... row, all 2s in the second row, and so on. In the “straight half” we see only prime numbers, with 2′ on the first diagonal, 3′ on the second diagona l, and so on. Lemma 6 tells us that every ... by moving the first letter of a wordto the end, and we show that the orbit structure of this action is encoded by the generating function for the major index on R(w0).1 Introduction and main ... of permutations of X, SX. Therefore, to find the cycle structure of the image of any bijection ω : X → X,it is enough to determine the order of the action of ω on X and find a polynomial X(q)such...