Content-Type: text/shitpost

More progress on counting paths on octahedra! Suppose you want to know how many paths of length !!n!! there are between two opposite vertices of an octahedron.

It turns out that it is the same as the number of ways to take !!n!! terms, each of which is either !!\pm1!! or !!\pm2!!, and add them up to get an odd multiple of 3. (Order matters.)

For example, there are 8 paths of length 3, which correspond to !!1+1+1, -1+2+2, 2-1+2, 2+2-1,!! and their negatives.

That actually seems like an improvement because it seems like counting those sequences will be a straightforward application of generating functions.