From David W. Wilson, May 17 2017: (Start)
Appears to equal the number of ways to partition Z into n residue classes. For example, a(4) = 10 since we can partition Z into 4 residue classes in 10 ways:
Z = ∪ {0(mod 2), 1(mod 4), 3(mod 8), 7(mod 8)}
Z = ∪ {0(mod 2), 3(mod 4), 1(mod 8), 5(mod 8)}
Z = ∪ {0(mod 2), 1(mod 6), 3(mod 6), 5(mod 6)}
Z = ∪ {1(mod 2), 0(mod 4), 2(mod 8), 6(mod 8)}
Z = ∪ {1(mod 2), 2(mod 4), 0(mod 8), 4(mod 8)}
Z = ∪ {1(mod 2), 0(mod 6), 2(mod 6), 4(mod 6)}
Z = ∪ {0(mod 3), 1(mod 3), 2(mod 6), 5(mod 6)}
Z = ∪ {0(mod 3), 2(mod 3), 1(mod 6), 4(mod 6)}
Z = ∪ {1(mod 3), 2(mod 3), 0(mod 6), 3(mod 6)}
Z = ∪ {0(mod 4), 1(mod 4), 2(mod 4), 3(mod 4)}
(End)
Unfortunately this conjecture is not correct; it fails at a(13).  Jeffrey Shallit, Nov 19 2017
The correct statement is that a(n) is the number of "natural exact covering systems" of cardinality n. These are covering systems (like the ones in David W. Wilson's comment) that are obtained by starting with the size1 system x == 0 (mod 1) and successively choosing a congruence and "splitting" it into r >= 2 new congruences.  Jeffrey Shallit, Dec 07 2017
