# Algebraic Combinatorics by Ulrich Dempwolff

By Ulrich Dempwolff

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**Example text**

This shows: √ √ 1− 5 1+ 5 , β= α= 2 2 Since the two factors are coprime we know that (by the main result on partial fractions) there exist A, B ∈ C with A B A + B − (αA + βB)X 1 = + = . 1 − X − X2 1 − αX 1 − βX (1 − αX)(1 − βX) Hence A + B = 1 and αA + βB = 0. The solutions are A = α/(α − β) and B = −β/(α − β) and therefore 1 β α f=√ . − 1 − βX 5 1 − αX On the other hand (1 − aX)−1 = 1 f=√ 5 n≥0 n≥0 an X n , so that we finally obtain (αn+1 − β n+1 )X n . We have reached the famous formula √ √ 1 1 + 5 n+1 1 − 5 n+1 Fn = √ ( .

N k=0 n B(k) X n . k Hence for n ≥ 1 B(n) (n − 1)! = [X n ] n≥1 = [X n ] n≥0 = 1 (n − 1)! n B(n)X n n! 1 n! n−1 k=0 n k=0 n B(k) X n+1 k n−1 B(k) k and the proposition is verified. ✷ Here is a version of the composition formula using permutations. 5 Define for the functions f, g : N → C, f (0) = 0, the EGF’s (n) n n F (X) = n≥1 f n! X , G(X) = n≥0 g(n) n! X . Define further h : N → C by h(0) = g(0) and h(n) = k≥1 π∈Sym(n),c(π)=k f (|S1 |) · · · f (|Sk |)g(k), 52 n ≥ 1, where {S1 , . . , Sk } are the supports of the cycles of π with c(π) = k.

2. Create a double sum S(X) = k≥0 ak X k ( n≥0 bn X n−k ) such that the inner sum has a useful description in a closed form B(X) = n≥0 bn X n−k . 3. Write S(X) = k≥0 ak X k B(X). 4. If one is lucky the coefficients of S can be evaluated. Examples will be helpful to explain the strategy. Examples (a) Evaluate: k , n−k sn = k≥0 n∈N According to step 1 we consider: k n−k Xn S(X) = n≥0 k≥0 k n−k For the second step we recall that = k≥0 n≥0 = 0 for n < k or n > 2k. So writing Xk S(X) = k≥0 k Xn n−k n≥0 k X n−k n−k we get for the inner sum k X n−k = n−k n≥0 k i=0 k X i = (1 + X)k .