Algebraic Combinatorics - download pdf or read online

By Ulrich Dempwolff

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Wilf in his book (p. 22) as follows: 1. Take the logarithm of both sides of the equation. 2. Differentiate both sides and multiply through by X. 3. Clear the equation of fractions. 51 4. For each n, find the coefficient of X n on both sides of the equation and equate them. We apply this method in the case of the Bell numbers. 4 For all n ≥ 1 n−1 B(n) = k=0 Proof. The formula B(x) = n≥0 after applying the logarithm to: log n≥0 n−1 B(k). k 1 n n! B(n)X = exp(exp(X) − 1) transforms 1 B(n)X n = exp(X) − 1 n!

K and roots a1 ∈ V (Γ1 ), . . , ak ∈ V (Γk ). We define a tree Γ0 on [n + 1] by adding n + 1 to the vertex set of Γ and the edges {a1 , n + 1}, . . , {ak , n + 1} to the set of edges. If on the other hand if ∆ is a tree on [n + 1] and N (n + 1) = {a1 , . . , ak } we delete form ∆ the vertex n + 1 and the edges {a1 , n + 1}, . . , {ak , n + 1} and obtain a rooted forest ∆0 on [n] with k connected components. Of course we take the ai ’s as roots and observe that the components are characterized by the ai ’s.

X . Thus Sk (X) = Gk (F (X)) = 1 (exp(X) − 1)k . k! n k=0 Recall that the Bell numbers are defined as B(n) = (as S(n, k) = 0 for k > n). Denote by B(X) = n≥0 S(n, k) = k≥0 S(n, k) B(n) n X n! the EGF of the Bell numbers. Then B(X) = n≥0 1 n! S(n, k) X n k≥0 = k≥0 = k≥0 n≥0 S(n, k) n X = n! Sk (X) k≥0 1 (exp(X) − 1)k = exp(exp(X) − 1). k! ✷ Remark Although no explicit formula for the Bell numbers is known we have a simple, explicit generating function for them. In cases where the generating function of a number series is explicitly known one can often find recurrences for these numbers following a recipe described by H.

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Algebraic Combinatorics by Ulrich Dempwolff


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