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Given a positive integer $n$, let $N(n)$ denote the number of groups of order $n$, up to isomorphism.

Question: Does $N(n)=n$ hold for some $n>1$?

I checked the OEIS-sequence https://oeis.org/A000001 as well as the squarefree numbers in the range $[2,10^6]$ and found no example. Since we have many $n$ with $N(n)<n$ and some $n$ with $N(n) \gg n$, I see no reason why $N(n)=n$ should be impossible for $n>1$.

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  • $\begingroup$ I asked this question also on math-stack-exchange. $\endgroup$ – Peter Nov 26 '15 at 23:35
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    $\begingroup$ math.stackexchange.com/questions/1547948/… $\endgroup$ – Peter Nov 26 '15 at 23:45
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    $\begingroup$ There are no square-free examples, because $N(n) \leq \phi(n)$, for square-free $n > 1$. $\endgroup$ – James Nov 27 '15 at 0:33
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    $\begingroup$ Note, though, that $N(2^k)$ grows much faster than $2^k$ (something like $2^{2k^3/27}$ if I remember right). Likewise for $N(p^k)$ for any fixed prime $p$ (when $N(p^k)$ grows like $p^{2k^3/27}$). $\endgroup$ – Noam D. Elkies Nov 27 '15 at 2:08
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    $\begingroup$ Also: wow, I didn't know (or didn't remember) that this was the very first OEIS sequence; what I expected would be the first is either #12 or #27. (Question edited only to fix spelling and change ">>" to "\gg" in the TeX.) $\endgroup$ – Noam D. Elkies Nov 27 '15 at 2:18
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A "near-miss" is $N(19328) = 19324$, while the only $n \leq 2000$ such that $|N(n)-n| \leq 25$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, $18$, $19$, $20$, $21$, $22$, $23$, $24$, $25$, $26$, $27$, $28$, $32$, $36$, $48$, and $72$.

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