Turning the necklace over doesn't count as a different necklace.
For k beads chosen from n distinct ones, we can have n!/(n-k)! possible permutations, then eliminate the cyclic permutations and reversals (2k). The case k=2 is special in that a cyclic permutation is equivalent to a reversal.
a(n) = 1 + n + n(n-1)/2 + sum_{3<=k<=n} n!/(2k(n-k)!).
EXAMPLE
For example for n=4 we have {}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}, {1,2,4,3}, {1,3,2,4}.
