# Why isn't $1$ a superior highly composite number?

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 $2, 6, 12, 60, 120, 360, 2520, 5040, 55440, \dots$

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Edit: Bump. I am still looking for answers.

• @Henry: A composite number and a highly composite number are two completely different things... And both 1 and 2 are highly composite numbers. – Omega Force May 31 '14 at 23:14
• +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. – Fred Kline Jul 3 '14 at 0:17
• 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. – Daniel Fischer Jul 18 '14 at 11:52
• 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.) – Daniel Fischer Jul 18 '14 at 12:37
• Geez, this phrasing is horrible. The scope of the quantifier are so confusing... – Gina Jul 18 '14 at 13:33

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.