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You are not logged in. #1 20060509 09:26:44
seriesThe 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. #2 20060510 00:03:58
Re: seriesIt 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 onedimesional pearl building on a piece of sand. How's that for a math metaphore! #3 20060514 22:03:33
Re: seriescan anyone else be able to find out the pattern of this series is please? #4 20060515 03:07:47
Re: seriesLet 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 i1 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..." #6 20060515 11:20:34
Re: seriess 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... Hahahaaaa!!! Last edited by krassi_holmz (20060515 11:21:50) IPBLE: Increasing Performance By Lowering Expectations. #7 20060515 11:27:26
Re: seriesG(1)=1; IPBLE: Increasing Performance By Lowering Expectations. #9 20060515 18:30:30
Re: seriesNo. You was right without being wrong. IPBLE: Increasing Performance By Lowering Expectations. 