Replies

Thank you for your post!

Indeed, the sequence is not random, therefore I strongly disagree with this statement:

"The sequence does generate every possible sequence of bits."

If that were true, Chaitin's Omega would be there as well.

For Lurkers, here is the Champernowne’s binary constant concatenated to base 10:

Champernowne's constant 0.1234567891011... is the number obtained by concatenating the positive integers and interpreting them as decimal digits to the right of a decimal point. It is normal in base 10 (Champernowne 1933, Bailey and Crandall 2003). Mahler (1961) showed it to also be transcendental. ..

The "binary" Champernowne constant is obtained by concatenating the binary representations of the integers

C₂=0.(1)(10)(11)(100)(101)(110)(111)…
=0.86224012586805…

You said:

Of course it generates Shakespeare with one typo, with two typos, three typos, etc.

Because more character representations of 8 bit ascii values 0-255 are unreadable than not, I continue to assert the Champernowne binary constant cannot be successfully concatenated to a readable character representation of Shakespeare.

...I continue to assert the Champernowne binary constant cannot be successfully concatenated to a readable character representation of Shakespeare.

Then you haven't understood the proof behind Champernowne's number.

Any finite segment of Chaitin's Omega is in the sequence.