2
$\begingroup$

A superior highly composite number is a positive integer $n$ for which there is an $\epsilon>0$ such that $\dfrac{d(n)}{n^\epsilon} \geq \dfrac{d(k)}{k^\epsilon}$ for all $k>1$, where the function $d(n)$ counts the divisors of $n$.

The first superior highly composite numbers are $2, 6, 12, 60, 120, 360, 2520, 5040, 55440, \dots$

.

Edit: Bump. I am still looking for answers.

$\endgroup$
  • 1
    $\begingroup$ @Henry: A composite number and a highly composite number are two completely different things... And both 1 and 2 are highly composite numbers. $\endgroup$ – Omega Force May 31 '14 at 23:14
  • 1
    $\begingroup$ +1 for interesting question. The only pattern difference I see is the parity (1 is odd, all others are even). However, I can't say that that is the reason. $\endgroup$ – Fred Kline Jul 3 '14 at 0:17
  • 1
    $\begingroup$ My best guess is that the definition actually says "there is an $\varepsilon \in (0,1)$" (and wikipedia lost the $< 1$ restriction), and then $1$ doesn't fit the bill. Or that $1$ is explicitly excluded, and wikipedia forgot that. $\endgroup$ – Daniel Fischer Jul 18 '14 at 11:52
  • 1
    $\begingroup$ Well, it wouldn't be the first time that I guess wrong. Problem is, however, that for $\varepsilon \geqslant 1$, the condition is fulfilled for $n = 1$, but all the pages agree that the first superior holy confessionless nutrient is $2$, so something would have to exclude $1$. (Yes, that is your question, I know.) $\endgroup$ – Daniel Fischer Jul 18 '14 at 12:37
  • 1
    $\begingroup$ Geez, this phrasing is horrible. The scope of the quantifier are so confusing... $\endgroup$ – Gina Jul 18 '14 at 13:33
0
$\begingroup$

The reason may be found in Ramanujan's original paper. He gives the following theorem:

If $N$ is a highly composite number, and $N = 2^{a_2} \times \dots \times p_1^{a_{p_1}}$ be its prime factorization, then $a_{p_1}=1$ except for $N=4$ and $N=36$.

This theorem fails if $N=1$ is considered to be highly composite (however theorem may not make sense if $N=1$). His definition of highly composite number is verbatim:

A number $N$ may be said to be a highly composite number, if $d(N') < d(N)$ for all values of $N’$ less than $N$.

With this definition $1$ should be highly composite, but $1$ does not appear in his list of highly composite numbers. So I suppose the author is implicitly considering numbers with at least one prime divisor (i.e. for which $p_1$ make sense in all his proofs). Same reason for why $1$ is not superior highly composite.

$\endgroup$
  • $\begingroup$ A version of the paper can be accessed without the paywall at ramanujan.sirinudi.org/Volumes/published/ram15.html (there appear to be some minor typos, however, in that "less than" is typeset twice as "less that" where "superior highly composite number" is defined). $\endgroup$ – Barry Cipra Jul 18 '14 at 15:01

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.