What is the Copeland-Erdős Number and Why is it Considered Normal?

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SUMMARY

The Copeland-Erdős number is constructed by concatenating prime numbers, resulting in a sequence such as 0.2357111317... This number is classified as a normal number, meaning every finite sequence of digits appears with equal frequency in its decimal expansion. Proving its normality is complex, but Dirichlet's theorem can be applied to demonstrate that any sequence of digits can yield prime numbers, which must then appear within the Copeland-Erdős number. The discussion highlights the mathematical intricacies involved in establishing the normality of this number.

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littlemathquark
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Homework Statement
Show that every finite sequence of digits is contained within the Copeland-Erdős number
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Show that every finite sequence of digits is contained within the Copeland-Erdős number
The Copeland-Erdős number is formed by writing prime numbers consecutively, one after the other. So ##0,2357111315171923...## is Copeland-Erdős number.

According to my net research; Copeland-Erdős number is a normal number. A number being normal means that any sequence of n digits in its decimal fraction part appears with asymptotically equal frequency. So that expression (definition of normal numbers) enough to say that "every finite sequence of digits is contained within the Copeland-Erdős number" ?
 
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I think so, but then how do you prove that it is normal? Have you just read that somewhere? How would you go about proving it?
 
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I have just read somewhere
mjc123 said:
I think so, but then how do you prove that it is normal? Have you just read that somewhere? How would you go about proving it?
 
And yet I saw this guy, IIRC Numberphile, use the (pairs of) digits in the Decimal Expansion of ##\pi## to illustrate Benford's Law ( "Real-world" Distribution of digits 0-9 in numbers ). Edit: Assuming normality of ##\pi## , we should expect a Uniform distribution, not a Benford.
 
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littlemathquark said:
We need an easy way.
If there were one, Erdös would have found it for sure.
 
fresh_42 said:
If there were one, Erdös would have found it for sure.
It seems possible to solve this by using Dirichlet's theorem.
 
According to Dirichlet's theorem, in an arithmetic sequence of the form a+n⋅d where gcd⁡(a,d)=1 there are infinitely many prime numbers. Now, consider any sequence of digits. To ensure the condition gcd⁡(a,d)=1append the digit 1to the end of this sequence and define the resulting number as a. Let k be the number of digits in this number, and set the common difference of the sequence as d=10^k. By Dirichlet's theorem, this sequence will contain at least one prime number, which, by definition, must appear as a substring of the Copeland–Erdős number.
For example, consider the number 684. By appending 1, we obtain a=6841, and by setting d=10^k we construct the sequence 6841,16841,26841,…By Dirichlet's theorem, this sequence must contain at least one prime.
 
  • #10
littlemathquark said:
According to Dirichlet's theorem, in an arithmetic sequence of the form a+n⋅d where gcd⁡(a,d)=1 there are infinitely many prime numbers. Now, consider any sequence of digits. To ensure the condition gcd⁡(a,d)=1append the digit 1to the end of this sequence and define the resulting number as a. Let k be the number of digits in this number, and set the common difference of the sequence as d=10^k. By Dirichlet's theorem, this sequence will contain at least one prime number, which, by definition, must appear as a substring of the Copeland–Erdős number.
For example, consider the number 684. By appending 1, we obtain a=6841, and by setting d=10^k we construct the sequence 6841,16841,26841,…By Dirichlet's theorem, this sequence must contain at least one prime.
It must contain infinitely-many primes according to Dirichlet's theorem. Edit: But I get your point, you just need to find one.
 
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  • #11
WWGD said:
It must contain infinitely-many primes according to Dirichlet's theorem. Edit: But I get your point, you just need to find one.
6841 must be prime.
 

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