[seqfan] Re: nxn grids colored black and white
Benoît Jubin
benoit.jubin at gmail.com
Mon Mar 26 07:21:07 CEST 2012
Using Burnside's lemma, I find
a(2n) = (4*2^(2n^2) + 4 * 2^(n^2) + 2* 2^(n*(2n+1)) + 2^(4n^2) ) / 16
The group acting on the grids is D_4 x C_2 and the terms above are the
cardinals of the fixators: for instance, to determine a grid fixed by
a quarter-rotation-and-flip, it is enough to give a coloring in a
quarter of the grid, hence the 2^(n^2).
a(2n+1) can be computed similarly
Benoît
On Sun, Mar 25, 2012 at 9:52 PM, Graeme McRae <g_m at mcraefamily.com> wrote:
> D'oh! I found a bug in my program, sorry. The sequence should be 1, 4, 51, 4324, 2105872, ...
>
> I'll stop posting on the list now, and begin emailing Isaac directly.
>
> --Graeme McRae,
> Palmdale, CA
>
> On Mar 25, 2012, at 8:25 PM, Graeme McRae <g_m at mcraefamily.com> wrote:
>
>> I agree that the first 3 terms are 1, 4, 51, and I calculate that the fourth is 5907.
>>
>> --Graeme McRae,
>> Palmdale, CA
>>
>> On Mar 25, 2012, at 5:57 PM, Isaac <isaacthegreat at gmail.com> wrote:
>>
>>> Neil-
>>>
>>> Thanks a lot! Yes, I meant 51. I would like to calculate the next term but
>>> ~2^16 4x4 grids would be way too much to check by hand. I have been sitting
>>> on this for a while, and since it is something I've looked at in my spare
>>> time I will definitely try to when I have the time.
>>>
>>> -Isaac
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan
mailing list