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There is an interesting irrational number, in binary it is,

.01101110010111011110001001101010111100110111101111...

It is made by appending the numbers 0,1,2,3,4,5,6,7 etc one after the other in binary.

0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111 etc

We can form a square matrix from this irrational number.

The first twenty-five binary digits would be placed in this order.

$ \begin{matrix} 1 & 4 & 9 & 16 & 25 \\ 2 & 3 & 8 & 15 & 24 \\ 5 & 6 & 7 & 14 & 23 \\ 10 & 11 & 12 & 13 & 22 \\ 17 & 18 & 19 & 20 & 21 \\ \end{matrix} $

So our matrix would begin with

$ \begin{matrix} 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 &1 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ \end{matrix} $

And the matrix would have an infinite number of digits to the right and bottom.

We can convert our matrix to a list like this.

    0.00010...  
    0.11000... 
    0.11111... 
    0.10110... 
    0.11100...
       ...

We will have a fixed, infinite eternal object. It will contain an infinite number of irrational numbers.

My question is, will this list contain numbers like the binary version of $\pi-3 = 0.141159 \ldots = 0.001001000011111100111110 \ldots$, and more important, will it contain any rational numbers like $ 0.10101010 \dots$

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    $\begingroup$ Notably, your list must miss most irrational numbers since your list is countable but the irrational numbers are uncountable. $\endgroup$ Jul 31, 2017 at 15:12
  • $\begingroup$ What have you tried, in terms of answering your two questions? Seems like a simple thing to try is to extend the numbers much further to the right to see if any patterns appear. If patterns appear that suggest a rational number you could prove/disprove whether the numbers are really rational. The third number in your list is a good candidate, even though it just has 5 digits. $\endgroup$
    – Χpẘ
    Jul 31, 2017 at 15:33
  • $\begingroup$ The first "interesting irrational number" in the question is half the base-2 Champernowne constant $C_2$. $\endgroup$ Jul 31, 2017 at 22:27
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    $\begingroup$ If the list contained $\pi-3$, that would imply a remarkably fast algorithm for computing binary digits of $\pi$ -- faster than I think anybody expects exists. $\endgroup$ Jul 31, 2017 at 22:39
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    $\begingroup$ @MichaelHardy: It would still mean that a fast algorithm for digits of $\pi$ exists; we just wouldn't know which algorithm that is. $\endgroup$ Aug 24, 2017 at 20:36

1 Answer 1

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NOT AT ANSWER - a long comment

Call the irrational number first mentioned in the OP, $B$. To generate digits of the numbers in the list, digits at specified indices must be selected from $B$. To calculate digit $n$ from $B$ it is first necessary to know the index of the first digit of the number in $B$ that has the same number of digits that $n$ points to. Let $f(n)$ be that index. Let $d(n)$ be the number of digits of the numbers that start at $f(n)$.

For example, take $n=500$. $f(500)=322$ and $d(500)= 7$. This means 500 indexes into the 7 digit numbers, which start at index 322. 500 corresponds to the $\lfloor\frac{500-322}7\rfloor=25$th (zero based) 7 digit number. That number is 64+25=89=1011001b. And the bit is the 3rd bit to the right of the MSB (most significant bit), namely a 1.

This formula gives the starting index for $d$ digit numbers within $B$: $s(d)=2^{\left(d-1\right)}\left(d-2\right)+2$. 1 digit numbers start at index 0, the 2 digit numbers begin at index 2, the 3 digit numbers at 6, 4 digit numbers at 18, etc.

The formula that gives the number of digits for a given index $n$ is $s$ inverted: $s^{-1}(n)=\left\lfloor\frac{W\left(\ln(2)\cdot\frac{n-2}2\right)}{\ln 2}\right\rfloor+2$, where $W$ is Lambert's W function. $s^{-1}(n)$ is the same as $d(n)$ mentioned earlier. (Note $d(n)$ is not a true inverse of $s(n)$ because it maps multiple input values to a single output value.)

Then $f(n)=s(d(n))$

Putting these together, the bit at index $n$ in $B$ is: $$\begin{align} \text{value of first d(n) digit number: }&a = 2^{d(n)-1} \\ \text{value of number that n points to: }&v =a + \left\lfloor\frac{n-f(n)}{d(n)}\right\rfloor\\ \text{index of bit to obtain: }&i = (n-f(n)) \bmod d(n)\\ \text{value of bit to obtain: }&b = \left\lfloor\frac{v}{2^{d(n)-i-1}}\right\rfloor \bmod 2\\ \end{align}$$

Using 500 as an example: $a=2^{7-1}=64, v = 64+\lfloor(178/7)\rfloor=89, i=178\bmod 7=3, b=\lfloor 89/2^{7-3-1}\rfloor \bmod 2=11\bmod 2=1$

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