## Binary Circles

### Problem 265

2^{N} binary digits can be placed in a circle so that all the N-digit clockwise subsequences are distinct.

For N=3, two such circular arrangements are possible, ignoring rotations:

For the first arrangement, the 3-digit subsequences, in clockwise order, are:

000, 001, 010, 101, 011, 111, 110 and 100.

Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29:

00010111

_{2}= 2300011101

_{2}= 29Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.

Find S(5).