I've seen the question for numbers like 50, 100 or 1000, but not for $n$. Although I found a formula that might be the answer, but I don't know the name of it or the proof for it. I couldn't find it anywhere on internet as well. Here it is: $$d\times(n+1) - \frac{10^d - 1}{9}$$ $d$ is the number of digits $n$ has.
For example, for $n = 16$:
Answer: $\displaystyle2\times (16 + 1) - \frac{10^2 - 1}{9} = 34 - 11 = 23$
Also: $12345678910111213141516\rightarrow 23 \text{ digits}$
Looking forward for replies!
Thanks!
Edit 1: For those who also want to know how $d$ is calculated: $d = \lfloor\log_{10}n\rfloor + 1$
Well, it's clear we write down a "ones" digit for each $n$. And we write down a "tens" digit for all but the first $9$ so we do $n-9$ "tens" digits. And $n-99$ hundreds digits and so on.
So if $d$ is the number of digits as you put it, we have the number of digits written is.
$n + (n-9) + (n-99) + ...... + (n-\underbrace{999....9}_{d-1})=$
$n\times d - (9 + 99 + ..... + \underbrace{999....9}_{d-1})=$
$n\times d- ((10-1) + (100-1) + ...... + (10^d -1))=$
$n\times d - [(10 + 100 + ..... + 10^{d-1}) -(d-1)]=$
$n\times d - [(10 + 100 + ..... + 10^{d-1})+1] + d=$
$n\times d + d - [1+10 + 100 + ..... + 10^{d-1}] =$
$(n+1)\times d - [\underbrace{111.....1}_d] =$
$(n+1)\times d - \frac {10^{d}-1}9$
Nice job!