[seqfan] Re: A025597 computation

Joseph S. Myers jsm at polyomino.org.uk
Sun Jun 24 22:42:24 CEST 2012

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.)

Joseph S. Myers
jsm at polyomino.org.uk

More information about the SeqFan mailing list