By K. Ribnikov

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**Example text**

This shows n−1 cn ≤ ci cn−1−i , i=0 and so in light of our earlier discussion, the two quantities must equal each other: n−1 cn = ci cn−1−i . i=0 Since the cn satisfy the same recurrence as the Catalan numbers with the same initial values, cn = Cn , and we have the following combinatorial characterization of Catalan numbers. 1. For n ≥ 1, |Sn (231)| = Cn . 3 Narayana numbers 23 via a bijection. 4 for another set of objects counted by Catalan numbers. 1. It is possible to exhibit bijections between the set Sn (231) and the set Sn (p), where p ∈ {123, 132, 213, 312, 321} is any pattern of length three.

A Schr¨ oder path of size n is a lattice path from (0, 0) to (n, n) that never passes below the line y = x and uses only steps “North” from (i, j) to (i, j +1), “East” from (i, j) to (i+1, j) and “Northeast” from (i, j) to (i + 1, j + 1). ) Let Rn denote the number of Schr¨oder paths of size n, with R0 = 1. We call the oder number. Here are the ﬁrst few values for Rn : number Rn a Schr¨ 1, 2, 6, 22, 90, 394, 1806, 8558, 41586, 206098, . .. Let R(z) = n≥0 Rn z n . Show that R(z) = 1−z− √ 1 − 6z + z 2 .

Rk } such that if {a, d} ⊆ Ri and {b, c} ⊆ Rj with a < b < c < d, then Ri = Rj . Here are the fourteen nonnesting partitions of {1, 2, 3, 4}: 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 ; 1 2 3 4 : Hint: Create a bijection between noncrossing and nonnesting partitions. Conclude that counting nonnesting partitions by number of blocks gives the Narayana numbers. 8 (Noncrossing matchings, balanced parenthesizations).