Combinatorial Analysis

Problem Set 7

Most of the problems are assigned from the required textbook Bona, Miklos. A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory. World Scientific Publishing Company, 2011. ISBN: 9789814335232. [Preview with Google Books]

A problem marked by * is difficult; it is not necessary to solve such a problem to do well in the course.

Problem Set 7

• Due in Session 19
• Practice Problems
  • Session 17: None from textbook
    • Let n≥1, and let f(n) be the number of partitions of n such that for all k, the part k occurs at most k times. Let g(n) be the number of partitions of n such that no part has the form i(i+1), i.e., no parts equal to 2, 6, 12, 20, …. Show that f(n)=g(n). Use generating functions.
    • Let f(n) denote the number of partitions of n with an even number of 1’s. Give a combinatorial proof and a generating function proof that f(n) + f(n-1) = p(n), the total number of partitions of n.
  • Session 18: Chapter 8: Exercises 20, 21
• Problems Assigned in the Textbook
  • Chapter 8: Exercises 27, 28, 32, 37*. In exercise 28, you can ignore the last sentence (about comparing with Exercise 4). Hint for 37. Consider the product 1/(1-qx)(1-qx²)(1-qx³)…
• Additional Problems
  • (A8*) Show that the number of partitions of n for which no part appears exactly once is equal to the number of partitions of n for which every part is divisible by 2 or 3. For instance, when n=6 there are four partitions of the first type (111111,2211,222,33) and four of the second type (222, 33, 42, 6). Use generating functions.
  • (A9) Show that the number of partitions of n for which no part appears more than twice is equal to the number of partitions of n for which no part is divisible by 3. For instance, when n=5 there are five partitions of the first type (5, 41, 32, 311, 221) and five of the second type (5, 41, 221, 2111, 11111). Use generating functions.
• Bonus Problems
  • (B2) Find the generating function G(x) = Σ_{_n_≥0} anxn/n!,    
    where a_n+1 = (n+1)a_n-{n\choose 2}a_n-2 for n≥0, and a₀=1.     
    Thus a₁=1, a₂=2, a₃=5. You don’t need to find a formula for a_n.