@article{oai:ynu.repo.nii.ac.jp:00007018,
 author = {Matsumoto, Keiichi and Nakamigawa, Tomoki and Watanabe, Mamoru},
 issue = {2},
 journal = {Yokohama Mathematical Journal = 横濱市立大學紀要. D部門, 数学},
 month = {},
 note = {application/pdf, In this paper we will study an alternative row version of Josephus problem. Suppose that ＄n＄numbers 1, 2, ＄￥cdot＄ . . , ＄n＄ are arranged in a line from left to right in this order. Starting with number 1, and counting each number from left to right, every second number is eliminated. Subsequently, starting with the right most number of the remains and counting each number in turn in the contrary direction, ＄i.e＄ . from right to left, every second number is eliminated. Repeat such a process by alternate changing the order of cunting and eliminating until only one number is left. Denote by ＄f_{t}(n)＄ the number of the ＄(n-t+1)＄-th element which is removed by the process described above. If ＄n' s＄ binary expansion is ＄￥sum_{k=0}^{￥infty}2^{k}n_{k}(n_{k}=1,0)＄ , let us denote ＄f(n)=￥sum_{k=0}^{￥infty}2^{2k+1}n_{2k+1}＄ . Let ＄g_{t}(n)＄ be either ＄0＄ for ＄t=1＄ , or ＄(-2)^{r}￥{f(￥sigma^{r}(2n-1))+f(￥sigma^{r}(n-1))-2t+3￥}＄ for ＄t￥geq 2＄ , where ＄ r=￥lfloor￥log_{2}￥frac{n-1}{t-1}￥rfloor＄ and ＄￥sigma(n)=L￥frac{n}{2}￥rfloor＄ , i.e. ＄￥sigma(n)＄ is one-bit shift right of ＄n' s＄ binary expansion. In this paper we prove that ＄f_{t}(n)=f(n-1)+1+g_{t}(n)＄ .},
 pages = {83--88},
 title = {ON THE SWITCHBACK VERTION OF JOSEPHUS PROBLEM},
 volume = {53},
 year = {2007}
}