Now why did I want to generate n-bracketings in the first place ... or what amounts to the same thing, ordered rooted trees (ORTs) with n+1 nodes? [Here each node corresponds to a matching pair of brackets, and its children to the pairs immediately within the parent pair. An ORT will be equipped with an extra half-edge ("bole"?) leading into the root node, which corresponds to a virtual outer pair of brackets imagined enclosing any given bracketing.] How many k-valent nodes occur among the set of all distinct (n+1)-ORTs? After spending all that time generating the damn' things, the answer is an anticlimax: apparently it's just (2n-k)_C_(n-1) the dear old binomial coefficient! Is this well-known? Can anyone suggest a proof? If so, should OEIS A007318 be amended to incorporate: " n_C_k = number of (2k+2-n)-valent nodes occurring among distinct ordered rooted trees with k+2 nodes, provided each root node is equipped with an extra inward half-edge " ? Example: n = 3 bracketing valency 1,...,4 ( ((())) ) [1, 3, 0, 0] ( (()()) ) [2, 1, 1, 0] ( ()(()) ) [2, 1, 1, 0] ( (())() ) [2, 1, 1, 0] ( ()()() ) [3, 0, 0, 1] totals [10, 6, 3, 1] Fred Lunnon