Problem 1824

The answer is $2^n - 3 \cdot 2^{n/3} + 2$ for $n = 3k$ and $2^n$ for $n \neq 3k$. The number of possible box choices is $2^n$. Let us examine cases when two or more different choices produce the same ball distribution. If different choices $C_1$ and $C_2$ produce the same distribution then these choices should not coincide on any two consecutive boxes. Otherwise $C_1$ and $C_2$ will coincide for all remaining boxes. Suppose that the boxes are numbered in clockwise order as $1, 2, \dots, n$ and the sum $i + j$ is defined in (mod $n$). If in choice $C$ box $m$ is chosen we write $C(m) = X$, otherwise we write $C(m) = Y$. Thus, for two consecutive boxes $m, m + 1$ there are four possibilities $XX, XY, YX, YY$. Note that if $C_1(m) = X$, $C_1(m+1) = X$ and $C_2(m) = Y$, $C_2(m+1) = Y$ then $C_1$ and $C_2$ can not produce the same distribution, since box $m$ will contain different number of balls for $C_1$ and $C_2$. for these two choices. Therefore, there are at most three pairwise different choices producing the same distribution. It can be readily shown that if $n$ is a multiple of $3$ then there are only two triples producing the same distribution:
$$C_1 = \dots X X X X X X X X X \dots$$
$$C_2 = \dots X X X X X X X X X \dots \quad (1)$$
$$C_3 = \dots X X X X X X X X X \dots$$
$$C_1 = \dots Y Y Y Y Y Y Y Y Y Y \dots$$
$$C_2 = \dots Y Y Y Y Y Y Y Y Y \dots \quad (2)$$
$$C_3 = \dots X Y Y Y Y Y Y Y Y Y \dots$$
In each of this two cases each choice can be obtained by shifting of any other one. Evidently if $n$ is not a multiple of $3$ there is no any such triple.

Now let us examine the cases when exactly two choices produce the same distribution. Note that for $C_1$ and $C_2$ there are two consecutive boxes $m$ and $m+1$ such that $C_1(m) \neq C_2(m)$ and $C_1(m+1) \neq C_2(m+1)$. Otherwise for some $m$ we have $C_1(m) = C_2(m)$, $C_1(m+2) = C_2(m+2)$, $C_1(m+4) = C_2(m+3)$, ... and consecutively $C_1$ will coincide with $C_2$. W.L.O.G. let $C_1(m) = X$, $C_1(m+1) = Y$ and $C_2(m) = Y$, $C_2(m+1) = X$. Then we get $C_1(m+2) = C_2(m+2) = Z$, where $Z$ is either $X$ or $Y$. Similarly, $C_1(m+3) = X$, $C_1(m+4) = Y$ and $C_2(m+3) = Y$, $C_2(m+4) = X$. Then we get $C_1(m+5) = C_2(m+5) = Z$, where $Z$ is either $X$ or $Y$. Continuing this way we get that starting from box $m$ the choices $C_1$ and $C_2$ are $XYZ_1XYZ_2XYZ_3XYZ_4$... and $YXZ_1YXZ_2YXZ_3YXZ_4$,... respectively. If $n$ is not multiple of $3$ then we can not cyclically close this chain and will get a contradiction.
Thus, if $n$ is not a multiple of $3$ there are no different choices producing the same distribution and the answer is $2^n$.
If $n$ is a multiple of $3$ then the positions of $Z_i$ can be determined in $3$ ways. If for some choice $Z_1 = Z_2 = Z_3$... then we get one of the six choices from (1) and (2). Therefore, for each of $3 \cdot (2^{n/3} - 2) \cdot 2$ choices there exists exactly one other choice producing the same distribution. Therefore, the answer is
$$2^n - 6 - 6 \cdot (2^{n/3} - 2) + \frac{6}{3} + \frac{6}{2} \cdot (2^{n/3} - 2) = 2^n - 3 \cdot 2^{n/3} + 2.$$