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. $\endgroup$ Mar 31, 2011 at 22:39
  • 2
    $\begingroup$ The OP agrees.... $\endgroup$ Apr 12, 2011 at 21:56

2 Answers 2


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.

  • 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. $\endgroup$ Oct 22, 2011 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. $\endgroup$ Oct 22, 2011 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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