At 10:14 PM 6/10/03, David Wilson wrote:
M is the 6x6 matrix
1 0 0 1 0 0 0 0 0 0 1/3 0 0 1 0 0 1 0 0 0 0 0 1/3 0 0 0 1 0 0 1 0 0 0 0 1/3 0
and w = M v, then w_k should give a reasonable approximation to the number of integers == k (mod 6) in c^-1(S). This means that for large n, we might expect A005186(n+1)/A005186(n) to be near the large eigenvalue of M.
By brute force, I found the large eigenvalue of M is about 1.26376261582, close to the reported 1.264. I believe that this is in truth (3+sqrt(21)/6 (thanks to Plouffe's inverter), but I don't have the matrix weaponry to prove it.
The characteristic polynomial of M is x^6 - x^5 - 4/3*x^4 + x^3 + 1/3*x^2 with roots 0 (twice), +-1, (3 +- sqrt(21))/6, so you're right about the largest eigenvalue. -- Fred W. Helenius <fredh@ix.netcom.com>