Josephus Nim

In this study, we propose a variant of Nim that uses two piles. In the first pile, we have stones with a weight of $a$, and in the second pile, we have stones with a weight of $-2a$, where $a$ is a natural number. Two players take turns to remove stones from one of the piles. The total weight of the stones to be removed should be equal to or less than half of the total weight of the stones in the pile. Therefore, if there are $x$ stones with weight $a$ and $y$ stones with weight $-2a$, then the total weight of the stones to be removed is less than or equal to $(ax-2ay)/2$. The player who removes the last stone is the winner of the game. The authors proved that when $(n,m)$ is the winning position of the previous player, $2m+1$ is the last remaining number in the Josephus problem, where there are $n+1$ numbers, and every second number is to be removed. For any natural number $s$, there are similar relationships between the position at which the Grundy number is $s$ and the $(n-s)$-th removed number in the Josephus problem with $n+1$ numbers.