Inspired by Lucia and Emma

I used a calculator

Let the equation be $a_n = 9b_n + n + 1$ , where $b_0 = 0$ and $b_n = 10b_{n-1} +n$ . Then we have:

$\begin{aligned} a_n & = 9b_n + n + 1 \\ \implies a_{n+1} & = 9b_{n+1} + n + 2 \\ & = 9(10b_n + n + 1) + n + 2 \\ & = 90b_n + 10n + 11 \\ & = 10(9b_n + n + 1) + 1 \\ & = 10a_n + 1 \end{aligned}$

Since $a_0 = 1$ , then $a_1 = 11$ , $a_2 = 111$ , $\cdots a_n = \underbrace{111\cdots 111}_{\text{\# of 1's} = n+1}$ .

Yes , $a_9 = 1111111111$ .

$\begin{aligned} \overline{123...n} &= 1 \times 10^{n-1} + 2 \times 10^{n-2} + ... n \times 10^0 \\ &= \sum_{r=1}^n r10^{n-r} \\ &= \dfrac{1}{81} (10^{n+1} - 9n - 10) ,\ \text{Sum of AP-GP} \\\\ \overline{123...n} \times 9 + (n+1) &= \dfrac{9}{81} (10^{n+1} - 9n - 10) + (n+1) \\ &= \dfrac{1}{9} (10^{n+1} - 9n - 10) + (n+1) \\ &= \dfrac{1}{9} (10^{n+1} - 1) \\ &= \overline{111...1}_{n \text{times}} \end{aligned}$

$\overline{12...n}\times (10-1)+(n+1)=\overline{12...n0}-\overline{12...n}+(n+1)$ for $n \leq 9$ .

$\overline{12...n0}$ is $\overline{12...n}$ , when digits are shifted one position to left. So, if you put the larger one on the top of the smaller one for subtraction, in all positions the digit on the top would be one more than the digit on the bottom, except the most right digits $0$ and $n$ . But $0$ would be added to $(n+1)$ and the problem is solved.