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The Champernowne constant in base $b \geq 2$ is obtained by concatenating the $b$-ary expansion of every integer. For example, in base $10$ this is $$ 0.123456789101112131415\dotsc $$ Question: Is the sequence of $b$-ary digits of the base $b$ Champernowne constant automatic?

My guess is no, but I don't know enough about this to give a definitive answer.

Note: Automatic numbers, i.e. numbers whose sequence of digits in some base $b$ is automatic, can very well be irrational. For example the $2$-ary number whose sequence of digits is the Thue-Morse sequence is irrational.

Even more, in 2007 Adamczewski and Bugeaud proved that every automatic number is either rational or transcendental.

A.P.
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1 Answers1

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A finite-state automaton that is not allowed to read the tape can only produce eventually periodic output (aka. rational numbers). This is because by pigeon-hole there must be some digit positions such that the machine is in the same state when wrting to these positions.

On the other hand, a Turing machine (i.e., an automaton that can also read the output tape and/or use it as scratchpad) is of course able to run in such a way that the output "converges" to the Champernowne constant (in the sense that each output place is eventually constant and one can predict from when on this is the case) just by doing some copy and increment runs at the "tail" of the growing output.

Hagen von Eitzen
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  • @A.P. I want to see a finite state automaton which generates a non-periodic sequence. Can you post an example (as an edit to your question)? – 5xum Jun 18 '15 at 08:56
  • I'm sorry, I realize that I wasn't precise enough. By sequence "generated by a finite-state automaton" I meant an automatic sequence. The Thue-Morse sequence isn't eventually periodic, but it is automatic. – A.P. Jun 18 '15 at 08:58