Re: [math-fun] Divisor chains

RIchard Guy wrote: << How do you know that there is such a `divisor chain' ? It took quite a while for me to realize that for {1,2,...,n}, if we let M = floor(n/2), the following permutations always work: For n even: M+1, 1, M+2, 2, ..., n-1, M-1, n, M. For n odd: M+1, 1, M+2, 2, ..., M-1, n-1, M, n The proof is both magical and completely obvious. E.g., for n = 10: 6, 1, 7, 2, 8, 3, 9, 4, 10, 5; for n = 37: 19, 1, 20, 2, ..., 35, 17, 36, 18, 37. Still don't know how to show there exists a 37-chain beginning with 37, 1. But since the final term must divide 37*19, the third term must be 2 in any such chain. D'uh! --Dan