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In number theory, a positive integer $n$ is called highly composite if it has more divisors than any smaller positive integer. This notion has been studied by several notable mathematicians; for instance, Ramanujan and Erdos.

We can define a group-theoretical analogue of this. Let's say a positive integer $n$ is groupy if there are more groups of order $n$ (up to isomorphism) than any smaller positive integer. Denote the $k$th groupy number by $g(k)$.

The first few groupy numbers are $1, 4, 8, 16, 24, 32, 48, 64, 128, 256, 512, 1024, 2048$ and indeed we have an OEIS sequence, namely A046059. Many powers of $2$. We have the natural conjecture that for $k \geq 8$, we have $g(k) = 2^{k-2}$. However, we do have "outliers" $24$ and $48$, and it's possible that there might be infinitely many such outliers. A weaker conjecture is that the set of indices $k$ for which $g(k)$ is not a power of $2$ has zero asymptotic density.

Has this sequence been studied in any depth before?

MathematicsStudent1122
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    Between the idea that "most" groups are nilpotent and the Higman-Sims estimate for the number of groups of order $p^n$, I'd be surprised if $g(k)$ isn't just $2^{k-2}$ for sufficiently large $k$ (though not necessarily $k\geq 8$). – anomaly Dec 28 '18 at 21:20
  • @anomaly [nitpick] I think you mean to say it would be $g(k) = 2^{k-M}$ for sufficiently large $k$ and some fixed $M$. If there are more outliers (beyond $k=8$), then we can't be sure it'll be $2^{k-2}$ anymore. – MathematicsStudent1122 Dec 28 '18 at 21:26
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    It's a folklore conjecture that the vast majority of groups are nilpotent of order a power of $2$; if this is the case then there aren't any more outliers. – Qiaochu Yuan Dec 28 '18 at 21:37
  • Related: https://math.stackexchange.com/q/241369/407165 – Chain Markov Apr 11 '19 at 16:25

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