A highly composite number (HCN) is a positive integer with more divisors than any smaller positive integer. The initial or smallest twenty-one highly composite numbers are listed in the table at right. 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 \( c_i \) 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 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 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.
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). [2] 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. Abundant number References ^ Srinivasan, A. K. (1948), "Practical numbers", Current Science 17: 179–180, MR0027799. Retrieved from "http://en.wikipedia.org/"
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