consider the strings (b,b+1,b+2,...,b+l) of consecutive natural numbers, all less than some fixed natural number n > b+l. Is there a way to find the longest string (length of a string = l+1) with gcd(b+i,n) > 1 for all 0<=i<=l?
Arrange arithmetic sequences with distances given by the prime factors of $n$ so as to cover as many consecutive numbers as possible; for instance, if the prime factors are $2$, $3$, $5$ and $7$, this could be
I don't know whether there's a systematic way of doing this, but it seems that just using them from the smallest to largest whenever a new one is required to fill a gap might be optimal. Then by the Chinese remainder theorem, using the associated algorithm you can find the residue mod $n$ where this sequence is realized.