Let $P(n)$ denote the largest prime factor of $n$. For any integer $x\ge2$, define the median $$ M(x) = \text{the median of the set }\{P(2), P(3), \dots, P(x) \}. $$ Classical results of Dickman and de Bruijn show that the median is roughly $x^{1/\sqrt{e}}$. More specifically, I think that the Dickman-de Bruijn rho-function approach can show the following: for any function $f(x)$ tending to infinity with $x$, the median $M(x)$ is between $x^{1/\sqrt{e}}/f(x)$ and $x^{1/\sqrt{e}}f(x)$ for all sufficiently large $x$.

But I got to thinking the other day: is there a way to determine how the median compares to $x^{1/\sqrt{e}}$ specifically? In other words, which one of the following is true?

  1. For all sufficiently large $x$, we have $M(x) \lt x^{1/\sqrt{e}}$.
  2. Each inequality $M(x) \lt x^{1/\sqrt{e}}$ and $M(x) \ge x^{1/\sqrt{e}}$ holds for arbitrarily large $x$.
  3. For all sufficiently large $x$, we have $M(x) \ge x^{1/\sqrt{e}}$.
  • $\begingroup$ This question has been answered. It is option #1, $M(x)<x^{1/\sqrt{e}}$. I spoke to the OP in person regarding the solution. $\endgroup$ – Eric Naslund Mar 31 '11 at 22:39
  • 2
    $\begingroup$ The OP agrees.... $\endgroup$ – Greg Martin Apr 12 '11 at 21:56

This is one of many questions that has been answered in the comments, so I will just summarize the answer with a CW posting: $M(x) < x^{1/\sqrt{e}}$ according to the poster Greg Martin. In a computer search, the ratio seems to converge to roughly $0.74$ for $x$ up to a million. On the other hand, it doesn't converge very quickly and might possibly not converge at all. All that was have from the original posting is that the lim sup of the ratio is finite (and at most 1 according to the answer) while the lim inf is positive.

| cite | improve this answer | |
  • 4
    $\begingroup$ Eric and I believe we can prove that the ratio $M(x)/x^{1/\sqrt e}$ tends to the constant $e^{(1-\gamma)/\sqrt e} \approx 0.7738$, where $\gamma$ is Euler's constant. $\endgroup$ – Greg Martin Oct 22 '11 at 5:49
  • $\begingroup$ @Greg Martin Cool! Let me suggest that you either accept this CW answer or post your own answer and accept it. $\endgroup$ – Greg Kuperberg Oct 22 '11 at 15:01

I posted a paper to the arXiv which deals with this question along with some other things.

Using some results regarding either the mean of $\omega(n)$, or integers without large prime factors, we can prove that $$M(x)=e^{(\gamma-1)/\sqrt{e}}x^{1/\sqrt{e}}\left(1+O\left(\frac{1}{\log x}\right)\right),$$ where $\gamma$ is the Euler Mascheroni constant.

More specifically, if we let $\text{li}_f(x)=\int_2^x \frac{\{x/t\}}{\log t} dt$, then $$M(x)=x^{\frac{1}{\sqrt{e}}\exp\left(-\frac{\text{li}_f(x)}{x}\right)}\left(1+O\left(e^{-c\sqrt{\log x}}\right)\right).$$ (In case the TeX is not readable, the exponent of $x$ in the above equation is $\frac{1}{\sqrt{e}}\exp\left(-\frac{\text{li}_f(x)}{x}\right)$.)

This function $\text{li}_f(x)$ has the asymptotic expansion $$\text{li}_f(x)=c_0 \frac{x}{\log x}+c_1 \frac{x}{\log^2 x}+\cdots+c_{k-1} \frac{(k-1)!x}{\log^k x}+O\left(\frac{x}{\log^{k+1}(x)}\right),$$

where $$c_k=1-\sum_{j=0}^k\frac{\gamma_k}{k!},$$ and $\gamma_k$ denotes the $k^{th}$ Stieltjes constant.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.