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Josephus revisited

Posted by ubpdqn on April 28, 2011

I have been puzzling over the Jospehus problem for step size 2. The wikipedia entry provides the solution for this and insights into its properties. It proves that the  number of the last man standing, f(n),  is 2 l+1 where n= 2^m +l, \quad 0\leq l<2^m.

A proof by induction is presented.   I sought a constructive approach to facilitate my understanding. I present my musings in the post: Two step Josephus.

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