When calculating how much time it takes to count from $1$ to $n$, it is normally used the approximation that it takes about $1s$ to say a number out loud, so it would take $n$ seconds, but there's a considerable difference in the time it takes to say $7$ and $85,346,753$ for example, so a more precise formula of how many seconds it takes to say a number out loud would be:
$$T(x) = \frac{d(x)}{3} \cdot 1.5s + \lceil\frac{d(x)}{3} - 1\rceil \cdot 0.5s$$ Where $d(x)$ is the number of digits of $x$ $(\lfloor log_{10}(x) \rfloor + 1)$
- $(\frac{d}{3} \cdot 1.5s)$:
Numbers in english are read in groups of 3, there are $\frac{d}{3}$ groups of 3, and it takes on average $1.5s$ to say a three-digit number out loud
- $(\lceil\frac{d}{3} - 1\rceil \cdot 0.5s)$:
This is for the magnitude words (thousand, million, trillion, etc.), these appear in every group of 3 digits, as said before there are $\frac{d}{3}$ groups of 3, the $-1$ there is because "units" is normally omitted, round up because even if there is a group of 1 or 2 digits, the magnitude word is still said, as in $25000$ (25 thousand), and it takes about $0.5s$ to say these words.
Now to calculate how many time it takes to count from 1 to $n$, the formula would be:
$$\sum_{k=1}^n \frac{d(k)}{3} \cdot 1.5s + \lceil\frac{d(k)}{3} - 1\rceil \cdot 0.5s$$
For comparison, this formula says that counting from 1 to 1 million would take about 39 days, the formula that consider 1s for each number says that it would take about 11 days.
How can i find a formula that can approximate this summation with less computation power needed? (like how $n!$ is approximated with $\sqrt{2\pi n}(\frac{n}{e})^n$)
For large enough $n$ for almost all $k \le n$, $d(k) \approx \log_{10}k + 1$ and $\lceil d(k)/3 - 1\rceil \approx \frac13\log_{10}k - \frac23$. So the sum approximates $$\frac 76n + \sum_{k=1}^n 2\log_{10} k = \frac 76n + 2\log_{10} n!$$ And by that same Sterling approximation, this is about $$\left(\frac76 - 2\log_{10}e\right)n + (2n+1)\log_{10}n$$ Where I've dropped the constant $\log_{10}2\pi$ as insignificant.
So you are looking at $O(n\log n)$ growth.