Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication

Theorem
Let $b \in \Z_{>1}$ be an integer greater than $1$.

Let $n = b^k - 1$ for some integer $k$ such that $k \ge 1$.

Then:
 * $2 n$ is the reversal of $\left({b - 1}\right) n$

when both are expressed in base $b$ representation.

Proof
Let $a = b - 1$.

Let $n$ be expressed in base $b$:
 * $\displaystyle n = \sum_{j \mathop = 0}^m {r_j b^j}$

that is:
 * $n = \left[{r_m r_{m-1} \ldots r_2 r_1 r_0}\right]_b$

Then:
 * $n = {\overbrace {\left[{aaa \ldots a}\right]}^k}_b$

Then:
 * $n + n$ expressed in base $b$ is the reversal

We have that:
 * $\displaystyle 2 n = 2 \sum_{j \mathop = 0}^k {\left({b - 1}\right) b^j}$

When $j = 0$ we have:
 * $\left({b - 1}\right) + \left({b - 1}\right) = b + \left({b - 2}\right)$

and so the least significant digit of $2 n$ is $b - 2$.

When $j = 1$:
 * $b \left({b - 1}\right) + b \left({b - 1}\right) + b = b^2 + b \left({b - 1}\right)$

and similarly for $j = 2, 3, \ldots, k - 1$.

When $j = k$:
 * $b^k \left({b - 1}\right) + b^k \left({b - 1}\right) + b^{k - 1} = b^{k + 1} + b^k \left({b - 1}\right)$

Thus:
 * $\displaystyle 2 b = b^{k + 1} \sum_{j \mathop = 1}^k \left({b - 1}\right) b^j + \left({b - 2}\right)$

and so $2 n$ has the representation base $b$:
 * $n = \left[{1 a a a \ldots a c}\right]$

where $c = b - 2$.

Now consider $\left({b - 1}\right) n$.

We have that:
 * $\displaystyle \left({b - 1}\right) n = \left({b - 1}\right) + \sum_{j \mathop = 0}^k {\left({b - 1}\right) b^j}$

When $j = 0$ we have:
 * $\left({b - 1}\right)^2 = b^2 - 2 b + 1 = b \left({b - 2}\right) + 1$

and so:
 * the least significant digit of $\left({b - 1}\right) n$ is $1$
 * the term for j = 1 has $b - 2$ to be added to it.

When $j = 1$:
 * $b \left({b - 1}\right) \left({b - 1}\right) + b - 1 = b^3 - 2 b^2 + b + \left({b - 2}\right) = b^2 \left({b - 2}\right) + \left({b - 1}\right)$

and so:
 * the next digit of $\left({b - 1}\right) n$ is $b - 1$
 * the term for j = 1 has $b - 2$ to be added to it

and similarly for $j = 2, 3, \ldots, k - 1$.

When $j = k$:
 * $b^k \left({b - 1}\right) \left({b - 1}\right) + b^{k - 1} \left({b - 2}\right) = b^{k + 2} - 2 b^{k + 1} + b^k \left({b - 2}\right) = b^{k + 1} \left({b - 2}\right) + b^k \left({b - 1}\right)$

and so:
 * the $k + 1$th digit of $\left({b - 1}\right) n$ is $b - 2$
 * the $k$th digit of $\left({b - 1}\right) n$ is $b - 1$.

Thus:
 * $\displaystyle \left({b - 1}\right) n = b^k \left({b - 2}\right) + \sum_{j \mathop = 1}^k \left({b - 1}\right) b^j + 1$

and so $2 n$ has the representation base $b$:
 * $n = \left[{c a a a \ldots a 1}\right]$

where $c = b - 2$.