[Alamo-Girl]

Thank you for your post!

Here is Doctor Stochastic’s formulation:

Simple Rule: concatenate the integers base two: 110111001001.... Treat blocks of eight as Ascii encoding. Everything ever written will be in the sequence: the works of Shakespeare, a description of design of a Sherman tank, this thread, a description of every living thing.

That produces this pattern:

1

10 11

100 101 110 111

1000 1001 1010 1011 1100 1101 1110 1111

10000 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111

100000 100001 100010 100011 100100 100101 100110 100111 101000 101001 101010 101011 101100 101101 101110 101111 110000 110001 110010 110011 110100 110101 110110 110111 111000 111001 111010 111011 111100 111101 111110 111111

1000000 1000001 1000010 1000011 1000100 1000101 1000110 1000111 1001000 1001001 1001010 1001011 1001100 1001101 1001110 1001111 1010000 1010001 1010010 1010011 1010100 1010101 1010110 1010111 1011000 1011001 1011010 1011011 1011100 1011101 1011110 1011111 1100000 1100001 1100010 1100011 1100100 1100101 1100110 1100111 1101000 1101001 1101010 1101011 1101100 1101101 1101111 1110000 1111000 1111001 1111010 1111011 1111100 1111101 1111110 1111111

etc.

If I treat blocks of 8 of the above binary as an ASCII byte, I will get numerous unprintable/unreadible characters throughout because of the pattern, the hanging high order 1's filling left to right. Thus, no Shakespeare, no Omega, etc.

You are starting in reverse, by taking a finite string of ASCII bytes at length of n and then figuring all binary encoded integers of bit-length n*8.

Eventually you will get the original string. Sure. Start with the length of Shakespeare work and then figure all the possible binary combinations in that length.

But that is not the pattern stipulated by Doctor Stochastic's formulation. Nor is your formulation compatible with the definition of Champernowne’s constant from Wolfram’s website and others I found, all of which are more like Doctor Stochastic’s where the increment for binary proceeds upward in a highly structured pattern.

[tortoise]

Your sequence "100 101 110 111" contains the bit pattern "001011" and so forth. There is no requirement to respect "byte boundaries" to locate the pattern i.e. you can effectively truncate the leading "1" you are using in your expression of it. My reformulation was an attempt to make it obvious why Champernownes number WILL contain Shakespeare, but apparently I failed.

Doctor Stochastic is correct, but it seems like you are making some subconscious assumptions when you are envisioning how it is expressed and interpreted that aren't required in the math.

[tortoise]

But that is not the pattern stipulated by Doctor Stochastic's formulation.

How I stated it may not be how Doctor Stochastic formulated it, but my formulation was intended for ease of demonstrating the point. My point was that once you view Shakespeare as some arbitrary bit-string, Doctor Stochastic's series will eventually append an integer to the sequence that happens to have the exact same bit-string as Shakespeare. You don't even have to view it as the concatenation of a lot of small integers. All you need is a sufficiently large single integer to have a finite bit pattern in the sequence that matches your original bit-string. And the thing about infinities is, eventually you'll come across that integer.