[seqfan] Re: nxn grids colored black and white

[Benoît Jubin]

To complete my previous message:
a(2n+1) = (2^((2*n+1)^2) + 2*2^(1+n*(n+1)) + 2*2^((n+1)*(2*n+1)) +
2^(n*(2*n+2)+1) + 2*2^((2*n+1)*(n+1)))/16

On Sun, Mar 25, 2012 at 10:21 PM, Benoît Jubin <benoit.jubin at gmail.com> wrote:
> 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
>>>>
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