Does the sequence $1, 2, 3, 4, 5, 6$ appear in the number of groups of order $n$ up to isomorphism?

Question

Let $\mathrm{gnu}(n)$ denote the number of groups of order $n$ up to isomorphism.

$\mathrm{gnu}(1), \mathrm{gnu}(2), \mathrm{gnu}(3), \dots$ is now a sequence of integers, and we may ask if and where the subsequence $1, 2, 3, \dots, k$ appears successively, for some integer $k$.

Just looking through the first few terms, it is not hard to find an instance of $1, 2, 3, 4$, namely $\mathrm{gnu}(73), \mathrm{gnu}(74), \mathrm{gnu}(75), \mathrm{gnu}(76)$. But $\mathrm{gnu}(77)=1$, so this is not a $1,2,3,4,5$.

To find $1,2,3,4,5$ is significantly harder. There are necessary and sufficient conditions on $n$ which determine whether $\mathrm{gnu}(n)=1,2,3,4$, listed for example in this paper on page 6. I implemented these to find the integers $n$ such that $\mathrm{gnu}(n+i)=i$ for $1 \leq i \leq 4$. I've written a computer program to find these - the sequence begins $n = 72, 20664, 66600, 84744, 89784, 141240, 175032, 232680, 271272, 288072, 378984,...$.

(Edit: I added this to the OEIS. See A373650.)

Note also that each term in this sequence is a multiple of $24$ - which helps us find these terms more easily. See here for details.

Hence our candidates for where we want to find a $5$ are at $77, 20669, 66605, 84749, 89789, 141245, 175037, 232685, 271277, 288077, 378989,...$.

I've tested these manually and, hoping that my calculations and program are correct, the earliest instance of a $1,2,3,4,5$ occurs at $\mathrm{gnu}(2814121), \mathrm{gnu}(2814122), \mathrm{gnu}(2814123), \mathrm{gnu}(2814124), \mathrm{gnu}(2814125)$. But $\mathrm{gnu}(2814126)=24$, so this is not a $1,2,3,4,5,6$.

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The obvious next challenge is to find a $1,2,3,4,5,6$. That's a lot harder as:

• I've struggled to find a necessary and sufficient condition for $\mathrm{gnu}(n)=5$ anywhere online (and am not sure on how to begin deriving it myself!)
• As $n$ gets large, the tests for whether $\mathrm{gnu}(n)=1,2,3,4$ tend to take longer, since they're based on the prime factorization of $n$

So, I'm wondering if anyone has any observations that'd make finding a $1,2,3,4,5,6$ easier:

• Is there a necessary and sufficient condition for $\mathrm{gnu}(n)=5$?
• Is there some smarter method for how we may search for this?

Answer 1

Found one!

$$\begin{aligned} \mathrm{gnu}(29436121) &= 1 \\ \mathrm{gnu}(29436122) &= 2 \\ \mathrm{gnu}(29436123) &= 3 \\ \mathrm{gnu}(29436124) &= 4 \\ \mathrm{gnu}(29436125) &= 5 \\ \mathrm{gnu}(29436126) &= 6 \end{aligned}$$

But $\mathrm{gnu}(29436127) = 1$, so this is not a $1,2,3,4,5,6,7$.

Magma is capable of calculating these orders, if you want to verify my answer, go here and enter

NumberOfSmallGroups(29436121);
NumberOfSmallGroups(29436122);
NumberOfSmallGroups(29436123);
NumberOfSmallGroups(29436124);
NumberOfSmallGroups(29436125);
NumberOfSmallGroups(29436126);

If all of my calculations are correct, then this is the first $1,2,3,4,5,6$.

Answer 2

Edit: Since writing this, I have found a $1,2,3,4,5,6$. See this other answer.

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This isn't an answer as to whether a $1,2,3,4,5,6$ occurs or not, but in my question I wrote:

$\mathrm{gnu}(1),\mathrm{gnu}(2),\mathrm{gnu}(3),\dots$ is now a sequence of integers, and we may ask if and where the subsequence $1,2,3,\dots,k$ appears successively, for some integer $k$.

I just realised that there's a fairly simple argument to show we can't get arbitrarily large $k$ which is probably worth noting.

Observe that $\mathrm{gnu}(32) = 51$. Thus, for all $n$, $\mathrm{gnu}(32n) \geq 51$.

But then, for any $N$, $\mathrm{gnu}(N+1), \dots, \mathrm{gnu}(N+32)$ can't be a $1,2,\dots,32$, as one of $N+1, \dots, N+32$ is a multiple of $32$ and hence has at least $51$ groups.

$32$ is the smallest $k$ such that $\mathrm{gnu}(k) > k$ hence why I picked it for this argument. It's possible we can lower this upper bound for $k$ further, though.

Another edit: Actually, we can lower this to show a $1,2,3,4,5,6,7,8$ must never appear. If $n$ has $\mathrm{gnu}(n+i) = i$ for $i=1,2,3,4$, it is possible to show $n=48m+24$ for some $m$. See here for some partial discussion on that. It follows that $16 \mid n+8$. One can then show that $\mathrm{gnu}(16l) \geq 14$ for any $l$. Hence $\mathrm{gnu}(n+8) \neq 8$. I’ve not found a $1,2,3,4,5,6,7$ yet - it’s possible $1,2,3,4,5,6$ is the longest we can get.

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As for finding a $1,2,3,4,5,6$, I've started running another brute force search. I've only found one other $1,2,3,4,5$:

$$\begin{aligned} \mathrm{gnu}(22411273) &= 1 \\ \mathrm{gnu}(22411274) &= 2 \\ \mathrm{gnu}(22411275) &= 3 \\ \mathrm{gnu}(22411276) &= 4 \\ \mathrm{gnu}(22411277) &= 5 \\ \end{aligned}$$

But $\mathrm{gnu}(22411278) \geq 16$ because e.g. $22411278 = 2 \cdot 3^2 \cdot 61 \cdot 20411 = 1098 \cdot 20411$, $\mathrm{gnu}(1098) = 16$ (https://oeis.org/A000001) so this is not a $1,2,3,4,5,6$.

A third $1,2,3,4,5$ does not occur below $2.9 \times 10^7$.

Edit: I just found the third $1,2,3,4,5$ which happens to also be a $1,2,3,4,5,6$. See my other answer.

Answer 3

Here's another one.

$$\begin{aligned} \mathrm{gnu}(117873721) &= 1 \\ \mathrm{gnu}(117873722) &= 2 \\ \mathrm{gnu}(117873723) &= 3 \\ \mathrm{gnu}(117873724) &= 4 \\ \mathrm{gnu}(117873725) &= 5 \\ \mathrm{gnu}(117873726) &= 6 \end{aligned}$$

But $\mathrm{gnu}(117873727) = 1$, so this is not a $1,2,3,4,5,6,7$.

Magma is capable of calculating these orders except $117873723$, if you want to verify most of my answer, go here and enter

NumberOfSmallGroups(117873721);
NumberOfSmallGroups(117873722);
NumberOfSmallGroups(117873723);
NumberOfSmallGroups(117873724);
NumberOfSmallGroups(117873725);
NumberOfSmallGroups(117873726);

and you'll get 1 2 0 4 5 6.

However, as I said, this paper has conditions for precisely when $\mathrm{gnu}(n)=3$, and you can verify $117873723$ passes them.

If all of my calculations are correct, then this is the second $1,2,3,4,5,6$.