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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 2 Solutions 1.6 In each part, a sequence { a n } n ≥ satisfies the given recurrence relation. Find the ordinary power series generating function of the sequence. (a) a n + 1 = 3 a n + 2 where ( n ≥ 0; a = ) . Solution. Let f ( x ) = ∑ n ≥ a n x n . Proceeding with the method outlined in Section 1.2.1 we see ∑ n ≥ a n + 1 x n = 3 ∑ n ≥ a n + 2 ∑ n ≥ x n and so f x = 3 f + 2 1 x so f ( x ) = 2 x ( 1 x )( 1 3 x ) . (c) a n + 2 = 2 a n + 1 a n where ( n ≥ 0; a = 0; a 1 = 1 ) . Solution. Let f ( x ) = ∑ n ≥ a n x n . Then ∑ n ≥ a n + 2 x n = 2 ∑ n ≥ a n + 1 x n ∑ n ≥ a n x n and so f x x 2 = 2 f x f f ( x ) = x ( 1 x ) 2 . 1.7 Give a direct combinatorial proof of the recurrence (1.44) as follows: given n , consider the collection of all partitions of the set [ n ] . There are b ( n ) of them. Sort out this collection into piles numbered k = , 1 ,..., n 1, where the k th pile consists of all partitions of [ n ] in which the class that contains the letter n contains exactly k other letters. Count the partitions in the k th pile, and you’ll be all finished. Proof. The recurrence we want to prove is b ( n ) = n 1 ∑ i = n 1 i b ( i ) . Using the hint in the problem, let’s count the k th pile. To do this we proceed as follows: The element n is in some class with k other letters. Choose those k elements from the remaining n 1 elements in ( n 1 k ) = ( n 1 n k 1 ) ways. Then partition the remaining n k 1 elements in...
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This note was uploaded on 05/19/2010 for the course MATH mat 146 taught by Professor Gregkuperberg during the Spring '10 term at UC Davis.
 Spring '10
 GregKuperberg

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