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

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 $$n = 72, 20664, 66600, 84744, 89784, 141240, 175032, 232680, 271272, 288072, 378984,...$$.

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 $$1,2,3,4,5,6$$.

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 get 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?
• I have to admit that I cannot really understand why lots of people find this question (and also a related question,"does there exits $n>1$ with ${\rm gnu}(n)=n$?") interesting. Is it just curiosity? Commented Jun 12 at 17:44
• @DerekHolt There are tons of boring questions and overmore heavily opinion-based , and then there occurs a really interesting question like this (namely whether the sequence of gnus contains $1,2,3,4,5,6$ and , if , when it occurs first) and you find this "boring" ? Seems that people have a very different opinion concerning "interesting" and "boring". The upvotes need not mean that the users consider this "interesting" , the excellent presentation alone deserves a high score. We need more of such questions ! Commented Jun 13 at 14:24
• The related question was posted by me but I was not the first. It is an old still open problem whether $gnu(n)=n$ has a nontrivial solution , but I did not know this. I independently arrived at this question which I consider both natural and fascinating. Commented Jun 13 at 14:30
• @Peter I did not use the word "boring", and I do not find the problem boring. I was hoping that someone would say why they find the question interesting. Commented Jun 13 at 16:07
• Similarly to many elementary questions in number theory (for example: are there infinitely many primes of the form $n^2+1$), it can be very difficult to say anything interesting about questions of this type. The question might become interesting if a solution is found.. Commented Jun 15 at 10:25