Recursive Description

The next number in the sequence is $312211$ . Every number following this one must contain a $1$ , a $2$ , and a $3$ at least once to properly describe the previous row. However, I argue that no digit greater than $3$ can appear in the sequence.

Assume that a digit greater than or equal to $4$ is in the sequence. This implies that there was a number at some point with at least $4$ repeating digits in a row. Let $n$ be the first such number. If this digit is $X$ , then $n=\overline{...AXXXXB...}$ where $A$ and $B$ could or could not be $X$ . What could the previous number have been?

We can divide $n$ into a series of descriptions, with each description containing a quantity value followed by a digit value. Currently, there have never been $10$ or more of a digit in a row, so a description will always have exactly $2$ digits. Depending on the location of $\overline{AXXXXB}$ in the number, the descriptions can be $A{\color{#D61F06}X}, X{\color{#D61F06}X}, XB$ or $X{\color{#D61F06}X}, X{\color{#D61F06}X}$ . In both cases, the digit value $X$ appears twice in a row (in red). Both of these situations are impossible since the same digit value cannot appear in two consecutive descriptions (if this did happen, they should be merged into one description). This is a contradiction, so $4$ digits in a row cannot appear in the sequence.

The $2018^{th}$ row contains only $1\text{s}$ , $2\text{s}$ , and $3\text{s}$ , so the sum of the unique digits is $1+2+3=\boxed{6}$