How does the Sequence work?

There are $n^2$ of each digit $n$ so the sum we for the digits $1$ to $9$ is:

$\sum_{n=1}^{9} (n^2)(n)=\sum_{n=1}^{9} n^3= \left (\dfrac{9(9+1)}{2} \right )^2=\boxed{2025}$

Now we look at when we introduce $10$ and $11$ . $10$ has a digit sum of $1$ and occurs $10^2=100$ times. $11$ has a digit sum $2$ and occurs $11^2=121$ times so the sum of these digits is:

$1 \times 100 + 2 \times 121=\boxed{342}$

The overall sum of the digits is therefore:

$342+2025=\boxed{\boxed{2367}}$

I have used the well-known formula:

$\sum_{n=1}^{i} n^3= \left (\dfrac{i(i+1)}{2} \right )^2$

Which can be proved by induction.

Thanks to Mateo Matijasevick for pointing out flaws in my original solution to this problem.