By Ulrich Dempwolff

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**Extra info for Algebraic Combinatorics**

**Example text**

N − k)! n k=0 n f (k)g(n − k) X n = H(X) k follows. ✷ Remark Let f1 , . . ,Tk ) n≥0 fj (n) n n! X . f1 (|T1 |) · · · fk (|Tk |) where (T1 , . . , Tk ) ranges over the ordered partitions of size k of [n] and let n H(X) = n≥0 h(n) n! X be the associated EGF. Then F1 (X) · · · Fk (X) = H(X). 1 by an obvious induction. With the next theorem we compute the composition of EGF’s. 2 (Composition formula for EGF,s) Define for the functions f, g : (n) n n N → C, f (0) = 0, the EGF’s F (X) = n≥0 f n! X , G(X) = n≥0 g(n) n!

We call the procedure associating. For n = 3 associating gives the two expression a1 (a2 a3 ) and (a1 a2 )a3 . Determine the number of expressions on can obtain by association. It turns out that all problems yield the same numbers. We do not prove this but determine these numbers in the last case. Definition Denote by Cn the number of ways of associating a free word of length n. The numbers Cn are called the Catalan numbers. 3 Cn = 1 2n − 2 (2n − 2)! (n − 1)! n n−1 n ≥ 1. Proof. We know C1 = C2 = 1 and C3 = 2.

Then f (|S1 |) · · · f (|Sk |)g(k) X n = g(k) F (X)k . k! Define H0 (X) = g(0). Then G(F (X)) = Hk (X) = g(0) + k≥0 k≥1 = g(0) + k≥1 n≥1 = g(0) + n≥1 1 n! 1 n! k≥1 g(k) F (X)k k! ,Sk } f (|S1 |) · · · f (|Sk |)g(k) X n f (|S1 |) · · · f (|Sk |)g(k) X n = H(X). 3 (Exponential formula) Define for the function f : N → C, (n) n f (0) = 0, the EGF F (X) = n≥0 f n! ,Sk } where {S1 , . . , Sk } ranges over the unordered partitions of [n]. Let H(X) = h(n) n n≥0 n! X be the associated EGF. Then exp(F (X)) = H(X).