
A highly composite number is a number, n, such that no number smaller than n has as many factors as n.
Example: The number 24 has 8 divisors. No smaller number has that many divisors, so we call 24 a highly composite number. All of the numbers in the first column of the table, below, are highly composite. By adding 1, we get squares of prime numbers. I checked to see if this happens again, and I was surprised to find another example. Can you find it?
24 25 5^{2} Source: http://www.mathpuzzle.com/ Graeme's thinking on this problem: Actually there are quite a few more highly composite numbers that aren't one less than a perfect square. But a lot of them are one less (or one more) than a prime. This table explains: Highly Composite Numbers

Highly composite numbers have sparked a number of interesting observations, some of which I will elaborate on here.
Bill McEachen pointed out to me that the offset from the HCN to the nearest prime (Sloane's A117825) is very often 1 i.e. adjacent to the HCN. All but 12 of the first 41 terms have a prime adjacent, and half of the 120 numbers listed here have a prime adjacent. Thanks to Bill for providing the data for the column of the table giving the offset to the nearest prime.
Reo Franklin Fortune, an anthropologist who was once married to Margaret Read, conjectured that all Fortunate numbers are prime. Fortunate numbers (OEIS A005235) are Pp_{k}#, where p_{k}# is the kth "primorial" number (i.e. product of the first k primes) and P is the smallest prime larger than p_{k}#+1. In other words, the distance from a primorial to the next larger prime (farther than one unit away) is a prime number. If H is an HCN, then the product of H's distinct prime factors, called the radical of H, or rad(H), is a primorial. That is, H is a multiple of a fairly large primorial, so it's no wonder that the distance from each HCN to the next larger prime also seems to be prime.
Here, P is used in an expression of the form Px to mean the next larger prime to x. Similarly, xp means the next prime smaller than x.
P(H+1), (no OEIS), 2, 3, 3, 5, 5, 5, 5, 5, 7, 7, 11, 11, 7, 7, 13, 17, 13, 11, 11, 13, 11, 11, 13, 19, 13, 17, 11, 17, 17, 19, 29, 13, 13, 47, 13, 17, 13, 23, 17, 19, 17, ...
(H1)p, (no OEIS), 2, 3, 5, 5, 5, 5, 7, 7, 7, 7, 7, 11, 11, 11, 11, 17, 17, 11, 11, 13, 11, 11, 19, 17, 13, 29, 23, 17, 17, 13, 17, 17, 13, 17, 13, 17, 19, 17, 23, ...
Pp_{k}#, (A005235), 3, 5, 7, 13, 23, 17, 19, 23, 37, 61, 67, 61, 71, 47, 107, 59, 61, 109, 89, 103, 79, 151, 197, 101, 103, 233, 223, 127, 223, 191, 163, 229, 643, ...
(p_{k}#1)p, (A055211), 3, 7, 11, 13, 17, 29, 23, 43, 41, 73, 59, 47, 89, 67, 73, 107, 89, 101, 127, 97, 83, 89, 97, 251, 131, 113, 151, 263, 251, 223, 179, 389, 281, ...
According to www.primepuzzles.net/problems/prob_004.htm, it was shown by Rosser & Shoenfeld in 1975 that ln(p_{k}#) < 1.001102p_{k}. (Rosser, J.B. & Schoenfeld, L. Sharper bounds for Chebyshev functions Teta(x) and PHI(x). Math .Comp., 29, 243269,1975)
Now, if there is a prime between x and x+ln(x)^{1.99}, (a conjecture a little finer than Schinzel's conjecture that there is always a prime between x and ln(x)^{2}), then there is a prime, P, between p_{k}# and p_{k}#+ln(p_{k}#)^{1.99}.
Combining these two ideas, we have P between p_{k}# and p_{k}#+(1.001102p_{k})^{1.99}.
Now, (1.001102p_{k})^{1.99} < p_{k}^{2} whenever p_{k} is at least 2, so there is a prime, P, between p_{k}# and p_{k}#+p_{k}^{2}.
