a nim-like game

Two players, starting with one pile of n stones. The players take turns. In one move, the current player can pick any pile of stones and split into 2 or 3 new piles with each new pile having at least 1 stone. The player cannot make a move loses. Given n, can you determine the winner, assuming both play optimally?

The answer is deceptively simple, so spend at least one day before you look at the solution below.





























Solution:
n = 1, first player loses
n >= 2, if n is even, then the first player can split into two equal piles and mimic the second players move, so the first player wins.
else n is odd, but then the first player can split into 1, k, k, assuming n = 2*k +1. Then the first player can again mimic the second players moves. So the first player wins as well.

In summary, the first player wins for all n >= 2. Simple, huh?