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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 $10^8$:

plot

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
Trevor
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  • I don't know if your result is correct, but its likely going to take a bit of work, considering that the ratio of adjacent primes tends to 1 in hte limit (https://math.stackexchange.com/q/2234978/208255) – user24142 Apr 16 '22 at 14:27
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    How did you think of $2^\pi$ and $2\pi$ – FShrike Apr 16 '22 at 14:43
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    @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. – Trevor Apr 16 '22 at 14:45
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    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 $[1,2]$. – Wojowu Apr 16 '22 at 15:08
  • Could you link to/post the code to generate the plot? – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Apr 17 '22 at 08:54
  • @TheSimpliFire Added... I think I used only standard library functions, but lemme know if it doesn't work. – Trevor Apr 17 '22 at 09:56
  • @Trevor Thanks. It might be interesting to obtain the limits for $S_{n,a}={(a-1)n< s< an : f(s)\leq an}$ or $f$ where it increases the prime index by an arbitrary $k>1$ to see if they have a similar closed form. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Apr 17 '22 at 10:29
  • @TheSimpliFire I had the same thought and checked a couple of them, but nothing immediately jumped out at me. – Trevor Apr 17 '22 at 11:19
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    Consider posting this to MO. This is a cool observation! – Clyde Kertzer Apr 19 '22 at 21:29
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    The Erdos-Wintner theorem https://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. – David E Speyer Apr 21 '22 at 14:26
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    To add a bit more detail, let the prime after $p$ be $p+g(p)$. There is a constant $\theta<1$ such that $g(p) < p^{\theta}$ https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds . So $\log \tfrac{f(p)}{p} < \log \tfrac{p+p^{\theta}}{p} < p^{-1+\theta}$. To check the hypotheses of the EW theorem, we must check that $\sum \tfrac{( p^{-1+\theta})}{p} = \sum p^{-2+\theta}$ and $\sum \tfrac{( p^{-1+\theta})^2}{p} = \sum p^{-3+2 \theta}$ converge, which is true since $-3+2 \theta$ and $-2+\theta$ are $< -1$. – David E Speyer Apr 21 '22 at 15:03

1 Answers1

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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 $x<n\le 2x$.) 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.

I think the assumption that either of the other two limits in the OP is a "plausible" constant is extremely unlikely. I would bet against either of the two proposed constants being correct; both are likely to be at least as complicated as the last infinite product above, and probably more complicated because (as David E Speyer describes) the mean values are measuring something even more particular about these functions' limiting distributions than their average values.

Greg Martin
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