Suppose the difference Pp_{k}# is composite. Then its smallest prime factor must be larger than p_{k}, or else P would also be a multiple of p_{k}, and it must have one other (not necessarily unique) prime factor larger than p_{k} (or else it wouldn't be composite), so this composite number must be larger than p_{k}^{2}, which would contradict the existence of a prime between p_{k}# and p_{k}#+p_{k}^{2}.
Now, I ask you: is this a proof? Absolutely not! First, it depends on Schinzel's conjecture, and then it goes further to depend on a finer conjecture which is really quite dubious. But it does serve as a kind of probabilistic argument that can help you understand why there are so few, if any, composite Fortunate numbers, and also so few, if any, nonprime differences to the prime nearest a Highly Composite Number.
Robert G. Wilson v commented (modified by Ray Chandler) that all powers of two through 2^{17} are present as a number of factors of an HCN, and no power of 2 above that is present at least through 2^{51}. I believe I will prove here that 2^{17} is the largest number that is both a power of two and the number of factors of an HCN  a property I'll call "special". In order for an HCN to be special, the exponents of all its prime factors must be one less than a power of 2. That is, the exponents in the prime factorization of such an HCN must be a sequence such as 3,1,0 or 7,3,3,1,0 or the like. This is because the number of factors is the product of one plus each exponent.
An effective way to show that a factorization can't be an HCN is to show that a different set of exponents gives a number smaller than the HCN but with as many or more factors. For example, no HCN's factorization can begin with equal exponents 2^{5} 3^{5}, or larger, because if it did, then 2^{8} 3^{3} would begin the factorization of a smaller number with exactly the same number of factors. My shorthand way of depicting this situation is:
5, 5  no good because 8, 3 gives smaller HCN with as many factors
Naturally, if 5, 5 is no good, then 6, 6 or 7, 7 is no good for the same reason  adding 3 to the first exponent and subtracting 2 from the second gives a smaller number with at least as many factors. So the shorthand implies that "higher" sets of exponents sharing the same relationship are no good for the same reason. Another shorthand I use from time to time is e_{p} to represent the exponent of the prime, p. For example, I'll refer to e_{2}, e_{3}, and the like.
Maybe it goes without saying, but just in case, I'll say it: the sequence of exponents in the prime factorization of an HCN must be nonincreasing. If, say 5, 4, appear as any two consecutive exponents then interchanging them makes a smaller number with the same number of factors.
It seems like there might be too many such sequences to nail them all down, but after plodding along for a while, I came to the conclusion there aren't really that many cases to consider, so here they are:
0  OK  gives a special HCN with 1 factor
1, 0  OK  gives a special HCN with 2 factors
1, 1, 0  OK  gives a special HCN with 4 factors
1, 1, 1  no good because 3, 1, 0 gives smaller HCN with as many factors
(e_{2}=e_{3}=e_{5} is no good when they're all 1 or bigger.)
3, 1, 0  OK  gives a special HCN with 8 factors
3, 1, 1...  OK  gives special HCNs with 16 and 32 factors
3, 1, 1, 1, 1  no good, because 3, 3, 1, 1 gives smaller HCN with as many factors
3, 3, 1...  OK  gives special HCNs with 64, 128, and 256 factors
3, 3, 1, 1, 1, 1, 1  no good because 7, 3, 1, 1, 1, 1 gives smaller HCN with as many factors
3, 3, 3  no good because 5, 3, 2 gives smaller HCN with as many factors
(see 1,1,1)
7, 1  no good because 5, 2 gives smaller HCN with more factors
7, 3, 1...  OK  gives special HCNs with 512, 1024, 2048, and 4096 factors
7, 3, and eight 1's  no good because that eighth 1 is the exponent of 29, which exceeds 25,
so 7, 3, 3, and six 1's gives smaller HCN with the same number of factors
7, 3, 3, 1...  OK  gives special HCNs with 8192, 16384, 32768, 65536, and 131072 factors
7, 3, 3, and eleven 1's  no good because that eleventh 1 is the exponent of 43, which exceeds 42,
so 8, 4, 3, 2 and nine 1's is smaller by a factor of 42/43, and has more factors
7, 3, 3, 3  no good because 8, 4, 3, 2 gives smaller HCN with more factors
7, 7  no good because 10, 5 gives smaller HCN with more factors
(e_{2} can't equal e_{3} unless they're smaller than 5)
15, 1  no good because 13, 2 gives smaller HCN with more factors
15, 3  no good because 13, 5 gives smaller HCN with more factors
15, 7, 1  no good because 10, 7, 3 gives smaller HCN with more factors
15, 7, 3  no good because 10, 7, 5 gives smaller HCN with more factors
(e_{2}+1 must be less than four times e_{5}+1 when e_{5} is at least 3)
15, 7, 7  no good because 14, 9, 6 gives smaller HCN with more factors
(e_{3} can't equal e_{5} unless they're smaller than 5 or e_{2} exceeds e_{3} by less than 4)
15, 15  no good (see 7, 7)
a, b, where a is any of {31, 63, 127, ...} and b is any number from 1 to a/2 
no good because a5, b+3 gives smaller HCN with more factors
a, a, where a is any of {31, 63, 127, ...}  no good (see 7, 7)
Whew! After reviewing 25 cases, all the sequences of exponents in the prime factorization of any "special" HCN have now been put to bed, and there are no others that might make an HCN "special". (By special, in case you have forgotten by now, I mean an HCN that has a power of 2 factors).
