Number of natural numbers that have primitive roots

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Discussion Overview

The discussion revolves around calculating the number of natural numbers between 2 and n that have primitive roots. It explores the conditions under which a number has a primitive root and the implications for counting such numbers, including the role of prime numbers and their density.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that a number m has a primitive root if it is cyclic, specifically when m=1,2,4 or m=p^k or m=2·p^k for prime p.
  • Participants discuss the need to calculate the number of primes between 2 and n^(1/k) to determine the count of numbers of the form p^k and 2·p^k.
  • There is a suggestion that the density of primes could be used to approximate the number of primes, but some participants express caution about relying on it for an exact count.
  • One participant raises the question of whether the actual number of primes is necessary to calculate the limit of the ratio a_n/n as n approaches infinity, while another suggests that the density of primes could suffice for this calculation.

Areas of Agreement / Disagreement

Participants generally agree on the conditions for a number to have a primitive root and the relevance of prime numbers. However, there is disagreement on the use of prime density for calculating exact numbers versus limits, indicating that the discussion remains unresolved.

Contextual Notes

Limitations include the reliance on approximations for prime density and the unresolved nature of whether actual counts are necessary for limit calculations.

mathmari
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Hey! :giggle:

How can we calculate the number of natural numbers between $2$ and $n$ that have primitive roots?

Let $m$ be a positive integer.
Then $g$ is a primitive root modulo $m$, with $(g,m)=1$, if the modulo of $g\in (Z/m)^{\star}$ is a generator of the group.

We have that $g$ is a primitive root modulo $m$ if it is a generator of a group, i.e. $m$ has a primitive root if $\mathbb{Z}_m$ is cyclic, right?

$\mathbb{Z}_m$ is cyclic if $m=1,2,4$ or $m=p^k$ or $m=2\cdot p^k$ for $p$ prime.

That means that the number of natural numbers that have a primitive root is $\#\{1,2,4,p^k, 2\cdot p^k\}$ for $p$ prime.

So we have to calculate the number of primes between $2$ and $n^{\frac{1}{k}}$ to calculate then the number of elements of the form $p^k$ and $2\cdot p^k$.

Have I understood that correctly? :unsure:
 
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Looks about right. :unsure:
 
Klaas van Aarsen said:
Looks about right. :unsure:

To calculate the number of these primes do we use the density of primes? :unsure:
 
mathmari said:
To calculate the number of these primes do we use the density of primes?
We can only approximate the density of primes.
So we cannot use it to find an actual number.
Assuming that we want a 'hard' number, I think we should express it in terms like 'the number of primes between $2$ and $n$'. :unsure:
 
Klaas van Aarsen said:
We can only approximate the density of primes.
So we cannot use it to find an actual number.
Assuming that we want a 'hard' number, I think we should express it in terms like 'the number of primes between $2$ and $n$'. :unsure:

Actually I want to calculate the limit $\displaystyle{\lim_{n\rightarrow \infty}\frac{a_n}{n}}$ where $a_n$ is the above number. So do we need the actual number to calculate this limit? :unsure:
 
mathmari said:
Actually I want to calculate the limit $\displaystyle{\lim_{n\rightarrow \infty}\frac{a_n}{n}}$ where $a_n$ is the above number. So do we need the actual number to calculate this limit?
No. I think we can use the density of primes to calculate that limit. :unsure:
 

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