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(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 $S_n$.

Trevor
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    I don't know whether the limit is as you suggest. $e$ and $\pi$ tend to pop up all over the place - see for example https://en.wikipedia.org/wiki/Prime_number_theorem#Prime-counting_function_in_terms_of_the_logarithmic_integral - so it wouldn't surprise me. To answer your main question - no - if your limit is correct is would not imply that $e/\pi$ is irrational. It's perfectly possible for a series which "fluctuates in finer pseudorandom detail" to tend to a rational limit: see the first equation in this link for a trivial example https://en.wikipedia.org/wiki/Prime_number_theorem#Statement. – Blitzer Apr 22 '22 at 06:25
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    Taking this up above $10^7$... there's a local minimum of $0.86475$ near $n = 4 \ 500 \ 000$, but then the curve turns upward and increases to over $0.08665$ so far, around $n = 70 \ 000 \ 000$. So I'd say it's just as likely to have a limit of $e/ \pi$ as it is to have a limit of $5/6$. – Eric Snyder Apr 22 '22 at 07:00
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    How close exactly did you get to $e/\pi$? The inverse symbolic calculator shows that there are many expressions close to that value with a precision of 6 decimal digits. – Martin R Apr 22 '22 at 07:04
  • @MartinR But how many of them are really meaningful and not just based on a minimal polynomial of an algebraic number near this number or pade-approximations or something like that ? – Peter Apr 22 '22 at 07:13
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    @Peter: Why should the limit be something “meaningful”? I know almost nothing about number theory, but it could be pure coincidence that the limit is close (but not equal) to some “meaningful” expression. – Martin R Apr 22 '22 at 07:17
  • @MartinR I ask because we can find such expressions for every given decimal digit string by various methods. What I mean are (probably) coincidental matches. – Peter Apr 22 '22 at 07:20
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    @Peter: Then we probably mean the same (and I might have expressed myself badly). The limit could be $e/\pi$ or $70000/80901$ or $(2+6^{1/4})^{1/2} \cdot 21^{1/2}/10$ or some other value for which no explicit symbolic is known. – My (unsubstantiated) bet would be on the latter. – Martin R Apr 22 '22 at 07:24
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    The author does not rule out this possibility. Eric Snyder is apparently on the way to decide it. The flucuations seem to be heavy , so it is not even clear whether the limit exists at all. I would also like to see a justification of the $\frac{1}{2}$-limit. – Peter Apr 22 '22 at 07:28
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    The limit of $\frac12$ comes from an answer to the linked earlier question. $f(n)$ here is completely multiplicative, as is $n/f(n)=:g(n)$. For any completely multiplicative function, the mean is $$\frac1x\sum_{n\le x} g(n)\sim\prod_p\biggl(1-\frac1p\biggr)\biggl(1-\frac{g(p)}p\biggr)^{-1}$$ so long as $g(p)\sim 1$, which it is for the $g(n)$ here. But for any given $p_i$, the partial product is $$\left(1-\frac{1}{p_i}\right)\left(1-\frac{1}{p_{i+1}}\right)^{-1}$$ Everything after $p = 2$ cancels out, leaving only $\frac12$ as $n \to \infty$. – Eric Snyder Apr 23 '22 at 00:18
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    (To be more precise: in the expression for the mean, the first term is $(1/2)(3/2)$, followed by $\cdot (2/3)(5/4) \cdot (4/5)(7/6) \cdots$, and each multiplication cancels two terms, leaving just $(1/2)(z)$ where $z \to 1$ as $n \to \infty$. – Eric Snyder Apr 23 '22 at 00:35
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    Having now computed up to $n = 4 \times 10^8$, $a/(n-a)$ continues increasing, though much much more slowly. The deceleration isn't enough to actually hint at a limit though, so I wouldn't counting on getting to one. At $n = 4 \times 10^8, a/(n-a) \approx 0.867015$, much greater than $e/\pi$, at least, "much" greater when we're discussing such small changes. – Eric Snyder Apr 23 '22 at 06:01

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