Infinite representation

By virtue of the $1 \rightarrow 4 \rightarrow 2 \rightarrow 1$ cycle, any of these bracketed representations is actually infinite. For example, $3 = \lbrace 5, 1, 0 \rbrace = \lbrace \ldots, 21, 19, 17, 15, 13, 11, 9, 7, 5, 1, 0 \rbrace$ ; which is to say,

$3 = \frac{2^5 - 2^1 - 3} {9} = \frac{2^7 - 2^5 - 2^1 \cdot 3 - 9} {27} = \frac{2^9 - 2^7 - 2^5 \cdot 3 - 2^1 \cdot 9 - 27} {81} = \ldots$

Conversely, numbers to the left of the bracket can be eliminated, if the gap to the next element is just

If is given by

$n = \frac{2^{a_m} - \displaystyle\sum_{k=0}^{m-1} 2^{a_k} \, 3^{m-1-k}} {3^m}$

multiplying by up and down, we get

$n = \frac{3 \cdot 2^{a_m} - 3 \cdot \displaystyle\sum_{k=0}^{m-1} 2^{a_k} \, 3^{m-1-k}} {3^{m+1}}$

then adding $0 = \displaystyle\frac{2^{a_m+2} - 2^{a_m+2}} {3^{m+1}} = \displaystyle\frac{2^{a_m+2} - 4 \cdot 2^{a_m}} {3^{m+1}}$ will produce

$n = n + \frac{2^{a_m+2} - 4 \cdot 2^{a_m}} {3^{m+1}} = \frac{2^{a_m+2} - 2^{a_m} - 3 \cdot \displaystyle\sum_{k=0}^{m-1} 2^{a_k} \, 3^{m-1-k}} {3^{m+1}} = \frac{2^{a_m+2} - \displaystyle\sum_{k=0}^m 2^{a_k} \, 3^{m-k}} {3^{m+1}}$

where the right-hand side expression corresponds to the alternative bracketed expression for

we were looking for, $\lbrace a_m+2, a_m, a_{m-1}, \ldots, a_2, a_1, a_0 \rbrace$ .

$\square$

The proof can be traced backwards, in order to show that the implication also holds in the other direction, making it possible to remove a_m+2 from the front of the brackets.