This question is essentially to come up with a solution for this problem in the case where there are $2^n$ coins. Let $C_n$ be the group $\{f : \mathbb Z / 2^n \mathbb Z \to \mathbb Z / 2 \mathbb Z\}$ equipped with the group operation pointwise addition. Let $R_n = \mathbb Z / 2 \mathbb Z \times \mathbb Z / 2^n \mathbb Z$, and let $R_n$ act on $C_n$ via $(\alpha, x) \cdot f(y) = \bar{\alpha} + f(x + y)$, where $\bar \alpha = (\alpha, \alpha, ..., \alpha)$. I would like to exhibit the existence of a series $\{0\} = G_0 < G_1 < G_2 < ... < G_{2^n} = C_n$ such that the following holds:
(1) $|G_i| = 2^i$
(2) $(G_i \setminus G_{i-1}) \cdot (G_i \setminus G_{i-1}) \subset G_{i-1}$. Edit: Sean Eberhard correctly pointed out that this condition is superfluous.
(3) Each $G_i$ is closed under the action of $R_n$ for $i > 0$.
I can exhibit such a series for $n = 1, 2, 3$. I'd like to know if such a series exists for all $n$. We will denote an element of $C_n$ as a subset of $\{0, 1, ..., 2^n\}$, so that $S$ denotes the function $f_S$ such that $f_S(x) = 1$ if and only if $x \in S$. For a set $T \subset C_n$, we will use the notation $gen(T)$ to denote the subgroup generated by elements of the form $r \cdot t$ for $r \in R_n, t \in T$.
For $n = 1$, we can take
- $G_1 = gen(\{\{0, 1\}\})$
- $G_2 = gen(G_1 \cup \{\{0\}\})$
For $n=2$, we can take
- $G_1 = gen(\{\{0, 1, 2, 3\}\})$
- $G_2 = gen(G_1 \cup \{\{0, 2\}\})$
- $G_3 = gen(G_2 \cup \{\{0, 1\})$
- $G_4 = gen(G_3 \cup \{\{0\}\})$
For $n=3$, we can take
- $G_1 = gen(\{\{0, 1, 2, 3, 4, 5, 6, 7\}\})$
- $G_2 = gen(G_1 \cup \{\{0, 2, 4, 6\}\})$
- $G_3 = gen(G_2 \cup \{\{0, 1, 4, 5\}\})$
- $G_4 = gen(G_3 \cup \{\{0, 4\}\})$
- $G_5 = gen(G_4 \cup \{\{0, 1, 2, 3\}, \{0, 1, 3, 6\}\})$
- $G_6 = gen(G_5 \cup \{\{0, 1, 2, 5\}, \{0, 2\}\})$
- $G_7 = gen(G_6 \cup \{\{0, 1\}, \{0, 3\}, \{0, 1, 3, 5\}, \{0, 1, 2, 4\}\})$
- $G_8 = gen(G_7 \cup \{\{0\}\})$