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

Let $n$ be an arbitrary HCN. It can easily be proven that the next HCN is at most $2n$ (which has more divisors than $n$). If there do not exist any other HCNs between $n$ and $2n$, then I will call $n$ a peak composite number.

Based on this list of the first 1200 HCNs, the first 7 peak composite numbers are 1, 2, 6, 12, 60, 360, and 2520. But there are no others less than $10^{88}$. This seems to suggest that 2520 is the largest peak composite number, but there may be a very large counterexample.

How can the (non)existence of such a counterexample be proven?

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Dan
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1 Answers1

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Let $n=2^b3^cm$ be an integer divisible by $6$ (where $\gcd(m,6)=1$). If $b>c+1$, then $3n/2$ is an integer with more divisors than $n$; if $b<2c-2$, then $4n/3$ is an integer with more divisors than $n$.

In particular, if $n$ is divisible by $2\cdot 3^4$, then there is always an integer strictly less than $2n$ that has more divisors than $n$ (since when $c\ge4$ it is always the case that either $b>c+1$ or $b<2c-2$).

Therefore the nonexistence of other counterexamples follows if all HCNs larger than the last one on your list are divisible by $2\cdot 3^4$. This is probably easy to show.

Greg Martin
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