# Have “groupy” numbers been studied before?

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?

• 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
• 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
• – Yanior Weg Apr 11 at 16:25