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I wonder if there is something like a general "prime component distribution pattern" of "the general natural number" $n$?

Using the following notation for the prime factorization $n = p_1^{\alpha_1}p_2^{\alpha_2}\dots p_{k-1}^{\alpha_{k-1}} p_k^{\alpha_k}$ with $k={\pi(n)}$ thus going through all primes from $2$ up to $p_{\pi(n)}$ and setting $\alpha_j = 0$ for all $p_j\nmid n$, I define a distribution density function

$\delta_n(x)=\begin{cases} \alpha_i \frac{\pi{(n)}}{\Omega{(n)}} & (i-1)/\pi(n) < x\le i/\pi(n), \forall i\in\{1,...,k\} \\ 0 & else \end{cases}$

Two examples $\delta_4$ and $\delta_{21}$ are shown below

<span class=$\delta_4$(x)" /> <span class=$\delta_{21}(x)$" />

The functions are normalized according to

$\int_{0}^{1}{\delta_n (x) \mathrm{d}x} = 1$

Now the question is if this limit

$$ \Delta_{\infty}(x) = \lim_{n\to\infty} \frac{1}{n}\sum_{k=2}^{n+1} \delta_k(x) \tag{1}$$

exists and in case how that might look like. I have calculated the average for $n = 250$ (my online Mathematica subscription doesn't allow for higher numbers) and it looks relatively interesting:

<span class=$\Delta_{250}(x)$" />

The increasing domain to the right is obviously due to the primes, the next peak to the left appears to be due to even numbers and so on.

I would be grateful for any comments, ideas on the primary question (first sentence in the post) and if this could be an appropriate way to adress it. Most of all if convergence of $(1)$ is to be expected or not.

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    $\begingroup$ This syntax $\begin{cases} 1 & x=0 \\ 0 & x\neq 0 \end{cases}$ lets you do cases. $\endgroup$
    – Will Sawin
    Commented Jul 16, 2021 at 21:32
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    $\begingroup$ en.wikipedia.org/wiki/Dickman_function may be relevant to your interests. $\endgroup$ Commented Jul 16, 2021 at 21:38
  • $\begingroup$ also related and with an interesting distribution shape, if you focus on the size of the divisors not their exponents is the “erdos arcsine law”. $\endgroup$
    – kodlu
    Commented Jul 16, 2021 at 22:09

1 Answer 1

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This will converge to $0$ for all $x$ except possibly $\frac{1}{d}$ for natural numbers $d$.

To check this, it suffices to show that $\delta_n(x)=0$ for all $n$ sufficiently large, for any such $x$. This happens as long as $p_{\lceil x \pi(n) \rceil}$ doesn't divide $n$. But by the prime number theorem, $p_{\lceil x \pi(n) \rceil}$ is asymptotic to

$$ \lceil x \pi(n) \rceil \log \lceil x \pi(n) \rceil \sim x \pi(n) \log (x \pi (n)) \sim x \pi(n) \log (\pi(n)) \sim x n $$

which can only divide $n$ if $x n+ o(n) = \frac{n}{d}$ for some $d$, or in other words if $x = \frac{1}{d} + o(1)$.

Any $x$ not of the form $\frac{1}{d}$ will be a positive distance far away from any $\frac{1}{d}$, so $p_{\lceil x \pi(n) \rceil}$ will not divide $n$ for $n$ sufficiently large.


For a typical number $n$, the numbers $ \frac{ \log \log p } { \log \log n}$, for $p$ dividing $n$, are evenly distributed between $0$ and $1$. Using this fact, one can likely define a similar average that converges to the constant function $1$, where each prime doesn't get an interval of the same width but rather $p_i$ has an interval of size $ \approx \frac{1}{ i \log i \log \log n }$, say.

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  • $\begingroup$ Very nice - thank you very much! I suppose then it will have poles at $\frac{1}{d}$? Can we say anything on the order of these poles? After all the integral has to be $1$ as well, in this case, I suppose. $\endgroup$ Commented Jul 16, 2021 at 22:03
  • $\begingroup$ The other alternative would be to divide the intervalls simply in $k$ pieces, such that all functions have $\delta(1)\ne0$ in that case it's immediately clear that it will converge to constant $1$ since all "combinations exist". $\endgroup$ Commented Jul 16, 2021 at 22:08
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    $\begingroup$ @RaphaelJ.F.Berger Actually I expect the function will behave like $f_n(x) = \begin{cases} 0 & x\leq0 \\ nx & 0 \leq x \leq 1/n \\ 2-nx & 1/n \leq x \leq 2/n \\ 0 & x \geq 2/n \end{cases}$ which pointwise converges to $0$ everywhere but has integral $1$. $\endgroup$
    – Will Sawin
    Commented Jul 16, 2021 at 22:11

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