In mathematics the Baum–Sweet sequence is an infinite automatic sequence of 0s and 1s defined by the rule:
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:
1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1 ... (sequence A086747 in the OEIS)
Contents
Historical motivation
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
F
2
(
(
X
−
1
)
)
{\displaystyle \mathbb {F} _{2}((X^{-1}))}
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
f
3
+
x
−
1
f
+
1
=
0.
{\displaystyle f^{3}+x^{-1}f+1=0.}
The authors show that by Hensel's lemma, there is a unique such root in
F
2
(
(
X
−
1
)
)
{\displaystyle \mathbb {F} _{2}((X^{-1}))}
because reducing the defining equation of
f
{\displaystyle f}
modulo
X
−
1
{\displaystyle X^{-1}}
gives
f
3
+
1
{\displaystyle f^{3}+1}
, which factors as
f
3
+
1
=
(
f
+
1
)
(
f
2
+
f
+
1
)
.
{\displaystyle f^{3}+1=(f+1)(f^{2}+f+1).}
They go on to prove that this unique root has partial quotients of degree
≤
2
{\displaystyle \leq 2}
. Before doing so, they state (in the remark following Theorem 2, p 598) that the root can be written in the form
f
=
∑
k
≥
0
f
i
X
−
k
{\displaystyle f=\sum _{k\geq 0}f_{i}X^{-k}}
where
f
0
=
1
{\displaystyle f_{0}=1}
and
f
k
=
1
{\displaystyle f_{k}=1}
for
k
≥
1
{\displaystyle k\geq 1}
if and only if the binary expansion of
n
{\displaystyle n}
contains only even length blocks of
0
{\displaystyle 0}
's. This is the origin of the Baum–Sweet sequence.
Mkaouar and Yao proved that the partial quotients of the continued fraction for
f
{\displaystyle f}
above do not form an automatic sequence. However, the sequence of partial quotients can be generated by a non-uniform morphism.
Properties
The Baum–Sweet sequence can be generated by a 3-state automaton.
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
b
n
=
{
1
if
n
=
0
0
if
m
is even
b
(
m
−
1
)
/
2
if
m
is odd
.
{\displaystyle b_{n}={\begin{cases}1&{\text{if }}n=0\\0&{\text{if }}m{\text{ is even}}\\b_{(m-1)/2}&{\text{if }}m{\text{ is odd}}.\end{cases}}}
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.