What is the limiting distribution of factors of natural numbers? For the first $k$ numbers I can calculate their scalar factors. Take $k= 10$, inducing the following table:
1 {1}
2 {1, 2}
3 {1, 3}
4 {1, 2, 4}
5 {1, 5}
6 {1, 2, 3, 6}
7 {1, 7}
8 {8, 1, 2, 4}
9 {1, 3, 9}
10 {1, 10, 2, 5}

As we increase $k$ we can further calculate the number of times that factor occurs:
1 10
2 5
3 3
4 2
5 2
6 1
7 1
8 1
9 1
10 1

We can plot these frequencies over factors as a histogram:

Taking this out to $k=10^5$, we obtain the histogram:

A naive guess is these finite distribution follow Zipf's law, and perhaps as $k \rightarrow \infty$ it converges to something like the zeta distribution.
What is the limiting distribution of this sequence?
 A: Asymptotically, $1/2$ of integers up to $X$ will have $2$ as a factor, $1/3$ of integers up to $X$ will have $3$ as a factor, and so on. In the limit, the curve formed from the points $(n, N)$, where $N$ is the number of times $n$ divides an integer up to $X$, will approach the curve $X/n$. This is lightly obscured in the plots due to the binning effect of the histograms.
Plotting directly the points $(n, N)$ will show a scaled version of the function $1/x$. This function is rather hard to recognize on large scales (since it decreases so rapidly). Note that $y = 1/x \implies \log y = - \log x$, and thus on a log-log plot we should see a line with slope $-1$. And we do!
Here is an example python implementation to verify this, after quickly and naively counting points up to $10^6$.
# simple python code
import matplotlib.pyplot as plt
import math


X = int(1e6 + 1)                 # Count up to 10^6
xs = list(range(X))
ys = [0] * X
for n in range(1, X):            # For each number from 1 to 10^6,
    for idx in range(n, X, n):   # count the number of times it divides a
        ys[n] += 1               # number up to 10^6.

logxs = [math.log(x) for x in xs[1:]]  # skip the first element, which is 0
logys = [math.log(y) for y in ys[1:]]  # skip the first element, which is 0

plt.scatter(logxs, logys)
pls.show()


We notice in this plot that this is very nearly a line with slope $-1$, except towards the end. The bounds of the plot are approximately $14$ as $\log(10^6) \approx 13.81$.
