A highly composite number is an integer greater than one which has more divisors than any positive integer below it. The first twenty highly composite numbers are

2, 4, 6, 12, 24, 36, 48, 60, 120, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 7560 and 10080

with 2, 3, 4, 6, 8, 9, 10, 12, 16, 18, 20, 24, 30, 32, 36, 40, 48, 60, 64 and 72 positive divisors respectively.

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 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, the necessary conditions for a number to be a highly composite number are that it has prime factors that are as small as possible, but not too many of the same: e.g. 2×3×3=18 can not be one because 2×2×3=12 has the same number of divisors while being smaller; similarly 2×5=10 cannot be one for the same reason, comparing with 2×3=6; however, with many of the same small prime factors, the number of divisors is relatively small, e.g. 2×2×2=8 is not a highly composite number, it has the same number of divisors as the smaller number 2×3=6.

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.

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 which are less than or equal to x, then there exist two constants a and b, both bigger than 1, so that

(lnx)aQ(x) ≤ (lnx)b.
Can we get approximations of a and b, and/or a proof of this fact?

See also

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