13
$\begingroup$

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.

$\endgroup$
1
25
$\begingroup$

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.

$\endgroup$
8
  • 7
    $\begingroup$ 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. $\endgroup$ – Ross Millikan Dec 21 '18 at 14:57
  • $\begingroup$ The associated OEIS sequence is A025487. $\endgroup$ – Charles Dec 21 '18 at 15:35
  • 3
    $\begingroup$ 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. $\endgroup$ – Darrel Hoffman Dec 21 '18 at 15:35
  • $\begingroup$ @DarrelHoffman I would love to have proofs of these. $\endgroup$ – Charles Dec 21 '18 at 15:36
  • 2
    $\begingroup$ @Charles: isn't the relevant sequence A002182 (oeis.org/A002182)? $\endgroup$ – Michael Lugo Dec 21 '18 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.