- 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 Leonard E. Baum and Melvin M. Sweet in 1976. In 1949, Khinchin conjectured that there does not exist a non-quadratic algebraic real number having bounded partial quotients in its continued fraction expansion. A counterexample to this conjecture is still not known. Baum and Sweet's paper showed that the same expectation is not met for algebraic power series. They gave an example of cubic power series in whose partial quotients are bounded. (The degree of the power series in Baum and Sweet's result is analogous to the degree of the field extension associated with the algebraic real in Khinchin's conjecture.)
One of the series considered in Baum and Sweet's paper is a root of
The authors show that by Hensel's lemma, there is a unique such root in because reducing the defining equation of modulo gives , which factors as
They go on to prove that this unique root has partial quotients of degree . Before doing so, they state (in the remark following Theorem 2, p 598) that the root can be written in the form
where and for if and only if the binary expansion of contains only even length blocks of 's. This is the origin of the Baum–Sweet sequence.
Mkaouar and Yao proved that the partial quotients of the continued fraction for above do not form an automatic sequence. However, the sequence of partial quotients can be generated by a non-uniform morphism.
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 (or is 0), 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.
More succinctly, by Cobham's little theorem the Baum–Sweet word can be expressed as a coding applied to the fixed point of a uniform morphism . Indeed, the morphism
generate the word in that way.
- Weisstein, Eric W. "Baum–Sweet Sequence". MathWorld.
- Baum, Leonard E.; Sweet, Melvin M. (1976). "Continued Fractions of Algebraic Power Series in Characteristic 2". Annals of Mathematics. 103 (3): 593–610. doi:10.2307/1970953. JSTOR 1970953.
- Waldschmidt, M. (2009). "Words and Transcendence". In W.W.L. Chen; W.T. Gowers; H. Halbertstam; W.M. Schmidt; R.C. Vaughan (eds.). Analytic Number Theory: Essays in Honour of Klaus Roth (PDF). Cambridge University Press. Section 31, p. 449–470.
- Khinchin, A.I. (1964). Continued Fractions. University of Chicago Press.
- Graham Everest, Alf van der Poorten, Igor Shparlinski, Thomas Ward Recurrence Sequences AMS 2003, p 236.
- Mkaouar, M. (1995). "Sur le développement en fraction continue de la série de Baum et Sweet". Bull. Soc. Math. France. 123 (3): 361–374. doi:10.24033/bsmf.2264.
- Yao, J.-Y. (1997). "Critères de non-automaticité et leurs applications". Acta Arith. 80 (3): 237–248. doi:10.4064/aa-80-3-237-248.
- Allouche and Shallit (2003) p 210.
- Allouche, J.-.P. (1993). "Finite automata and arithmetic". Séminaire Lotharingien de Combinatoire: 23.
- Allouche and Shallit (2003) p 176.