It is well-known that the interated sum-of-digits function equally distributes the numbers from $1$ to $10^k-1$ to the digits $1,\ldots,9$. And this holds true for any base $b$. For example, see the nearly decade-old MO question Sum of digits iterated. I want to ask a similar question for the product of digits.

Let $\pi(n)$ be the product-of-digits function, mapping $n$ to the product of the digits of $n$, and repeating until a single digit is reached. So $\pi(13579) = 0$ because $$ 13579 \rightarrow 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 = 945 \rightarrow 9 \cdot 4 \cdot 5 = 180 \rightarrow 0 \;. $$ If $\pi( )$ is applied to all the numbers from $1$ to $m$, the distribution naturally heavily favors $0$, but is otherwise (apparently) irregular:

^{ $\pi(n)$ for $n=1,\ldots,10^5-1=99999$, base $10$. }

The distribution for base $5$ seems to have a more regular distribution, with the frequency of each non-zero digit monotonically increasing:

^{ $\pi_5(n)$ for $n=1,\ldots,5^8-1=44444444_5$, base $5$. }

The limit distribution for each base appears to be just the $0$-digit bin approaching 100%, although there are arbitrarily large numbers that avoid mapping to $0$, for example, $\pi(111\ldots111d111\ldots111) = d$. Many details remain unclear to me:

. What explains the differently shaped distributions for $\pi_b(n)$ in different bases $b$? Can anything general be said?QIs it the case that, in base $10$, the bins for digits $3$ and $7$ are equal, both $15$ in the example above? Why does $5$ occur more frequently than $3$ and $7$?

Why are the digit frequencies increasing with digit value in base $5$, but not, say, in base $7$?

It may be that these questions have been previously explored, in which case pointers would be welcomed.

2more comments