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The integers 1,2,...,n are arranged in order a1,a2,...an, so that each term ak is either larger than all aj with j<k or else it is smaller than all aj with j<k. For example, the order 3,2,4,1 satisfies these conditions for n=4.
It seems like, if you have your n integers arranged in a row, you will have to choose some starting number, and then pick a number that borders on it for the second number in the series. Then pick a number than is a neigbor on the side of that group. It just continues to build up on the original number picked, like a one-dimesional pearl building on a piece of sand. How's that for a math metaphore!
Let A be the set of numbers already in the sequence. Let s be the sup of A, and i be the inf of A. Then the next number in the sequence must be either i-1 or s+1. So it would seem to be 2^n. But there lies a problem with your starting number. If you pick say 2 as your first number, and then 1, there is only one way to place the rest of the numbers, no matter what size n.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
s is sup of A if s>all members of A.
Then the number we're looking for is:
Good. Now just have to simplify it...
Last edited by krassi_holmz (2006-05-15 11:21:50)
IPBLE: Increasing Performance By Lowering Expectations.