Unformatted text preview: AMS 301 (Fall 2009) Homework Set # 10 – Solution Notes # 6, 7.1: (a). a n = a n − 1 + a n − 2 . If the first digit is a 0, there are a n − 1 ways to complete the sequence. If the first digit is a 1, then the second digit must be a 0, and then there are n 2 digits left, so a n − 2 ways to complete the sequence. a 1 = 2, a 2 = 3 are the initial condition. (b). a n = 2 a n − 1 + 2 a n − 2 . If the first digit is a 0, or a 2, there are a n − 1 ways to complete the sequence. If the first digit is a 1, then the second digit must be a 0, or 2, and then there are n 2 digits left, so a n − 2 ways to complete the sequence. a 1 = 3, a 2 = 8 are the initial condition. # 24, 7.1: In this case once a digit 2 appears, all digits to its right must be 0 and 2’s (no 1’s allowed). Our recursion conditions on the first digit. If it is a 0 or 1 (two choices) then any sequence of length n 1 that has no 1 to the right of a 2 can follow. However, if the first digit is a 2, the remaining1 that has no 1 to the right of a 2 can follow....
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This note was uploaded on 08/08/2010 for the course AMS 301 taught by Professor Arkin during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 ARKIN

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