By Richard Stanley,

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Tn )|ti =1 . ∂tk Hence Ek (n)xn = n≥0 ∂ x2 x3 exp t1 x + t2 + t3 + · · · ∂tk 2 3 ti =1 x2 x3 xk exp x + + +··· k 2 3 xk = exp log(1 − x)−1 k xk 1 = k 1−x xk = xn . k n≥0 = It follows that Ek (n) = 1/k for n ≥ k. Can the reader think of a simple explanation (Exercise 118)? Now define c(n, k) to be the number of permutations w ∈ Sn with exactly k cycles. The number s(n, k) := (−1)n−k c(n, k) is known as a Stirling number of the first kind, and c(n, k) is called a signless Stirling number of the first kind.

11: (a) The min-max tree M = M(5, 10, 4, 6, 7, 2, 12, 1, 8, 11, 9, 3); (b) The transformed tree ψ7 M = M(5, 10, 4, 6, 7, 2, 1, 3, 9, 12, 11, 8) so that they keep their same relative order. Again this defines ψi M(w). 11(b) shows that ψ7 M(5, 10, 4, 6, 7, 2, 12, 1, 8, 11, 9, 3) = M(5, 10, 4, 6, 7, 2, 1, 3, 9, 12, 11, 8). We have a7 = 12, so ψ7 permutes vertex 12 and the vertices on the right subtree of 12. Vertex 12 is replaced by 1, the smallest vertex of the right subtree. The remaining elements 1, 3, 8, 9, 11 get replaced with 3, 8, 9, 11, 12 in that order.

Let w = 683941725 ∈ S9 . Then γ1 = 6. It is irrelevant at this point whether 6 < w2 or 6 > w2 since there can be only one compartment, and γ2 = 68. Now 8 > w3 = 3, so we split 68 after numbers greater than 3, getting 6 | 8. Cyclically shifting the two compartments of length one leaves them unchanged, so γ3 = 683. Now 3 < w4 = 9, so we split 683 after numbers less than 9. We get 6 | 8 | 3 and γ4 = 6839. Now 9 > w5 = 4, so we split 6839 after numbers greater than 4, giving 6 | 8 | 39. The cyclic shift of 39 is 93, so γ5 = 68934.