Highly composite number

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A highly composite number (HCN) is a positive integer with more divisors than any smaller positive integer. The term was coined by Ramanujan (1915), who showed that there are infinitely many such numbers. The related concept of largely composite number refers to a positive integer which has at least as many divisors as any smaller positive integer.

The initial or smallest 38 highly composite numbers are listed in the table at right. 10 of them are superior highly composite numbers.

Order	HCN n	prime factorization	prime exponents	prime factors	d(n)	primorial factorization
1	1			0	1
2*	2	$2$	1	1	2	$2$
3	4	$2^{2}$	2	2	3	$2^{2}$
4*	6	$2\cdot 3$	1,1	2	4	$6$
5*	12	$2^{2}\cdot 3$	2,1	3	6	$2\cdot 6$
6	24	$2^{3}\cdot 3$	3,1	4	8	$2^{2}\cdot 6$
7	36	$2^{2}\cdot 3^{2}$	2,2	4	9	$6^{2}$
8	48	$2^{4}\cdot 3$	4,1	5	10	$2^{3}\cdot 6$
9*	60	$2^{2}\cdot 3\cdot 5$	2,1,1	4	12	$2\cdot 30$
10*	120	$2^{3}\cdot 3\cdot 5$	3,1,1	5	16	$2^{2}\cdot 30$
11	180	$2^{2}\cdot 3^{2}\cdot 5$	2,2,1	5	18	$6\cdot 30$
12	240	$2^{4}\cdot 3\cdot 5$	4,1,1	6	20	$2^{3}\cdot 30$
13*	360	$2^{3}\cdot 3^{2}\cdot 5$	3,2,1	6	24	$2\cdot 6\cdot 30$
14	720	$2^{4}\cdot 3^{2}\cdot 5$	4,2,1	7	30	$2^{2}\cdot 6\cdot 30$
15	840	$2^{3}\cdot 3\cdot 5\cdot 7$	3,1,1,1	6	32	$2^{2}\cdot 210$
16	1260	$2^{2}\cdot 3^{2}\cdot 5\cdot 7$	2,2,1,1	6	36	$6\cdot 210$
17	1680	$2^{4}\cdot 3\cdot 5\cdot 7$	4,1,1,1	7	40	$2^{3}\cdot 210$
18*	2520	$2^{3}\cdot 3^{2}\cdot 5\cdot 7$	3,2,1,1	7	48	$2\cdot 6\cdot 210$
19*	5040	$2^{4}\cdot 3^{2}\cdot 5\cdot 7$	4,2,1,1	8	60	$2^{2}\cdot 6\cdot 210$
20	7560	$2^{3}\cdot 3^{3}\cdot 5\cdot 7$	3,3,1,1	8	64	$6^{2}\cdot 210$
21	10080	$2^{5}\cdot 3^{2}\cdot 5\cdot 7$	5,2,1,1	9	72	$2^{3}\cdot 6\cdot 210$
22	15120	$2^{4}\cdot 3^{3}\cdot 5\cdot 7$	4,3,1,1	9	80	$2\cdot 6^{2}\cdot 210$
23	20160	$2^{6}\cdot 3^{2}\cdot 5\cdot 7$	6,2,1,1	10	84	$2^{4}\cdot 6\cdot 210$
24	25200	$2^{4}\cdot 3^{2}\cdot 5^{2}\cdot 7$	4,2,2,1	9	90	$2^{2}\cdot 30\cdot 210$
25	27720	$2^{3}\cdot 3^{2}\cdot 5\cdot 7\cdot 11$	3,2,1,1,1	8	96	$2\cdot 6\cdot 2310$
26	45360	$2^{4}\cdot 3^{4}\cdot 5\cdot 7$	4,4,1,1	10	100	$6^{3}\cdot 210$
27	50400	$2^{5}\cdot 3^{2}\cdot 5^{2}\cdot 7$	5,2,2,1	10	108	$2^{3}\cdot 30\cdot 210$
28*	55440	$2^{4}\cdot 3^{2}\cdot 5\cdot 7\cdot 11$	4,2,1,1,1	9	120	$2^{2}\cdot 6\cdot 2310$
29	83160	$2^{3}\cdot 3^{3}\cdot 5\cdot 7\cdot 11$	3,3,1,1,1	9	128	$6^{2}\cdot 2310$
30	110880	$2^{5}\cdot 3^{2}\cdot 5\cdot 7\cdot 11$	5,2,1,1,1	10	144	$2^{3}\cdot 6\cdot 2310$
31	166320	$2^{4}\cdot 3^{3}\cdot 5\cdot 7\cdot 11$	4,3,1,1,1	10	160	$2\cdot 6^{2}\cdot 2310$
32	221760	$2^{6}\cdot 3^{2}\cdot 5\cdot 7\cdot 11$	6,2,1,1,1	11	168	$2^{4}\cdot 6\cdot 2310$
33	277200	$2^{4}\cdot 3^{2}\cdot 5^{2}\cdot 7\cdot 11$	4,2,2,1,1	10	180	$2^{2}\cdot 30\cdot 2310$
34	332640	$2^{5}\cdot 3^{3}\cdot 5\cdot 7\cdot 11$	5,3,1,1,1	11	192	$2^{2}\cdot 6^{2}\cdot 2310$
35	498960	$2^{4}\cdot 3^{4}\cdot 5\cdot 7\cdot 11$	4,4,1,1,1	11	200	$6^{3}\cdot 2310$
36	554400	$2^{5}\cdot 3^{2}\cdot 5^{2}\cdot 7\cdot 11$	5,2,2,1,1	11	216	$2^{3}\cdot 30\cdot 2310$
37	665280	$2^{6}\cdot 3^{3}\cdot 5\cdot 7\cdot 11$	6,3,1,1,1	12	224	$2^{3}\cdot 6^{2}\cdot 2310$
38*	720720	$2^{4}\cdot 3^{2}\cdot 5\cdot 7\cdot 11\cdot 13$	4,2,1,1,1,1	10	240	$2^{2}\cdot 6\cdot 30030$

