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I've tagged this as reference request as surely this question must be very well investigated, I just don't know how to look for it. Most likely the perfect answer will be in form of a keyword for googling :)

I would like to know about dynamics of relative frequencies of prime factorizations. More precisely, let $[k_1,k_2,...]$ be any multiset of natural numbers, and for $x>0$ let $f_x[k_1,k_2,...]$ be the number of those natural $n\le x$ with prime factorization of the form $p_1^{k_1}p_2^{k_2}\cdots$, divided by $x$. How does the "champion" (i. e. $[k_1,k_2,...]$ with largest $f_x[k_1,k_2,...]$) change with $x$? Do all "champions" have some common features? (Say, are they all of the form $[1,1,...]$?) At which $x$es do "champion changes" occur?

More generally, how does the list of all multisets $[k_1,k_2,...]$ ordered according to $f_x[k_1,k_2,...]$ vary with $x$?

Here is the plot for $x$ from $100000$ to up to ten million of the $f_x[k_1,k_2,...]$ for five most frequent (at ten million) $[k_1,k_2,...]$, namely, for $[1,1,1]$, $[1,1]$, $[1,1,1,1]$, $[1,1,2]$ and $[1]$.

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Indeed such problems have been studied before. I claim that the champion multisets will just be a string of $r$ ones (for a suitable $r$), and in this case one is counting square-free integers up to $x$ with exactly $r$ prime factors. This case has been widely investigated. For example, Balazard establishing a conjecture of Erdős showed that for each large $x$ this sequence is unimodal in $r$ (see Theorem E on page 24), and (as one can see by Hardy-Ramanujan/Erdős-Kac) attains its maximum for $r$ close to $\log \log x$. Balazard's paper will have more references -- in particular to work of Hildebrand and Tenenebaum giving asymptotics in wide ranges. In particular, it follows that the maximum number of integers up to $x$ having a factorization of type $p_1 \cdots p_r$ is $$ \sim \frac{6}{\pi^2} x \frac{1}{\sqrt{2\pi \log \log x}}, \tag{1} $$ which is attained for $r$ around $\log \log x$. (The $6/\pi^2$ comes from counting square-free numbers.)

Now consider a general factorization type $[k_1,\ldots, k_s, 1, 1,\ldots ,1]$ where $k_1$, $\ldots$, $k_s$ are all at least $2$ (and the number of $1$'s in the multiset could possibly be zero). We'll show that these factorization types will contribute an amount smaller than the largest possibility for all ones given above. Write every number of this factorization type as $n=ab$ where $a$ is square-full and $b$ is square-free and coprime to $a$ (thus $a$ has type $[k_1,\ldots, k_s]$ and $b$ is of type $[1,\ldots, 1]$). Thus the number of integers of this factorization type is (with $a$ and $b$ having the above meaning) $$ \sum_{1< a=p_1^{k_1}\cdots p_s^{k_s}} \sum_{b \le x/a} 1 \le \sum_{a >\sqrt{x}} \frac{x}{a} + \sum_{a\le \sqrt{x}} \Big(\frac{6}{\pi^2} +o(1)\Big) \frac{x}{a \sqrt{2\pi \log \log x}}, \tag{2} $$
where we bounded the sum over $b$ trivially in the first sum, and using (1) in the second sum (note $\log \log (x/a) \sim \log \log x$ there, and we also ignored the condition that $b$ is coprime to $a$). Since the number of square-full integers up to $x$ is $\ll \sqrt{x}$, the first sum in (2) may be bounded by $\ll x^{\frac 34}$. We may of course evaluate the second term exactly, for any given choice of $[k_1,\ldots,k_s]$. Alternatively, we could bound that term by letting $a$ run over all square-full integers $>1$: since $$ \sum_{1<a, a\text{ square-full}} \frac 1a =\prod_{p} \Big(1+\frac{1}{p^2}+\frac{1}{p^3}+\ldots \Big) -1 = \frac{\zeta(2)\zeta(3)}{\zeta(6)}- 1 =0.943\ldots . $$ This completes our proof that factorization types having at least one number strictly larger than $1$ contribute an amount that is strictly smaller than the maximum attained by all ones.

One can be extract more information than given above. For example, after types involving all $1$'s the next champion will be $[2,1,1,1,\ldots,1]$, and here one can give an asymptotic like (1) for the maximum such number etc.

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  • $\begingroup$ Thank you! Should probably figure out the part about $1$'s myself. This allows for simplification in my calculations, maybe I will add more stuff here. In any case obviously this is the right answer to accept. $\endgroup$ Commented Sep 4, 2015 at 17:46
  • $\begingroup$ ...on the afterthought though, - sorry, seems I still don't understand even the simplest part. I mean, with this argument then also if $p_1p_2\le x$ then also $p_1\le x$, but this does not mean that there are more primes below $x$ than numbers of the form $p_1p_2$, right? $\endgroup$ Commented Sep 5, 2015 at 5:08
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    $\begingroup$ I edited to make that step a little clearer. $\endgroup$
    – Lucia
    Commented Sep 5, 2015 at 17:35
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    $\begingroup$ That argument was a little too glib, and I need to be a bit more careful. The problem is that if we count numbers of the form $p^{2}q^3$ say then this will be different from $p^3q^2$, but both map down to $pq$. What I wrote is still true, but requires a bit more work. I'll edit it later today if I get a chance. $\endgroup$
    – Lucia
    Commented Sep 5, 2015 at 18:12
  • $\begingroup$ You're very welcome! $\endgroup$
    – Lucia
    Commented Sep 6, 2015 at 6:28

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