As you are no doubt aware, having read each of the 25 cases in the last proof, an HCN is "happy" when its exponents e_{2}, e_{3}, e_{5}, etc. form certain ratios. The exponents can't be too close to each other, or too far away, either. If e_{2}=e_{3}, for example, and they are bigger than a certain size, then the HCN isn't happy, and it's not an HCN at all. If e_{2} is more than twice e_{3}, and bigger than a certain size, then it's not happy, either.
What are these happy ratios?
Let H be a Highly Composite Number. Let's imagine for the moment that the exponents in the prime factorization are real numbers, and let's further restrict our scope to just the primes 2 and 3. Let's suppose the exponent of 2 is x1 and the exponent of 3 is y1. Together, these two prime numbers "contribute" xy factors of H. Now, let's further suppose that the contribution of these two primes is a constant, a, and that xy=a. This means that y=a/x, so we have
2^{x1} 3^{a/x}
in the prime factorization of H. H, we remember is a minimal number with this many factors, so we need to minimize H. 2^{x1} 3^{a/x1} is minimized when the derivative of ln(2)(x1)+ln(3)(a/x1) is minimized. The derivative of
ln(2)(x1)+ln(3)(a/x1)
is
ln(2)a ln(3)/x^{2},
which is zero when x=sqrt(a ln(3)/ln(2)), and y=a/x=sqrt(a ln(2)/ln(3)), so the ratio of x:y is ln(3):ln(2). In round numbers, for a given number of factors of H, H is minimized when the ratio of one plus the exponents of 2 and 3 is close to ln(3):ln(2), or about 1.585.
I'll call this the "exponent ratio" (e_{2}+1)/(e_{3}+1)  the ratio of one plus the exponent of 2 to one plus the exponent of 3. Looking at Flammenkamp's table of HCNs, we see this exponent ratio fluctuate from 1 to more than 2, but the geometric mean of this ratio over all 1200 HCNs is 1.58946, which is very close to the expected value of 1.585. However, it shows no signs settling down to that value very quickly. In fact, three of the last ten HCNs listed have exponent ratios of 2 or larger. I would still expect to find an "N" such that for n larger than N, the exponent ratio lies properly between 1 and 2.
This concept generalizes to other primes as well. We even see empirical evidence the generalization is right. The geometric mean of the ratio (e_{3}+1)/(e_{5}+1) in Flammenkamp's data is 1.47278, very close to the expected ln(5)/ln(3) = 1.465.
A. Flammenkamp, Table of 1200 Highly Composite Numbers
The Sigma Function  σ(n) is the sum of the factors of n, and σ_{k}(n) is the sum of the n's factors, each to the power k. So, in particular, σ_{0}(n), sometimes called τ(n), tau(n), is the number of factors of n.
Factoring Polynomials, a topic where factoring whole numbers comes up as one step of a procedure.
The Totient Function  the number of coprimes of a number
The webmaster and author of this Math Help site is Graeme McRae.