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F (k − 1) can be chosen arbitrarily while the other values of f are determined by the initial r conditions. We observe i=1 di = k and for each polynomial Pi one can choose 0 the coefficients of X , X, . . , X di −1 arbitrarily. So dim V3 = k too. Similar in V4 the coefficients of the Gi ’s can chosen arbitrarily which forces dim V4 = k. For f ∈ V1 we have f (n)X n P (X) = Q(X) n≥0 k f (n)X n + = n≥0 bi f (n)X n+i i=1 n≥0 k = ∗ + ∗X + · · · + ∗X k−1 + f (n + k) + n≥0 39 i=1 bi f (n + k − i) X n+k .

Sk } are the supports of the cycles of π with c(π) = k. Let n H(X) = n≥0 h(n) n! X be the associated EGF. Then H(X) = G( n≥1 f (n) n X ). n Proof. The number of k–cycles of a k–set is (k − 1)!. So if Π = {S1 , . . , Sk } is a partition of [n] we have precisely (|S1 | − 1)! · · · (|Sk | − 1)! permutations whose supports of the cycles induce the partition Π. f (|Sk |)g(k) where {S1 , . . , Sk } ranges over the unordered partitions of [n] of size k. ,Sk } Defining F ∗ (X) = ∗ f (n) n n! X f ∗ (|S1 |) · · · f ∗ (|Sk |)g(k).

N−k k≥0 (b) Problem: Show the identity k≥0 m k n+k m = k≥0 46 m k n k 2 , k m, n ≥ 0. We multiply both sides by X n . Summing over n we have to verify the identity L(X) = R(X) of power series with Xn L(X) = n≥0 k≥0 m k n+k , m Xn R(X) = n≥0 k≥0 We first rearrange the LHS and get using 1/(1 − X)k+1 = n+k n n≥0 k X L(X) = k≥0 = k≥0 = = m X −k k n≥0 m k n k 2 . k n≥0 n+k n Xn = n+k X n+k m Xm m X −k k (1 − X)m+1 Xm 1 1 + )m (1 − X)m+1 X (1 + X)m (1 − X)m+1 In a similar way we treat the RHS. R(X) = k≥0 = = = m k 2 k 1 (1 − X) n≥0 k≥0 m k m Xn k 2X (1 − X) 2X 1 1+ (1 − X) 1−X (1 + X)m (1 − X)m+1 k m Hence L(X) = R(X) and the identity follows.