[seqfan] Observation on A050385

[David Wilson]

It looks as if A050385(n) counts the partitions of Z into n residue classes.
For example, we can partition Z into 4 residue classes in A050385(4) = 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)}

I have verified this programmatically for n = 1..10.
I leave the proof to someone more combinatorically inclined than myself.