theorem either i(w) ≥ p + 1 or d(w) ≥ q + 1. We can evaluate each f λ appearing in Schensted’s theorem by the hooklength formula. Hence the theorem is most interesting when there are few partitions λ satisfying (λ) = p and λ1 = q.

Nevertheless, we hope that our brief description will take some of the mystery out of equation (24). We first color the squares of the Aztec diamond AZn black and white in the usual chessboard fashion, with the first (leftmost) square in the top row colored white. Here is a tiling of AZ3 with the chessboard coloring shown. 46 Each domino will have one white square and one black square. There are four possible colorings and orientations of a domino, shown in the illustration below. With each of these four possible colored dominos we associate a direction: up, down, right, and left, as indicated below by an arrow.