# [seqfan] Re: nxn grids colored black and white

Benoît Jubin benoit.jubin at gmail.com
Mon Mar 26 08:02:44 CEST 2012

```You're right Graeme, and it's actually the formula I had found: to summarize,
a(2n) = (6*2^(2*n^2) + 4*2^(n^2) + 2*2^(n*(2*n+1)) + 2^(4*n^2)) / 16
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

Benoît

On Sun, Mar 25, 2012 at 10:53 PM, g_m at mcraefamily.com
<g_m at mcraefamily.com> wrote:
> Your formula for odd n gives a(5)=2105872, which is the same number I got
> by just counting them all.  But your formula for even n gives a(4)=4292,
> while I got 4324 by brute force.  I think your even formula should be
> a(2n) = (6*2^(2*n^2) + 4*2^(n^2) + 2*2^(n*(2*n+1)) + 2^(4*n^2)) / 16
>
>
> Original Message:
> -----------------
> From: Benoît Jubin benoit.jubin at gmail.com
> Date: Sun, 25 Mar 2012 22:37:05 -0700
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Re: nxn grids colored black and white
>
>
> 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
>
>
> --------------------------------------------------------------------
> mail2web.com - Microsoft® Exchange solutions from a leading provider -