A normal number is a number where no number is favored to appear in the digits. Does this definition imply that all whole numbers appear in its digits? Because the definition involves notions from probability, I was wondering if it might happen that a certain number would not appear in the digits of a normal number without contradicting the definition.
Do all natural numbers appear in the digits of a normal number?
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elementary-number-theory
1 Answers
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Yes it does. Working in base B, a string with A digits should have the natural density $ \frac {1}{ B^A} $ by definition of a normal number. If the string doesn't appear at all, then it has natural density 0.
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0That should work, if all you want is for the string to appear once. But since you talked about natural numbers, I was looking at the density of any string. For example, it is possible that the number of times that 12 appears is greatly changed since we're inserting 0's in many many places, so the initial 12, or 2122, etc would not contribute the count any more, and this could make the natural density 0. My argument was that the difference is linear / exponential, which tends to 0, hence the density is $ \frac {1}{2 \times 10^A}$. – 2012-12-29