# Are all highly composite numbers even?

A highly composite number is a positive integer with more divisors than any smaller positive integer. Are all highly composite numbers even (excluding 1 of course)? I can't find anything about this question online, so I can only assume that they obviously are. But I cannot see why.

Yes. Given an odd number $$n$$, choose any prime factor $$p$$, and let $$k\geq 1$$ be the number such that $$p^k\mid n$$ but $$p^{k+1}\not\mid n$$. Then $$n\times\frac{2^k}{p^k}$$ has the same number of factors, and is smaller.

• The same idea extends to show the primes dividing a highly composite number must be the smallest primes and the exponents must decrease as the primes get larger. – Ross Millikan Dec 21 '18 at 14:57
• The associated OEIS sequence is A025487. – Charles Dec 21 '18 at 15:35
• Furthermore, from 6 on, they are all multiples of 3. From 12 on, they are all multiples of 4. From 60 on, they are all multiples of 5. I believe that for any given factor N, there is a point after which all numbers in the sequence are multiples of N. I don't have a proof for this, but it seems like it's probably the case. – Darrel Hoffman Dec 21 '18 at 15:35
• @DarrelHoffman I would love to have proofs of these. – Charles Dec 21 '18 at 15:36
• @Charles: isn't the relevant sequence A002182 (oeis.org/A002182)? – Michael Lugo Dec 21 '18 at 16:39