# Why does the limit $\log{2^\pi}/\log{2\pi}$ emerge as a ratio of a certain set of adjacent prime factors?

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. $$f(20)=f(2^2\cdot 5)=f(p_1^2\cdot p_3)=p_{2}^2\cdot p_{4}=3^2\cdot 7=63$$.

Let the set $$S_n=\{n< s< 2n : f(s)\leq 2n\}$$, for any $$n\in\mathbb N$$.

If you take the mean of $$\frac{f(s)}{s}$$ for each element $$s \in S_n$$, empirical results for $$n \leq 10^8$$ strongly suggest that this mean converges:

$$\lim_{n\to\infty} \left(\frac{1}{|S_n|}\sum_{s \in S_n} \frac{f(s)}{s}\right)=\frac{\log 2^{\pi}}{\log {2\pi}}\approx 1.18484.$$



Can anyone explain why this should be so? My motivation in asking is as part of a larger study of the behavior of $$f$$, and this result seems too nice and tidy to ignore without investigation. If anyone can explain the origin of the RHS term, or show that it is not a limit after all, I'll consider this question answered.

As a rough picture of what empirical results I'm referring to, here's a plot of this mean as $$n$$ increases by a factor of $$1.1$$ each step to nearly $$10^8$$:

It appears convincingly centered around $$\frac{\log{2^\pi}}{\log{2\pi}}$$, which seems much more plausible than any other nearby constants I could find, and it's varying by around $$10^{-6}$$ at the tail end there.

Per request, Mathematica code to generate plot. This was quick 'n dirty and I'm certain it could be optimized to run much faster. Use Alt+. to halt execution.

ClearAll[f, dyn];

f[1] = 1;
f[n_, k_ : 1] := Times @@ (NextPrime[#1, k]^#2 &) @@@ FactorInteger@n;
SetAttributes[f, Listable];
SetAttributes[dyn, HoldAll];

dyn[expr_, symbols_List : {}, interval_ : \[Infinity]] :=
PrintTemporary[
Dynamic[Refresh[expr, TrackedSymbols :> symbols,
UpdateInterval -> interval]]]

tab = {};
disp := Column[{Length@tab, n, ListLinePlot[Last/@tab, ImageSize -> Large],
Grid[Prepend[
Reverse@tab, {Style["n", Bold], Style["S mean", Bold]}],
Dividers -> All, Alignment -> {Left, Top}]}]
dyn[disp, {tab}];

n = 100;
While[True,
s = Select[f[Range[n + 3 - Mod[n, 2, 1], 2 n, 2], -1], # > n &];
m = Mean[N[f@#, 8]/# & /@ s];
AppendTo[tab, {n, m}];
n = Ceiling[1.1 n];
]
disp

• How did you think of $2^\pi$ and $2\pi$ Commented Apr 16, 2022 at 14:43
• @FShrike Well, I cheated... I plugged in 1.1848 to Wolfram Alpha, and it gives a handful of plausible expressions for nearby constants, of which that one seemed the best fit. Commented Apr 16, 2022 at 14:45
• More generally one can ask about the distribution of $f(s)/s$. If the limiting distribution has a nice density function, then the answer to your question should be possible to read off from this function restricted to $[1,2]$. Commented Apr 16, 2022 at 15:08
• Consider posting this to MO. This is a cool observation! Commented Apr 19, 2022 at 21:29
• The Erdos-Wintner theorem encyclopediaofmath.org/wiki/Erd%C5%91s%E2%80%93Wintner_theorem applied to the function $\log \tfrac{f(n)}{n}$ shows that $f(s)/s$ has a limiting distribution. But it is almost never possible to get explicit formulas for the limiting distributions coming this theorem, so your formula is a mystery to me. Commented Apr 21, 2022 at 14:26

It is a general result in analytic number theory that if $$g(n)$$ is any completely multiplicative function (meaning $$g(mn)=g(m)g(n)$$ always) such that $$g(p)$$ is asymptotically $$1$$, then $$\frac1x\sum_{n\le x} g(n) \sim \prod_p \biggl( 1-\frac1p \biggr) \biggl( 1-\frac{g(p)}p \biggr)^{-1},$$ where the product is a convergent product over all primes. (The same holds if the range of summation is $$x.) If we apply this to $$g(n) = n/f(n)$$, using the notation $$p_\rightarrow$$ for the prime following $$p$$, then we get $$\frac1x\sum_{n\le x} \frac n{f(n)} \sim \prod_p \biggl( 1-\frac1p \biggr) \biggl( 1-\frac{p/p_\rightarrow}p \biggr)^{-1} = \prod_p \biggl( 1-\frac1p \biggr) \biggl( 1-\frac1{p_\rightarrow} \biggr)^{-1} = \frac12$$ since the product telescopes. On the other hand, applying this to $$g(n) = f(n)/n$$ yields $$\frac1x\sum_{n\le x} \frac{f(n)}n \sim \prod_p \biggl( 1-\frac1p \biggr) \biggl( 1-\frac{p_\rightarrow}{p^2} \biggr)^{-1} \approx 4.128,$$ which is unlikely to have anything to do with any constant we've seen before.