Quote:
Originally Posted by MiniGeek
I'll take k=349 and k=353 from 1M to 1.5M.
Edit: Files sent

Files received and work has begun. I don't really want to calculate an ETA, (might give me second thoughts about this
) but for the most part it will have the full attention of my quad, and I'll report on it periodically.
By a remarkable coincidence, there are exactly as many candidates for each k in the range I reserved! Each has 14561 candidates in n=1M1.5M. I originally (for n=600K+) picked them because they had similar weights, but this is still a very odd coincidence. (over 1M2M, 353 has 120 more candidates than 349, which is actually opposite from what you'd expect from their Nash weights: 349's 1463 vs 353's 1416)