I initially thought in this manner: Assume there is only one pile having n stones for now

• $n = 0$: winning state $\implies g(0) = 1$ (Can I even do this? I am not following the MEX rule but it is a winning state so g(0) can't be 0 so just assigned it 1)
• $n = 1$: losing state $\implies g(1) = 0$
• $n = 2$: winning state since you can go to a losing state, $g(2) = mex(g(0), g(1)) = mex(1, 0) = 2$
• $n = 3$: $g(3) = 3$ (winning state)
• $n = 4$: $g(4) = 4$ (winning state) and so on...

This is wrong probably because of my assumption in $n = 0$. An Example of where it fails:

• Say the game when there are 2 piles having 1 stone each: {1, 1}
• $g(1) = 0 \implies$ using Sprague-Grundy theorem, $g(Game) = g(1) \oplus g(1) = 0 \oplus 0 = 0$
• $g(Game) = 0$ but it is not a losing state