By Paul Zeitz

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The first terms of the sequence then are 0,1,0,1,0,1,1,0, etc. We need to know if the subsequence 1,1,1,1,1 can ever appear. If it does, then it must appear for the first time. Suppose that 1,1,1,1,1 is such a subsequence. ). So now we have 0,1,1,1,1,1 to work with. Since the righmost term is 1, and it is the sum (modulo 2) of the 6 previous terms, the term to the left of the 0 must also be 1. By the same reasoning, the term to the left of this 1 must also be 1, etc. It is not hard to deduce that there must, in fact, be 5 1’s to the left of the 0: we now have 1,1,1,1,1,0,1,1,1,1,1.

3 1 Since 1 ⊥ 1, we have f (1) = f (1 · 1) = f (1) f (1), which means that f (1) equals 0 or 1. The first possibility is excluded because the range of f is N. 6 Let d|n, where n = ab with a ⊥ b. By the FTA, we can write d = uv, with u|a and v|b. The trickier part is showing that this is a one-to-one correspondence. It suffices to show that if the u, u and v, v are respectively divisors of a and b and uv = u v , then u = u , v = v . This follows by mimicking the final steps of the proof of the FTA on p.

We have s2 = 3, s3 = 5 (including the empty set). Now let us compute sn+1 . The subsets under consideration partition into two classes: those which contain the element n + 1, and those that do not. The second class consists of the sn subsets of {1, 2, . . , n} which contain no two consecutive elements. The first class consists of subsets of {1, 2, . . , n − 1} which contain no two consecutive elements, each with the element n + 1 appended. There are sn−1 such subsets, so sn+1 = sn + sn−1 , yielding yet another Fibonacci recurrence.

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The Art And Craft Of Problem Solving Instructors Manual by Paul Zeitz

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