- bn = 1 if the binary representation of n contains no block of consecutive 0s of odd length;
- bn = 0 otherwise;
for n ≥ 0.
For example, b4 = 1 because the binary representation of 4 is 100, which only contains one block of consecutive 0s of length 2; whereas b5 = 0 because the binary representation of 5 is 101, which contains a block of consecutive 0s of length 1.
Starting at n = 0, the first few terms of the Baum–Sweet sequence are:
The properties of the sequence were first studied by L.E. Baum and M.M. Sweet in 1976.
The value of term bn in the Baum–Sweet sequence can be found recursively as follows. If n = m·4k, where m is not divisible by 4, then
Thus b76 = b9 = b4 = b0 = 1, which can be verified by observing that the binary representation of 76, which is 1001100, contains no consecutive blocks of 0s with odd length.
The Baum–Sweet word 1101100101001001..., which is created by concatenating the terms of the Baum–Sweet sequence, is a fixed point of the morphism or string substitution rules
- 00 → 0000
- 01 → 1001
- 10 → 0100
- 11 → 1101
- 11 → 1101 → 11011001 → 1101100101001001 → 11011001010010011001000001001001 ...
From the morphism rules it can be seen that the Baum–Sweet word contains blocks of consecutive 0s of any length (bn = 0 for all 2k integers in the range 5.2k ≤ n < 6.2k), but it contains no block of three consecutive 1s.