A289677 a(n) = A289671(n)/2^f(n), where f(n) = 2*floor((n-1)/3) + ((n+2) mod 3) = A004523(n).
0, 1, 1, 2, 2, 3, 4, 4, 5, 11, 12, 13, 22, 19, 20, 43, 46, 44, 85, 88, 89, 171, 185, 192, 366, 380, 396, 774, 793, 814, 1586, 1589, 1610, 3136, 3106, 3130, 6123, 6078, 6103, 12088, 12147, 12229, 24283, 24534, 24736, 49040, 49482, 49879, 99031, 99792, 100747, 200444, 201892, 203765, 405931, 408478, 411403
Offset: 1
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Python
from _future_ import division def A289677(n): c, k, r, n2, cs, ts = 0, 1+(n-1)//3, 2**((n-1) % 3), 2**(n-1), set(), set() for i in range(2**k): j, l = int(bin(i)[2:],8)*r, n2 traj = set([(l,j)]) while True: if j >= l: j = j*16+13 l *= 2 else: j *= 4 l //= 2 if l == 0: ts |= traj break j %= 2*l if (l,j) in traj: c += 1 cs |= traj break if (l,j) in cs: c += 1 break if (l,j) in ts: break traj.add((l,j)) return c # Chai Wah Wu, Aug 03 2017
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