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.
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3$\begingroup$ The exponents of the prime factors are nonincreasing and the last one is $1$. $\endgroup$ – Shaun Dec 21 '18 at 14:56
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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.
answered Dec 21 '18 at 14:52
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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
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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
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$\begingroup$ @DarrelHoffman I would love to have proofs of these. $\endgroup$ – Charles Dec 21 '18 at 15:36
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2$\begingroup$ @Charles: isn't the relevant sequence A002182 (oeis.org/A002182)? $\endgroup$ – Michael Lugo Dec 21 '18 at 16:39
