Pattern in number of digits in $(10^n)!$

Question

I was observing the OEIS sequence A061010 and its values up to $n=1000$, and it seems that the pattern is as follows. For $n$, $f(n)$ is for the form string of $(n-1)$+(pattern of repeating digits that grow by $1$ every iteration)+(last 3-4 digits with no seeming particular pattern)

• Like if $n=13$, $f(n)$ is 12 56570551809 75

• Like if $n=14$, $f(n)$ is 13 565705518096 83

• Like if $n=15$, $f(n)$ is 14 5657055180967 57

Is there a pattern to this madness or am I barking up the wrong tree? Any help is much appreciated.

Answer 1

Your first link gives an approximation

a(n) = 10^n*(n - 1/log(10)) + n/2 + O(1). [Arkadiusz Wesolowski, Jan 21 2012]

Note that $1-\frac{1}{\log_e(10)} \approx 0.5657055180967481723488710810833949177056\ldots$ so this gives the pattern you are observing

Compare this to $a(40) = 395657055180967481723488710810833949177077$