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(Here, $\mathrm{gnu}(n)$ denotes the number of groups of order $n$ up to isomorphism.)

I wonder whether the number of groups of order $2304=2^8\times 3^2$ is known. GAP exited because of the memory. $\mathrm{gnu}(2304)$ must be greater than $1,000,000$ because of $\mathrm{gnu}(768)=1,090,235$ and $768=2^8\times 3|2^8\times 3^2=2304$.

Is $\mathrm{gnu}(2304)$ known or at least a tight upper bound?

What is the smallest number $n$, such that it is infeasible to calculate $\mathrm{gnu}(n)\ ?$ I think, $\mathrm{gnu}(2048)$ will be known in at most ten years, probably much earlier.

Could $n=3072=2^{10}\times 3$ be the smallest too difficult case ?

Robin
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Peter
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1 Answers1

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There are indeed $112\,184+1\,953+15\,641 993 = 15\,756\,130$ groups of order 2304, computed using an algorithm developed by Bettina Eick and myself. As Olexandr Konovalov already kindly pointed out, you can find this number in our paper "The construction of finite solvable groups revisited", J. Algebra 408 (2014), 166–182, also available on the arXiv.

This is part of an on-going project to catalogue all groups up to order 10,000 (with a few orders excepted, e.g. multiples of 1024, as there are simply to many of these). So in particular, we skip groups of order 3072. There are already $49\,487\,367\,289$ groups of order 1024, and I expect the number of groups of order 3072 to be several orders of magnitude larger.

To maybe slightly motivate why I think so, consider this the proportion of number of (isomorphism clases of) groups of order $2^n$ vs $3\cdot 2^n$, computed here using GAP:

gap> List([0..9], n -> NrSmallGroups(3*2^n)/NrSmallGroups(2^n)*1.0);
[ 1., 2., 2.5, 3., 3.71429, 4.52941, 5.77903, 8.66366, 19.4366, 38.9397 ]

If you plot $n$ against $gnu(3\cdot 2^n)/gnu(2^n)$, you'll see a roughly exponentially looking curve. Of course that is a purely empiric argument, not a proof of anything.

Olexandr Konovalov
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Max Horn
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  • why is the number you've stated (15755894) different from 15756130 given in http://arxiv.org/abs/1306.4239 ? – Olexandr Konovalov Jan 04 '16 at 11:28
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    Oops, because I copied the answer from my own slides (the ones you gave a link to already, in fact), but these contain a mistake in the computation of the sum (i.e. very, very basic and stupid). I should have looked at the paper, were we of course double checked evertything. Thank you very much for pointing this out to me. – Max Horn Jan 04 '16 at 11:33