

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

If $d(n)$ denotes the number of divisors of $n \in \mathbb{N}$, we may define the function $$C(n) = \frac{\log(d(n)) \cdot \log(\log n)}{\log 2 \log n}.$$

According to Wikipedia, the Swedish mathematician Carl Severin Wigert proved that $\displaystyle \limsup C(n) = 1.$ $C$ can therefore reasonably be used as a measure of degree of compositeness. However, the natural log base is somewhat arbitrary, and arguably a better choice would be to use logs base 2, $\log_2 n$, and set $$\displaystyle C_b(n) = \frac{\log_2(d(n)) \cdot \log_2(\log_2 n)}{\log_2 n}.$$

This also has $\displaystyle \limsup$ equal to 1, and has the further nice property that $C_b(2) = 0$, so that the minimal degree of compositeness is zero.

Suppose we define an "ultracomposite number" to be an $n \in \mathbb{N}$ such that $C_b(n) > 1$, and for every $m < n$, we have $C_b(m) < C_b(n).$

It follows that the set of ultracomposite numbers is finite. Two questions now arise:

(1) What is the largest ultracomposite number, or failing that, what is a bound or estimate?

(2) How large is the set of ultracomposite numbers, or at least what is a bound or estimate for the size?

(3) More specifically, I conjecture there are seventeen ultracomposite numbers, of which 55440 is the largest.

It should be noted that the interest of this question depends on the acceptance of log base 2 as the privileged choice for this problem--natural logs, for example, lead to far more ultracomposites.

share|cite|improve this question
It would be interesting to know how the number of ultracomposites varies with the choice of logarithm base. – Noah Schweber Aug 9 '13 at 21:25
short version of my answer: calculate your quantity for the numbers listed here: – Will Jagy Aug 9 '13 at 22:18
I also emailed you Nicolas and Robin, (1983) in the Canadian Mathematical Bulletin, volume 26, number 4, pages 485-492, giving full details of the proof of the first upper bound I quote. In French. – Will Jagy Aug 14 '13 at 21:07
up vote 7 down vote accepted

EDIT, 11:36 pm. The conjecture is correct. From the first result of Nicolas and Robin, $$ C_2(n) \leq 1.53793986 + \frac{0.56367483}{\log \log n}. $$ This bound decreases with $n.$ Furthermore this gives $ 1.743557976 $ when $n = 5433960.$ This says that we need only confirm the value of the ratio $C_2(n)$ for $n \leq 5433960,$ by any means that brings joy, because $$ C_2(55440) = 1.743557976.$$

ORIGINAL: Did it myself. For logs base 2 as described, 55440 now seems a good bet. For natural logarithms, Nicolas and Robin's number 6,983,776,800. Note where the "ratio" 1.537939861 appears in the first Nicolas-Robin result in my earlier answer. That is, it is a theorem that the relevant ratio attains its maximum at 6,983,776,800 and no other place. It may well be that the apparent list for the logs base 2 come as a corollary of one of the Nicolas-Robin results.

  logs base 2:

  n   2 divisors 2 ratio   0   2 bump   2^1
  n   6 divisors 4 ratio   1.060087604   6 bump   3^1
  n   12 divisors 6 ratio   1.32815683   12 bump   2^2
  n   60 divisors 12 ratio   1.555150513   60 bump   5^1
  n   120 divisors 16 ratio   1.614640529   120 bump   2^3
  n   360 divisors 24 ratio   1.666250961   360 bump   3^2
  n   2520 divisors 48 ratio   1.729062351   2520 bump   7^1
  n   5040 divisors 60 ratio   1.73879969   5040 bump   2^4
  n   55440 divisors 120 ratio   1.743557976   55440 bump   11^1

natural logs:

  n   2 divisors 2 ratio   -0.5287663729   2 bump   2^1
  n   6 divisors 4 ratio   0.6509780925   6 bump   3^1
  n   12 divisors 6 ratio   0.9468861067   12 bump   2^2
  n   60 divisors 12 ratio   1.234235881   60 bump   5^1
  n   120 divisors 16 ratio   1.308415107   120 bump   2^3
  n   360 divisors 24 ratio   1.380756857   360 bump   3^2
  n   2520 divisors 48 ratio   1.467704178   2520 bump   7^1
  n   5040 divisors 60 ratio   1.484851215   5040 bump   2^4
  n   55440 divisors 120 ratio   1.51180374   55440 bump   11^1
  n   720720 divisors 240 ratio   1.525218379   720720 bump   13^1
  n   1441440 divisors 288 ratio   1.527797982   1441440 bump   2^5
  n   4324320 divisors 384 ratio   1.531905251   4324320 bump   3^3
  n   21621600 divisors 576 ratio   1.534731579   21621600 bump   5^2
  n   367567200 divisors 1152 ratio   1.537551622   367567200 bump   17^1
  n   6983776800 divisors 2304 ratio   1.537939861   6983776800 bump   19^1
share|cite|improve this answer

What you want to look at are Ramanujan's "superior highly composite" numbers, described, for example, in a 1988 survey in English by Jean-Louis Nicolas. Email we if you wish a copy.The shc numbers can be readily produced, in order, on a computer. I recently improved by software for this type of thing owing to a related question by Igor Rivin. See and Furthermore, since the prime factorizations are figured out before multiplying out the number itself (put in order, the next one is just a small prime times the current one), various ratios of the type you want can be found without needing to print out the number itself. Maybe the best way to convey the advantage of working with these numbers is to point out that no factorization is required; you know the prime factorization before you know the number.

Anyway, the shc numbers give the biggest ratios of your type. Indeed, they provide bounds for all numbers, such as

$$ d(n) \leq n^{\frac{1.537939860675 \log 2}{\log \log n}} $$

$$ d(n) \leq n^{ \left( \frac{ \log 2}{\log \log n} \right) \left( 1 + \frac{1.93485096797}{ \log \log n} \right) } $$


$$ d(n) \leq n^{ \left( \frac{ \log 2}{\log \log n} \right) \left( 1 + \frac{1}{ \log \log n} + \frac{4.762350121}{ (\log \log n)^2} \right) } $$

I'm not entirely sure what happens if you drop the 4.76235 term in the third one, I think the conjecture would be that infinitely many shc numbers make the revised inequality false, but the numbers will be really huge.

Hmmm. Nicolas does not conjecture that, so let's call that a guess on my part. Meanwhile, he says that the Riemann Hypothesis implies that, with $$ \theta = \frac{\log (3/2)}{\log 2} \approx 0.5849625, $$ there is a constant $c$ such that (result by Ramanujan) $$ \frac{\log d(n)}{\log 2} \leq \operatorname{li} (\log n) + c (\log n)^\theta. $$ He points out that the methods being described would allow one to find the best possible $c$ in this formulation but that is not in this particular survey. If you think about the usual asymptotic expansion for the logarithmic integral function, you see how Ramanujan's RH implication closely resembles the unconditional results above it.

EEEDDDIIITTT: I made a jpeg of some info I once printed out, the specific s.h.c numbers that give equality in the Nicolas-Robin bounds above. Note that $$ 1.537939860675 \log 2 \approx 1.066018678297. $$ Given some $\delta > 0,$ the exponent of the prime $p$ in the corresponding s.h.c. number, call it $N_\delta,$ is $$ \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor. $$ So, in the jpeg, we see the four real coefficients indicated, then the three s.h.c numbers, each with its number of divisors, then a value for $\delta$ that will produce that s.h.c. number written by hand. Note that the three s.h.c. numbers depicted are lines numbered 15, 39, 87 in THIS LIST.

enter image description here

share|cite|improve this answer

Your Answer


