Let us label $(a, b)$ the position meaning "the player who has to move finds $a$ stones in the heap and can take at most $b$ stones".

We prove that the losing positions are those of the forms $(0, r)$ and $(2^k(2m+1), r)$, where $k, m \in \mathbb{N}$, $r \in \{1, 2, \dots, 2^k - 1\}$.

The initial position being $(n, n-1)$, the losing initial positions are those with $m=0$, i.e. the powers of $2$. The conclusion will be that if $n$ is a power of $2$, the second player has a winning strategy, while in the other cases, the first player has a winning strategy. Let us notice that:

- the game ends only in positions of the form $(0, r)$

- from a position of the form $(2^k(2m+1), r)$, $0 < r < 2^k$, there is no move that could leave the opponent in a position of similar form

- from any position that is not one of the above form, there exists a move that leaves the opponent in a position of the form $(2^k(2m+1), r)$, $0 < r < 2^k$.

• If from $(2^k(2m+1), r)$, $0 < r < 2^k$, one could move to $(2^s(2p+1), t)$, $0 < t < 2^s$, that move would consist in removing $t$ stones, hence $2^k(2m+1) - 2^s(2p+1) = t$. As $t \le r < 2^k$, it follows that $s < k$, hence $2^s \mid 2^k(2m+1) - 2^s(2p+1)$. Then $2^s \mid t$, which contradicts $0 < t < 2^s$. In conclusion, such a move does not exist.

Obviously, there is no move to a position $(0, x)$ either.

• From any of the remaining positions, that is $(2^k(2m+1), r)$ with $r \ge 2^k$, there is a move that leaves the opponent in position $(2^k \cdot 2m, 2^k)$, obviously a losing one. Simply take $2^k$ stones.

From the above, it is easy to see what the appropriate winning strategy is for the player that has such a strategy: always take $2^k$ stones, where $2^k$ is the largest power of $2$ that divides the number of the remaining stones.