It seems the asymptotic formula,

$\sum_{n=2}^{x} P(n)$ ~ $\frac{\pi^2}{12}\frac{x^{2}}{log(x)}$ as $x \rightarrow \infty$

where $P(n)$ is the largest prime factor of the positive integer $n$, cannot be used to find an asymptotic formula for $P(x)$ simply taking the difference, $\sum_{n=2}^{x} P(n) - \sum_{n=2}^{x-1} P(n)$.

I have found papers which elaborate the asymptotic formula for the sum, (e.g. "The Average Largest Prime Factor", Naslund, Integers 13 (2013)), however I have not been able to find any papers which derive an asymptotic formula for $P(x)$ for any subset of the Naturals.

Is such a formula known for any non-trivial subset of the Naturals and in particular for the subset consisting of all Naturals that have exactly k prime factors and are square-free? If so, I would appreciate any references.

Reply (Nov 5) to Greg's and mixedmath's answers/comments

Thanks for your answers/comments. $P(n)$ has the bounds you mention and it is irregular but I do not understand how boundedness and irregularity imply the non-existence of an asymptotic formula for $P(n)$, if that is what you are saying.

The boundedness of $P(n)$ does not determine its behavior within those bounds and $P(n)$ may have quite a lot of regularity depending on its domain.

For example, a domain of $P(n)$ can be constructed as follows:

  1. Let $X$ be the set of square-free Naturals with $k$ prime factors.

  2. Let $X_j $ be the subset of $X$ such that $P(n) = p_j $, the j-th prime, and where $j \geq k$. The cardinal number of this set will be finite and will depend on $j$ and $k$ .

  3. Let $ X'_j $ be the subset of $X_j$ consisting of members of $X_j$ that are greater than any member of the sets $X_1, X_2,\dots,X_{j-1} $ . This ensures the sets $ X'_j $ are disjoint and the members of each set of the sequence are greater than the members of any previous set. The set $ X'_j$ will be non-empty and have a finite cardinal number.

  4. Let $ X''_j $ be the set $ X'_j $ ordered by size.

Then $P(n)$ on the domain {$ {X''_1, X''_2 \dots}$} will be an increasing step function. The length of a step will depend on $j$ and $k$ and the height of a step will be the $ j $ th prime gap. $ P(n)$ on this domain has quite a lot of regularity.

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    $\begingroup$ Since $P(n)=2$ for infinitely many $n$, how can there be an asymptotic formula? $\endgroup$ – Barry Cipra Nov 4 '14 at 0:11
  • $\begingroup$ Barry: I will need to edit my question. I was thinking of n belonging to a certain subset of the Naturals but I failed to state what this subset was. On this subset, P(n) tends to infinity as n tends to infinity. $\endgroup$ – gjh Nov 4 '14 at 0:31
  • $\begingroup$ For your last paragraph: no, $P(n)$ is too rough even on the set of squarefree numbers with $k$ prime factors. On this set, $P(n)$ will range from around $n^{1/k}$ all the way to close to $n$ itself. $\endgroup$ – Greg Martin Nov 4 '14 at 18:22
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    $\begingroup$ (Indeed, if there were an asymptotic formula for $P(n)$ itself, nobody would be looking at $\sum_{n\le x} P(n)$!) $\endgroup$ – Greg Martin Nov 4 '14 at 18:22
  • $\begingroup$ As a matter of experience, useful, meaningful asymptotics as $n\to\infty$ involve regular smooth functions (hence only apply to sufficiently regular functions, e.g. $\pi(x)\sim\frac{x}{\log x}$), and precise asymptotics of irregular functions tend to be trivially equivalent to the original function (e.g. $P(n)\sim P(n)$), else asymptotics of irregular functions are uselessly crude (since upper and lower bounds are separated by such a large chasm, e.g. $2\le P(n)\le n$). $\endgroup$ – whacka Nov 4 '14 at 22:20

No, there is no asymptotic. In short, $P(x)$ is simply too rough. Most importantly, $P(n) = 2$ for infinitely many $n$, and $P(n) = n$ for infinitely many $n$. So the only bounds for individual $P(x)$ that you can get are

$$ 2 \leq P(n) \leq n,$$

which is not useful.

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    $\begingroup$ For your last paragraph: no, $P(n)$ is too rough even on the set of squarefree numbers with $k$ prime factors. On this set, $P(n)$ will range from around $n^{1/k}$ all the way to close to $n$ itself. $\endgroup$ – Greg Martin Nov 4 '14 at 0:49
  • $\begingroup$ @GregMartin I think you meant to comment on the question, and not my answer. But spot on comment. $\endgroup$ – davidlowryduda Nov 4 '14 at 0:54
  • $\begingroup$ hm, maybe I'd better copy it up there.... $\endgroup$ – Greg Martin Nov 4 '14 at 18:21

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