# [seqfan] Re: A025597 computation

David Wilson davidwwilson at comcast.net
Mon Jun 25 03:43:08 CEST 2012

```I computed the original sequence by effective repeated application of
the adjacency matrix. I would have guessed that the eigenvalues were so
many and unwieldy that a formula would be huge and ugly. What surprised
me is that it could be expressed so compactly.

On 6/24/2012 4:42 PM, Joseph S. Myers wrote:
> On Sun, 24 Jun 2012, David Wilson wrote:
>
>> I have been sent the following ostensible formula for A025597.
>> *
>> f(n) = (4/81) * sum(j=1...8, k=1...8) [(-1)^(j+k)] *
>> [sin(j*pi/9)*sin(k*pi/9)]^2 * [(1+2cos(j*pi/9))*(1+2cos(k*pi/9))-1]^n
> The [(1+2cos(j*pi/9))*(1+2cos(k*pi/9))-1]^n part certainly seems
> plausible.  (The following is a sketch of why it's plausible, not a full
> proof of anything.)
>
> Those numbers are going to be the eigenvalues of the process that at each
> step replaces the number on each square by the sum of the numbers on the
> up to eight surrounding squares - multiplication each time of a length-64
> vector by the same 64x64 matrix.  The boundary conditions of the 8x8 board
> make this not especially nice to analyze, but you can change them to
> periodic boundary conditions with an 18x18 period (every ninth row and
> column has 0 kings on all squares; if say the original board has
> coordinates from 1 to 8, then a king on square (x, y) is paired with one
> on (-x, -y) and negative kings on (-x, y) and (x, -y); each step the
> process then preserves all the symmetries including the zero rows and
> columns).  This process clearly has eigenvector solutions with
> e^(i(jx+ky)pi/9) kings on square (x, y), easily leading to the above
> eigenvalues.  (j and k should be thought of mod 18 here.  Negating j or k
> does not change the eigenvalue; you have four eigenvectors you need to
> combine appropriately to get a real eigenvector with the required
> symmetries and rows and columns of zeroes.  j or k = 0 or 9 presumably do
> not lead to any real eigenvectors with the required properties.)
>

```