Credit: A challenge from my professor.
two people are playing nim and the win condition is taking the last sticks. Let x be some number greater than 1. the first person takes 1 stick. when it is a persons turn and the person before them took n sticks, they can take anywhere from 1 to floor(nx) sticks. what amount of sticks would guarantee a win for person 1? for person 2?
I don't know the full answer, but I solved it for n = 2 earlier. It was known by the name NIM2.
The losing positions are usually the Fibonacci numbers. Except where $F_n \geq F_{n-1} * 1.5.$ In this case, instead of $F_n$, the losing position is $F_{n-1}$ plus the largest losing number which is less than $F_n-F_{n-1}$. It is a recursive process, so we are essentially using a Fibonacci number system.
So the winning strategy for n = 2 is to step to a losing position. If you can't, substract the largest possible Fibonacci number $F_k$ from it, repeat the process again, and add $F_k$ back to the result.