(Related to this prior post.)
Let $f(n)$ for $n\in\mathbb N$ be a function that increases the prime index of each prime factor of $n$ (with multiplicity) by $1$.
e.g. $f(20)=f(2^2\cdot 5)=f(p_1^2\cdot p_3)=p_{2}^2\cdot p_{4}=3^2\cdot 7=63$.
Consider $\frac{n}{f(n)}$ as $n\to\infty$. The mean of these terms has a clear limit:
$$\lim_{n\to\infty} \left(\frac{1}{n}\sum_{i=1}^n \frac{i}{f(i)}\right)=\frac{1}{2}.$$
But more interesting is that if you let $a$ be a count of how many terms through $n$ have $\frac{i}{f(i)}>\frac{1}{2}$, and therefore $n-a$ be the count of terms with $\frac{i}{f(i)}<\frac{1}{2}$, then it appears that
$$\lim_{n\to\infty} \frac{a}{n-a}=\frac{e}{\pi}.$$
This is a strictly empirical result, but it seems to hold within a margin of error of around $\frac{1}{\sqrt{a}}$. So far, I've checked this for $n\leq 10^7$, and while the results aren't strong enough to rule out coincidence yet, it seems too suggestive to ignore.
Does anyone know why this might arise? And if it's correct, might it imply that $\frac{e}{\pi}$ is irrational? It seems to me that $\frac{a}{n-a}$ can't be rational in the limit since it will never cease to fluctuate in finer pseudorandom detail based on the distribution of the primes, but perhaps the limit negates that somehow.
If anyone can explain why the limit $\frac{e}{\pi}$ would appear here (or can show that I'm mistaken), I'll consider this answered.
Update
The more I look, the less convinced I am that I had this right; I still believe there's a limit, but not $e/\pi$, as it seems to run about $0.002$ higher pretty consistently.
As for the $1/2$ limit, I'm quite sure that's solid. Empirical data leaves essentially no doubt, and it's very much in line with the behavior of $f$, which has properties like for any $n$, you'll get $f(k)> n$ for exactly half of the naturals $k\in [1,n]$. (See this post for more.)
That said, I don't know how to prove the $1/2$ limit, so I'd accept that as an answer here as well if someone can demonstrate it.
Basic notes on $\frac{i}{f(i)} > \frac{1}{2}$ properties
Let $S_n = \{1 \leq s \leq n : f(s) > 2s\}$.
(Note $f(s) > 2s$ is equivalent to $\frac{s}{f(s)} < \frac{1}{2}$, but cleaner looking.)
Regardless of choice of $n$, $|S_n|$ will be slightly larger than $n/2$.
Example
$S_{128}=\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27, 28, 30,\\ 32, 35, 36, 39, 40, 42, 44, 45, 48, 49, 50, 52, 54, 56, 57, 60, 63, 64, 66,\\ 68, 69, 70, 72, 75, 76, 78, 80, 81, 84, 88, 90, 91, 92, 96, 98, 99, 100,\\ 102, 104, 105, 108, 110, 112, 114, 116, 117, 120, 124, 125, 126, 128\}.$
$|S_{128}|=67$.
This set can be substantially compressed by eliminating multiples, leaving $\{4, 6, 9, 10, 14, 15, 21, 35, 39, 49, 57, 69, 91, 125\}$. Any integer divisible by any of these values will also have the property $f(i) > 2i$.
It is clear with a little inspection that there will be infinitely many of these values as $n$ is made arbitrarily large. Various patterns abound, e.g. $5^2 \cdot p_i$ will be one of these minimal elements for all $i \geq 3$. The overall distribution appears complex, similar to the primes themselves in many ways.
All integers in these $S$ sets are necessarily composite, as having a prime would violate Bertrand's postulate. On the whole, elements in these sets are "especially" composite, with larger factor counts and smaller factors. Contrast this to the complementary set $R_n$ comprising those $r\leq n$ such that $f(r) < 2r$:
Example
$R_{128}=\{1, 2, 3, 5, 7, 11, 13, 17, 19, 22, 23, 25, 26, 29, 31, 33,\\ 34, 37, 38, 41, 43, 46, 47, 51, 53, 55, 58, 59, 61, 62, 65, 67, 71, 73,\\ 74, 77, 79, 82, 83, 85, 86, 87, 89, 93, 94, 95, 97, 101, 103, 106,\\ 107, 109, 111, 113, 115, 118, 119, 121, 122, 123, 127\}.$
$|R_{128}|=61.$
$R_n$ necessarily contains all the primes $\leq n$ as well as most $2p$, $p^2$, and other semiprime forms; in general, it is populated by high-prime-index low-factor-count integers, as expected from the complement to $S_n$.