 Divisor function d(n) up to n = 250 Prime-power factors

In mathematics, a superior highly composite number is a natural number which has the highest ratio of its number of divisors to some positive power of itself than any other number. It is a stronger restriction than that of a highly composite number, which is defined as having more divisors than any smaller positive integer.

The first 10 superior highly composite numbers and their factorization are listed.

# prime
factors
SHCN
n
prime
factorization
prime
exponents
# divisors
d(n)
primorial
factorization
1 2 2 1 2 2 2
2 6 2 ⋅ 3 1,1 22 4 6
3 12 22 ⋅ 3 2,1 3×2 6 2 ⋅ 6
4 60 22 ⋅ 3 ⋅ 5 2,1,1 3×22 12 2 ⋅ 30
5 120 23 ⋅ 3 ⋅ 5 3,1,1 4×22 16 22 ⋅ 30
6 360 23 ⋅ 32 ⋅ 5 3,2,1 4×3×2 24 2 ⋅ 6 ⋅ 30
7 2520 23 ⋅ 32 ⋅ 5 ⋅ 7 3,2,1,1 4×3×22 48 2 ⋅ 6 ⋅ 210
8 5040 24 ⋅ 32 ⋅ 5 ⋅ 7 4,2,1,1 5×3×22 60 22 ⋅ 6 ⋅ 210
9 55440 24 ⋅ 32 ⋅ 5 ⋅ 7 ⋅ 11 4,2,1,1,1 5×3×23 120 22 ⋅ 6 ⋅ 2310
10 720720 24 ⋅ 32 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 4,2,1,1,1,1 5×3×24 240 22 ⋅ 6 ⋅ 30030 Plot of the number of divisors of integers from 1 to 1000. Highly composite numbers are labelled in bold and superior highly composite numbers are starred. In the SVG file, hover over a bar to see its statistics.

For a superior highly composite number n there exists a positive real number ε such that for all natural numbers k smaller than n we have

${\frac {d(n)}{n^{\varepsilon ))}\geq {\frac {d(k)}{k^{\varepsilon )))$ and for all natural numbers k larger than n we have

${\frac {d(n)}{n^{\varepsilon ))}>{\frac {d(k)}{k^{\varepsilon )))$ where d(n), the divisor function, denotes the number of divisors of n. The term was coined by Ramanujan (1915).

For example, the number with the most divisors per square root of the number itself is 12; this can be demonstrated using some highly composites near 12: ${\frac {2}{\sqrt {2))}\approx 1.414,{\frac {3}{\sqrt {4))}=1.5,{\frac {4}{\sqrt {6))}\approx 1.633,{\frac {6}{\sqrt {12))}\approx 1.732,{\frac {8}{\sqrt {24))}\approx 1.633,{\frac {12}{\sqrt {60))}\approx 1.549$ .

The first 15 superior highly composite numbers, 2, 6, 12, 60, 120, 360, 2520, 5040, 55440, 720720, 1441440, 4324320, 21621600, 367567200, 6983776800 (sequence A002201 in the OEIS) are also the first 15 colossally abundant numbers, which meet a similar condition based on the sum-of-divisors function rather than the number of divisors. Neither set, however, is a subset of the other.

## Properties All superior highly composite numbers are highly composite. This is easy to prove: if there is some number k that has the same number of divisors as n but is less than n itself (i.e. $d(k)=d(n)$ , but $k ), then ${\frac {d(k)}{k^{\varepsilon ))}>{\frac {d(n)}{n^{\varepsilon )))$ for all positive ε, so if a number "n" is not highly composite, it cannot be superior highly composite.

An effective construction of the set of all superior highly composite numbers is given by the following monotonic mapping from the positive real numbers. Let

$e_{p}(x)=\left\lfloor {\frac {1}((\sqrt[{x}]{p))-1))\right\rfloor \quad$ for any prime number p and positive real x. Then

$\quad s(x)=\prod _{p\in \mathbb {P} }p^{e_{p}(x)}\quad$ is a superior highly composite number.

Note that the product need not be computed indefinitely, because if $p>2^{x)$ then $e_{p}(x)=0$ , so the product to calculate $s(x)$ can be terminated once $p\geq 2^{x)$ .

Also note that in the definition of $e_{p}(x)$ , $1/x$ is analogous to $\varepsilon$ in the implicit definition of a superior highly composite number.

Moreover, for each superior highly composite number $s^{\prime )$ exists a half-open interval $I\subset \mathbb {R} ^{+)$ such that $\forall x\in I:s(x)=s^{\prime )$ .

This representation implies that there exist an infinite sequence of $\pi _{1},\pi _{2},\ldots \in \mathbb {P}$ such that for the n-th superior highly composite number $s_{n)$ holds

$s_{n}=\prod _{i=1}^{n}\pi _{i)$ The first $\pi _{i)$ are 2, 3, 2, 5, 2, 3, 7, ... (sequence A000705 in the OEIS). In other words, the quotient of two successive superior highly composite numbers is a prime number.