By Robin's theorem

$$G(n)=\frac{\sigma(n)}{n \log \log n}$$

is bounded by $e^\gamma \approx 1.78107241799$ for $n>5040$ assuming Riemann hypothesis .

For $n=\mathrm {lcm} (1,2 \dots k)$, $G(n)$ appears generally increasing as $k$ increases reaching $\approx 1.781063$ for $n=\mathrm {lcm} (1,2 \dots10^8)$ and this is relatively close to the bound.

Empirically slightly better choice appears $n=\mathrm {lcm} (1,2 \dots k) \prod_{\text{prime } p, p<\log k}{k}$ -- again $G(n)$ generally increases as $k$ increases.

Using pari/gp reals with moderate precision so far agrees with integer results and is significantly faster. With reals got the (possibly incorrect) value of $\approx 1.781072152947513062$

Which $n$ maximize $G(n)=\frac{\sigma(n)}{n \log \log n}$?

  • $\begingroup$ That's not what Robin's theorem says, and in any case the opening paragraph is nonsensical as $a$ is not defined. Possibly, what you are trying to ask is this: define $G(n)$, and then set $a(0)=3$ and $a(k)$ to be the least integer $x$ with $x>a(k-1),G(x)>G(a(k-1))$. Give other descriptions of the sequence $a(k)$. Please clarify! $\endgroup$ – Kevin O'Bryant Nov 3 '11 at 14:19
  • $\begingroup$ @Kevin thank you, will try to clarify. I mean $a(n)$ is some sequence and I gave example with $a(n)=lcm(1..n)$. What's wrong with Robin's theorem, $G(n)$ must be $<e^\gamma$. $\endgroup$ – joro Nov 3 '11 at 14:33
  • $\begingroup$ ...assuming RH. $\endgroup$ – joro Nov 3 '11 at 14:34
  • $\begingroup$ Thank you again Kevin, tried to clarify. a(n) was indeed confusing. $\endgroup$ – joro Nov 3 '11 at 15:11
  • 3
    $\begingroup$ Anyway, $\lim_{k\to\infty}G(\mathrm{lcm}(1,\dots,k))=e^\gamma$ follows from Mertens’ theorems. $\endgroup$ – Emil Jeřábek Nov 3 '11 at 16:11

First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 13,$ smaller than $$ e^\gamma + \frac{0.64821364942...}{(\log \log n)^2},$$ with the constant in the numerator giving equality for $n=12.$ from which it follows that your supremum is achieved for some $n,$ perhaps 5040 itself, but almost certainly with $n \leq 5040.$ Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.

Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS

There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.

EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is $$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$

I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(1/6) = 12, \; f(1/10) = 60, \; f(1/12) = 120,$$ then $$ f(1/14) = 360, \; f(1/17) = 2520, \; f(1/25) = 5040, \; f(1/31) = 55440, \; f(1/39) = 720720,$$ and so on as $\delta$ decreases.

If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let $$ \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p} $$

| cite | improve this answer | |
  • 2
    $\begingroup$ I don't believe your argument in the first paragraph is correct (or maybe I don't understand it). Just because there's an upper bound for the ratio that's decreasing to $e^\gamma$ doesn't mean that the supremum is achieved for some $n$. For example, $1+1/n$ is an upper bound for the sequence $\{1-1/n\}$, but the supremum is not achieved. $\endgroup$ – Greg Martin Nov 4 '11 at 8:06
  • $\begingroup$ I may have coded CA numbers wrong, but lcm gives better results for the ranges I can test. $\endgroup$ – joro Nov 4 '11 at 15:55
  • $\begingroup$ If joro is allowing numbers below 5040, as Emil points out, there are several numbers with large $G$ such as $G(60) \approx 1.986369.$ As Emil points out, this means the maximum is achieved, for $n=3$ if we permit such small $n.$ I can add that, if RH is wrong, there exists a CA number $n$ larger than 5040 with $G(n) > e^\gamma.$ This does not mean that CA numbers give the "best" easily-calculated sequence to get large values of $G(n)$ when $n > 5040.$ But $G(n)$ can be calculated for, essentially, arbitrarily large CA numbers. $\endgroup$ – Will Jagy Nov 4 '11 at 20:09
  • $\begingroup$ Better implementation of CA gave much better results. $\endgroup$ – joro Nov 7 '11 at 12:39
  • $\begingroup$ Keith Briggs did experiments with CA of size $\log{n}=10^{11}$. $\endgroup$ – joro Nov 8 '11 at 8:29

The question is still unclear to me, therefore I will outline what the options are.

The function $G(n)$ (well-defined for integers $n>1$, and positive for $n>3$) has the following properties:

  • Grönwall’s theorem: $$\limsup_{n\to\infty}G(n)=e^\gamma.$$ For a concrete infinite sequence $\{n_k\}$ for which $G(n_k)$ tends to $e^\gamma$, one may take $n_k=\mathrm{lcm}(1,\dots,k)$.

  • There are a bunch of numbers $n\le5040$ for which $G(n)>e^\gamma$.

  • Robin’s theorem: If the Riemann hypothesis holds, then $G(n)< e^\gamma$ for all $n>5040$. If the Riemann hypothesis fails, there are infinitely many $n$ such that $G(n)>e^\gamma$ (and even $G(n)>e^\gamma+c/(\log n)^\beta$ for some constants $c,\beta>0$).

  • Another Robin’s theorem: $G(n)< e^\gamma+0.6482/(\log\log n)^2$ for all $n>1$ (unconditionally).

Therefore, there are the following possibilities:

  1. Joro wants to find $n>1$ where $G(n)$ is maximal.

    The answer is $\mathbf{n=3}$, which gives $G(3)\approx14.177183749182$. All other $G(n)$ are smaller by the unconditional Robin’s theorem above.

  2. Joro wants to find $n>5040$ where $G(n)$ is maximal.

    Case A: Riemann hypothesis holds.

    The supremum of the sequence is $e^\gamma$, but all its elements are strictly smaller. Thus, there is no maximum.

    Case B: Riemann hypothesis fails.

    Let $n_0>5040$ be such that $G(n_0)>e^\gamma$. Then $G(n)< G(n_0)$ for all but finitely many $n$ by Grönwall’s theorem, hence the sequence does have a maximum, which is greater than $e^\gamma$. The point $n$ where it is achieved must be quite large, and cannot be exhibited explicitly at present, since this would amount to disproving the Riemann hypothesis. There may be more than one $n$ achieving the maximal $G(n)$, but in any case there are only finitely many such $n$.

  3. Joro wants to exhibit an infinite sequence of $n$ on which $G(n)$ tends to $e^\gamma$.

    (He already said he does not, but I think it’s actually a more natural question than the other two readings above.) There is no unique answer, one possibility is to take the numbers $n=\mathrm{lcm}(1,\dots,k)$ as above.

  4. Joro wants something else,

    in which case he should state the question more clearly.

| cite | improve this answer | |
  • 1
    $\begingroup$ I believe that Joro wants an infinite sequence of n, such that G(n) tends to $e^{\gamma}$ 'as quickly as possible' $\endgroup$ – Woett Nov 4 '11 at 15:59
  • $\begingroup$ What is “as quickly as possible”? I don’t see how this is a well-defined problem. $\endgroup$ – Emil Jeřábek Nov 4 '11 at 19:08

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.