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As some may know, Champernowne's constant is one of the only known constants proven to be normal. The number is constructed by concatenating whole numbers as you count up and appending them behind a decimal point. In base 10, the constant is 0.12345678910111213...

My main question here is that in order to be "simply normal", each digit must be equally likely to appear throughout the entire number. This, however, does not seem to hold for 0 since it's skipped in the counting. For instance, since we don't start counting at 0, the first time it appears is in the back of 10. The first time 00 appears, is in the back of 100. I used some Python code to write out the first 68888897 digits of the constant and then calculated the probabilities of each digit occurring.

Here is the output:

The number 0 occurs 5888896 times giving a probability of 0.085483964128501

The number 1 occurs 7000001 times giving a probability of 0.10161290577783529

The number 2 occurs 7000000 times giving a probability of 0.10161289126170796

The number 3 occurs 7000000 times giving a probability of 0.10161289126170796

The number 4 occurs 7000000 times giving a probability of 0.10161289126170796

The number 5 occurs 7000000 times giving a probability of 0.10161289126170796

The number 6 occurs 7000000 times giving a probability of 0.10161289126170796

The number 7 occurs 7000000 times giving a probability of 0.10161289126170796

The number 8 occurs 7000000 times giving a probability of 0.10161289126170796

The number 9 occurs 7000000 times giving a probability of 0.10161289126170796

As you can see, with this counting style, 0 seems underrepresented. That being said, I know the number has been proven to be normal so my question is this: if we were to take the constant and remove 1, 11, 111, 1111, etc. from the counting, would the number that remains still be normal? If not, why is the lack of zeros not an issue in the original constant? If it is, then how do the digits still get represented equally as the limit tends to infinity?

Connor James
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    Champernowne's constant is not the only known normal number. Correct is that no normal number is known that has not been constructed for this purpose. – Peter Sep 07 '23 at 20:22
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    @Peter Yes, that was my bad. I had forgotten that other numbers like the Copeland-Erdos constant have also been proven normal. Sorry about that. – Connor James Sep 07 '23 at 20:23
  • Imagine the point where we start to concatenate , say , $10^{10}$ , $10^{10}+1$ and so on. This gives an extreme high ratio of zeros. If we replace $10^{10}$ by $10^{20}$ , this gets worse. – Peter Sep 07 '23 at 20:28
  • Do you only remove the numbers $1,11,111,1111$ ? Then , the ratio deos not change if $n$ tends to $\infty$ since the ratio of rep-units quickly decreases to $0$. – Peter Sep 07 '23 at 20:34
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    What was the last number you concattenated ? Maybe this explains the bias. – Peter Sep 07 '23 at 20:34
  • The bias is present regardless of where I choose to stop. For this code, I stopped at 10000000. Here's the code I used in case you want to play around with it but I was never able to get 0 to be the most commonly occurring, let alone equal in occurrence to the other digits:

    temp = "" for i in range(1, 10000001): temp += str(i)

    print(len(temp)) for k in range(10): print(f"The number {k} occurs {temp.count(str(k)1)} times giving a probabilitiy of {temp.count(str(k)1)/len(temp)}")

     – Connor James Sep 07 '23 at 22:04
  • The bias comes because no number starts with $0$. This only impacts the first digit of a number. In your experiment it shows up easily because each number only has $7$ digits. If you went up to numbers with $10^{100}$ digits the effect would be tiny. – Ross Millikan Sep 08 '23 at 00:04
  • @RossMillikan So I take it that implies that removing 1, 11, 111, etc. wouldn't keep the constant form being simply normal? – Connor James Sep 08 '23 at 00:45
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    No, those are such a small fraction of the $1$s they won't matter in the long run. – Ross Millikan Sep 08 '23 at 12:47

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