Erdős Problem #687
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
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`Y x` is \(Y(x)\), the maximal \(y\) such that there is a choice of congruence classes \(a_p\), one for each prime \(p \leq x\), such that every integer in \([1, y]\) is congruent to at least one of the \(a_p \pmod{p}\). The covering condition is `Green46.IsCoveredByResidues x y`. We take \(x\) to be a natural number, since \(Y(x)\) only depends on \(\lfloor x \rfloor\). The set of such \(y\) contains \(0\) and is bounded by \(\prod_{p \leq x} p\) (`Erdos687.lt_primorial_of_isCoveredByResidues`), so `sSup` is its maxim
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Independently certified \(Y(x)=h(\pi(x))-1\) and the exact values through \(x<43\) (in particular \(Y=73\) on \(41\le x<43\)); a fixed positive interval-sieve survivor bound (5.3) would prove \(Y=o(x^
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