﻿

# Highly composite number

A highly composite number (HCN) is a positive integer with more divisors than any positive integer smaller than itself.

The initial or smallest twenty-one highly composite numbers are listed in the table at right.

n number of
divisors of n
1 1
2 2
4 3
6 4
12 6
24 8
36 9
48 10
60 12
120 16
180 18
240 20
360 24
720 30
840 32
1,260 36
1,680 40
2,520 48
5,040 60
7,560 64
10,080 72

The sequence of highly composite numbers (sequence A002182 in OEIS) is a subset of the sequence of smallest numbers k with exactly n divisors (sequence A005179 in OEIS).

There are an infinite number of highly composite numbers. To prove this fact, suppose that n is an arbitrary highly composite number. Then 2n has more divisors than n (2n itself is a divisor and so are all the divisors of n) and so some number larger than n (and not larger than 2n) must be highly composite as well.

Roughly speaking, for a number to be highly composite it has to have prime factors as small as possible, but not too many of the same. If we decompose a number n in prime factors like this:

$n = p_1^{c_1} \times p_2^{c_2} \times \cdots \times p_k^{c_k}\qquad (1)$

where $p_1 < p_2 < \cdots < p_k$ are prime, and the exponents ci are positive integers, then the number of divisors of n is exactly

$(c_1 + 1) \times (c_2 + 1) \times \cdots \times (c_k + 1).\qquad (2)$

Hence, for n to be a highly composite number,

• the k given prime numbers pi must be precisely the first k prime numbers (2, 3, 5, ...); if not, we could replace one of the given primes by a smaller prime, and thus obtain a smaller number than n with the same number of divisors (for instance 10 = 2 × 5 may be replaced with 6 = 2 × 3; both have 4 divisors);
• the sequence of exponents must be non-increasing, that is $c_1 \geq c_2 \geq \cdots \geq c_k$; otherwise, by exchanging two exponents we would again get a smaller number than n with the same number of divisors (for instance 18 = 21 × 32 may be replaced with 12 = 22 × 31; both have six divisors).

Also, except in two special cases n = 4 and n = 36, the last exponent ck must equal 1. Saying that the sequence of exponents is non-increasing is equivalent to saying that a highly composite number is a product of primorials. Because the prime factorization of a highly composite number uses all of the first k primes, every highly composite number must be a practical number.[1]

Highly composite numbers higher than 6 are also abundant numbers. One need only look at the three or four highest divisors of a particular highly composite number to ascertain this fact. It is false that all highly composite numbers are also Harshad numbers in base 10. The first HCN that is not a Harshad number is 245,044,800, which has a digit sum of 27, but 27 does not divide evenly into 245,044,800.

Many of these numbers are used in traditional systems of measurement, and tend to be used in engineering designs, due to their ease of use in calculations involving vulgar fractions.

If Q(x) denotes the number of highly composite numbers less than or equal to x, then there are two constants a and b, both greater than 1, such that

$\ln(x)^a \le Q(x) \le \ln(x)^b \, .$

The first part of the inequality was proved by Paul Erdős in 1944 and the second part by Jean-Louis Nicolas in 1988.

## Examples

 The highly composite number :  10,080. 10,080  =  (2 × 2 × 2 × 2 × 2)  ×  (3 × 3)  ×  5  ×  7 By (2) above, 10,080 has exactly seventy-two divisors. 1 × 10,080 2 × 5,040 3 × 3,360 4 × 2,520 5 × 2,016 6 × 1,680 7 × 1,440 8 × 1,260 9 × 1,120 10 × 1,008 12 × 840 14 × 720 15 × 672 16 × 630 18 × 560 20 × 504 21 × 480 24 × 420 28 × 360 30 × 336 32 × 315 35 × 288 36 × 280 40 × 252 42 × 240 45 × 224 48 × 210 56 × 180 60 × 168 63 × 160 70 × 144 72 × 140 80 × 126 84 × 120 90 × 112 96 × 105
 Note:  The bolded numbers are themselves highly composite numbers. Only the twentieth highly composite number 7560 (=3×2520) is absent. 10080 is a so-called 7-smooth number, (sequence A002473 in OEIS).

The 15,000th highly composite number can be found on Achim Flammenkamp's website. It is the product of 230 primes:

$a_0^{14} a_1^9 a_2^6 a_3^4 a_4^4 a_5^3 a_6^3 a_7^3 a_8^2 a_9^2 a_{10}^2 a_{11}^2 a_{12}^2 a_{13}^2 a_{14}^2 a_{15}^2 a_{16}^2 a_{17}^2 a_{18}^{2} a_{19} a_{20} a_{21}\cdots a_{229}$,

where an is the sequence of successive prime numbers, and all omitted terms (a22 to a228) are factors with exponent equal to one (i.e. the number is $2^{14} \times 3^{9} \times 5^6 \times \cdots \times 1451$). [2]

## Prime factor subsets

For any highly composite number, if one takes any subset of prime factors for that number and their exponents, the resulting number will have more divisors than any smaller number that uses the same prime factors. For example for the highly composite number 720 which is 24 × 32 × 5 we can be sure that

• 144 which is 24 × 32 has more divisors than any smaller number that has only the prime factors 2 and 3
• 80 which is 24 × 5 has more divisors than any smaller number that has only the prime factors 2 and 5
• 45 which is 32 × 5 has more divisors than any smaller number that has only the prime factors 3 and 5

If this were untrue for any particular highly composite number and subset of prime factors, we could exchange that subset of primefactors and exponents for the smaller number using the same primefactors and get a smaller number with at least as many divisors.

This property is useful for finding highly composite numbers.

## References

1. ^ Srinivasan, A. K. (1948), "Practical numbers", Current Science 17: 179–180, MR0027799 .
2. ^ Flammenkamp, Achim, Highly Composite Numbers .

Wikimedia Foundation. 2010.