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).

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. By the fundamental theorem of arithmetic, every positive integer n has a unique prime factorization:

$n=p_{1}^{c_{1}}\times p_{2}^{c_{2}}\times \cdots \times p_{k}^{c_{k}}\qquad (1)$

where $p_{1}<p_{2}<\cdots <p_{k}$ are prime, and the exponents $c_{i}$ are positive integers.

Any factor of n must have the same or lesser multiplicity in each prime:

$p_{1}^{d_{1}}\times p_{2}^{d_{2}}\times \cdots \times p_{k}^{d_{k}},0\leq d_{i}\leq c_{i},0<i\leq k$

So the number of divisors of n is:

$d(n)=(c_{1}+1)\times (c_{2}+1)\times \cdots \times (c_{k}+1).\qquad (2)$

Hence, for n to be a highly composite number,

• the k given prime numbers p_i 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 four 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 c_k must equal 1. It means that 1, 4, and 36 are the only square highly composite numbers. 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.

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 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

$\ln(x)^{a}\leq Q(x)\leq \ln(x)^{b}\,.$

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

$1.13862<\liminf {\frac {\log Q(x)}{\log \log x}}\leq 1.44\$

and

$\limsup {\frac {\log Q(x)}{\log \log x}}\leq 1.71\ .$

Examples

The highly composite number: 10080 10080 = (2 × 2 × 2 × 2 × 2)  ×  (3 × 3) ×  5  ×  7 By (2) above, 10080 has exactly seventy-two divisors.
1 × 10080	2 × 5040	3 × 3360	4 × 2520	5 × 2016	6 × 1680
7 × 1440	8 × 1260	9 × 1120	10 × 1008	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:  Numbers in bold 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 $a_{n}$ 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 $2^{14}\times 3^{9}\times 5^{6}\times \cdots \times 1451$).

Plot of the number of divisors of integers from 1 to 1000. The first 15 highly composite numbers are in bold.

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 prime factors and exponents for the smaller number using the same prime factors and get a smaller number with at least as many divisors.

This property is useful for finding highly composite numbers.

Largely composite numbers

A positive integer n is a largely composite number if d(n) ≥ d(m) for all m ≤ n. The counting function Q_L(x) of largely composite numbers satisfies

$(\log x)^{c}\leq \log Q_{L}(x)\leq (\log x)^{d}\$

for positive c,d with $0.2\leq c\leq d\leq 0.5$.

See also

• Abundant number
• Highly totient number
• Superior highly composite number
• Table of divisors
• Euler's totient function

References

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• Paul Erdős: On Highly Composite Numers. Journal of the London Mathematical Society, 1944
• Paul Erdős, L. Alaolglu: On Highly Composite and Similar Numbers. Transaction of the Americal Mathematical Society, Vol. 56, No 3, November 1944, pp. 448–469
• Srinivasa Ramanujan, Jean-Louis Nicolas, Guy Robin: Highly Composite Numbers. The Ramanujan Journal I, 1997, pp. 119–153

External links

• Weisstein, Eric W., "Highly Composite Number", MathWorld.
• Algorithm for computing Highly Composite Numbers
• First 10000 Highly Composite Numbers
• Achim Flammenkamp, First 779674 HCN with sigma,tau,factors
• Online Highly Composite Numbers Calculator