Proving the Formula for the n-th Prime Number using Number Theory

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Homework Help Overview

The discussion revolves around proving a formula related to the n-th prime number, specifically in the context of number theory. The original poster presents a formula involving a summation and floor function, expressing difficulty in proving its validity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the formula, with one questioning the correctness of the output for specific values of n, particularly n=3. There are also humorous remarks about the interpretation of the primes.

Discussion Status

The conversation is ongoing, with participants exploring the formula's validity and implications. Some guidance is offered regarding the interpretation of the summation, but no consensus has been reached on the correctness of the formula.

Contextual Notes

There appears to be confusion regarding the formula's structure, particularly the upper limit of the summation and the identification of prime numbers based on the formula's output.

eljose
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Hello i need help to prove the formula:

[tex]p_n =\sum_{k=2}^{2^{n}}[ |n/(1+ \pi (k))|^{1/n}][/tex]

where []=floor function and |x| the modulus of x...this appear in number theory to evaluate the n-th prime however i am not able to prove it...:frown: :frown: :frown:
 
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So according to this formula, with n= 3, 2 is the third prime? I did not know that!
 
HallsofIvy said:
So according to this formula, with n= 3, 2 is the third prime? I did not know that!

:smile: Pure comedy!:smile:
 
Though in fact, that's 2^n at the top of the summation. Even then, something's wrong, I get the third prime as being three according to the summation. Need to check again since I rushed thru it, but it doesn't look right.
 

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