Ternary rule for comma-numbers

A complete proof and corrected boundary classification for ternary comma-numbers

In The Comma Sequence: A Simple Sequence With Bizarre Properties, The Fibonacci Quarterly 62 (2024), 215--232, Eric Angelini, Michael S. Branicky, Giovanni Resta, N. J. A. Sloane, and David W. Wilson state Conjecture 9.2 concerning the comma-number of every positive integer written in base three. We prove the conjectured comma-number values for all digit lengths and make its exceptional families disjoint, correcting several boundary annotations in the printed table.

Write a positive integer in its canonical ternary representation

$$n=(d_1d_2\cdots d_m)_3, \qquad d_1\in\{1,2\}, \qquad d_m\in\{0,1,2\}. \tag{1}$$

Throughout, a superscript in a digit word indicates repetition, so that $2^i$ denotes a string of $i$ copies of the ternary digit $2$, and $2^0$ denotes the empty string. Define the uncorrected comma-number by

$$c_0(n)=(d_md_1)_3=3d_m+d_1. \tag{2}$$

The complete, disjoint classification is

$$\begin{array}{c|c|c} \text{ternary word for }n&\operatorname{cn}(n)& \operatorname{cn}(n)-c_0(n)\\ \hline 12^i1,\quad i\geq1 &(12)_3&+1\\ 12^ij2,\quad i\geq0,\ j\in\{0,1,2\}&(22)_3&+1\\ 2\text{ or }22&(21)_3&-1\\ 2^i1,\quad i\geq1&(11)_3&-1\\ 2^ij2,\quad i\geq1,\ j\in\{0,1,2\}&(21)_3&-1\\ 2^i11,\quad i\geq0&\text{no successor}&\text{undefined}\\ \text{every other canonical word}&c_0(n)&0 \end{array} \tag{3}$$

In particular, this gives every value asserted by the source conjecture, including all positions with no comma-successor.

Reduction to two leading digits. Let $r=d_m=n\bmod3$. A comma-child whose leading ternary digit is $e\in\{1,2\}$ must equal

$$N_e=n+(re)_3=n+3r+e. \tag{4}$$

Conversely, $N_e$ is a comma-child if and only if its actual leading ternary digit equals $e$. Thus

$$N_e\text{ is a comma-child} \quad\Longleftrightarrow\quad \delta_3(N_e)=e, \tag{5}$$

where $\delta_3$ denotes the leading ternary digit. Because $N_1<N_2$, the comma-successor uses $e=1$ whenever (5) holds for $e=1$, and otherwise uses $e=2$ if (5) holds for $e=2$.

Suppose first that $m\geq3$, and put

$$P=3^{m-1}\geq9. \tag{6}$$

The increments in (4) are at most $8<P$. Consequently, starting from a number with leading digit $1$ can encounter only the boundary $2P$, and starting from a number with leading digit $2$ can encounter only the boundary $3P$. At either boundary $B$, define

$$h=B-n>0. \tag{7}$$

Since $3\mid B$, the trailing digit satisfies

$$h\equiv-r\pmod3. \tag{8}$$

Leading digit $1$. Here $P\leq n<2P$ and $B=2P$. The candidate $N_1$ has leading digit $1$ precisely when it remains below $B$, namely when

$$h>3r+1. \tag{9}$$

In this case the successor has comma-number $3r+1=c_0(n)$. If $h\leq3r+1$, then $N_1$ crosses the boundary and has leading digit $2$, whereas $N_2=N_1+1$ also has leading digit $2$. Hence the successor instead has comma-number

$$\operatorname{cn}(n)=3r+2=c_0(n)+1. \tag{10}$$

Combining the inequality $h\leq3r+1$ with (8) gives precisely

$$\begin{array}{c|c|c} r&h&\text{ternary word}\\ \hline 1&2&12^{m-2}1\\ 2&1,4,7&12^{m-3}j2,\quad j\in\{2,1,0\}. \end{array} \tag{11}$$

There is no exceptional case when $r=0$. The two rows of (11) are exactly the first two families in (3).

Leading digit $2$. Here $2P\leq n<3P$ and $B=3P$. The smaller candidate $N_1$ has leading digit $1$ precisely when it crosses $B$, that is, when

$$h\leq3r+1. \tag{12}$$

In this case the successor has

$$\operatorname{cn}(n)=3r+1=c_0(n)-1. \tag{13}$$

If $h\geq3r+3$, then $N_1$ still has leading digit $2$ and is invalid, whereas $N_2$ also remains below $B$ and has leading digit $2$. Consequently $N_2$ is the successor and

$$\operatorname{cn}(n)=3r+2=c_0(n). \tag{14}$$

The remaining possibility is

$$h=3r+2. \tag{15}$$

Then $N_1=B-1$ has leading digit $2$, while $N_2=B$ has leading digit $1$. Neither candidate has its required leading digit, so there is no comma-successor.

Using (8), the exceptional and terminal possibilities become

$$\begin{array}{c|c|c|c} r&h&\text{ternary word}&\text{outcome}\\ \hline 1&2&2^{m-1}1&-1\text{ correction}\\ 2&1,4,7&2^{m-2}j2,\quad j\in\{2,1,0\}&-1\text{ correction}\\ 1&5&2^{m-2}11&\text{no successor}. \end{array} \tag{16}$$

No other value of $h$ changes the usual comma-number. In particular, the terminal family agrees with the independently established no-successor classification in Theorem 5.1 of the source.

One- and two-digit boundary cases. For the remaining eight positive integers below $9$, the defining criterion (5) gives

$$\begin{array}{c|cccccccc} n\text{ in base }3&1&2&10&11&12&20&21&22\\ \hline \operatorname{cn}(n)\text{ in base }3&11&21&1& \text{none}&21&2&11&21. \end{array} \tag{17}$$

The words $1$, $10$, $12$, and $20$ are ordinary; the words $2$, $21$, and $22$ have correction $-1$; and $11$ has no successor. Combining these boundary cases with (11) and (16) proves (3) for every positive integer.

The original table gives the correct comma-number values but has overlapping endpoint conventions in its correction column. In particular, $2^0 1=1$ and $2^0 12=12$ have correction $0$, not $-1$; $2^0 02=02$ is not a canonical ternary word; and both $2$ and $22$ have correction $-1$. The revised index ranges and separate boundary row in (3) remove these ambiguities while proving the full comma-number statement of Conjecture 9.2.