2. A total of $2^{n}$ coins are distributed among several children. If during the distribution of the coins, one of the children has at least half of the coins, then that child gives to each of the remaining children as many coins as they already have. In the case where all the coins are with one child, there is no redistribution of the coins, and if two children have more than half of the coins, then one of them gives all the coins to the other. Determine the maximum possible number of consecutive redistributions of the coins. (For example, if 32 coins are initially distributed among 6 children as $17,2,9,1,2,1$, then after the first redistribution, the children will have $2,4,18,2,4,2$ coins, and after the second redistribution, the children will have $4,8,4,4,8,4$ coins, so there are only two redistributions.)
Solution. First, we will give an example when we have $n$ consecutive redistributions. Let the given $2^{n}$ coins be distributed among 3 children as follows:

$$
1,2^{n-1}+2^{n-2}+\ldots+2,1
$$

The subsequent $n$ redistributions will be:

$$
2^{1}, 2^{n-1}+2^{n-2}+\ldots+2^{2}, 2^{1}
$$

$$
\begin{aligned}
& 2^{n-2}, 2^{n-1}, 2^{n-2} \\
& 2^{n-1}, 0,2^{n-1} \\
& 0,0,2^{n}
\end{aligned}
$$

$$
2^{2}, 2^{n-1}+2^{n-2}+\ldots+2^{3}, 2^{2}
$$

Assume that there exists an initial distribution for which at least $n+1$ redistributions are possible. Clearly, after each redistribution, the number of coins each child has is an even number. Specifically, the number of coins for the children who receive coins is doubled, and the number of coins for the child who gives coins is the difference between the number $2^{n}$ and the sum of even numbers, so it is also an even number. This means that after the $k$-th redistribution, the number of coins each child has is divisible by $2^{k}$. Therefore, after $n+1$ redistributions, the number of coins each child has is divisible by $2^{n+1}$, which is not possible since there is a child with a non-zero number of coins, and that number is less than or equal to $2^{n}$.