I'm a bit disappointed I didn't see this quicker back when I did this problem like 15 years ago or whatever it is now. My solution was to use matrix exponentiat

I'm a bit disappointed I didn't see this quicker back when I did this problem like 15 years ago or whatever it is now. My solution was to use matrix exponentiation on the adjacency matrix of the path graph, which is a heck of a lot better than brute force enumeration of all paths, but still roughly cubic rather than linear like computing a bunch of factorials. Even as a math major, I'd done a lot more linear algebra than combinatorics.