On the table, there are 2023 tokens. Taking turns, Adrien and Iris must remove at least one token and at most half of the remaining tokens at the time they play. The player who leaves a single token on the table loses the game. Adrien starts. Determine which of the two players has a winning strategy and describe such a strategy.
We start from the end to find the first losing positions: $2,5,11,23$. It seems that the losing positions are exactly the terms of the sequence $\left(u_{n}\right)_{n \geqslant 1}$ satisfying $u_{1}=2$ and $u_{n+1}=2 u_{n}+1$ (i.e., the terms obtained by successively multiplying by 2 and adding 1).

Let's prove by induction on $n$ that if the initial number of tokens $k$ satisfies $u_{n} \leqslant k < u_{n+1}$, then the position is winning, and if $k = u_{n}$, then the position is losing.

**Base Case:** For $n=1$, $u_1 = 2$. If $k = 2$, the first player can only take 1 token, leaving 1 token for the second player, who then wins. Thus, $k = 2$ is a losing position.

**Inductive Step:** Assume that for some $n \geq 1$, if $u_{n} \leqslant k < u_{n+1}$, the position is winning, and if $k = u_{n}$, the position is losing. We need to show that if $u_{n+1} \leqslant k < u_{n+2}$, the position is winning, and if $k = u_{n+1}$, the position is losing.

If $k = u_{n+1}$, the first player can only take a number of tokens $m$ such that $1 \leq m \leq \frac{k}{2}$. Since $k = u_{n+1} = 2u_{n} + 1$, the maximum number of tokens the first player can take is $\frac{k}{2} = u_{n} + \frac{1}{2}$. Therefore, the first player can take at most $u_{n}$ tokens, leaving $k - u_{n} = u_{n} + 1$ tokens for the second player. Since $u_{n} + 1 < u_{n+1}$, the second player is in a winning position by the inductive hypothesis. Thus, $k = u_{n+1}$ is a losing position.

If $u_{n+1} < k < u_{n+2}$, the first player can take $m = k - u_{n+1}$ tokens, leaving $u_{n+1}$ tokens for the second player. Since $k - u_{n+1} \leq \frac{k}{2}$, this move is valid. We need to show that $k - u_{n+1} \leq \frac{k}{2}$, which is equivalent to $\frac{k}{2} \leq u_{n+1}$, and thus to $k \leq 2u_{n+1}$, which is true because $k < 2u_{n+1} + 1$. The second player is then in a losing position, so if $u_{n+1} < k < u_{n+2}$, the position is winning, which completes the inductive step.

Thus, the losing positions are exactly the terms of the sequence, which is strictly increasing: they are $2, 5, 11, 23, 47, 95, 191, 383, 767, 1535, 3071$. Therefore, 2023 is not a term of the sequence, so Adrien has a winning strategy.