To address the questions:

Question 1:

The function ( f(n) ) measures the "primeness" of ( n ). For primes, ( f(n) = 1 ), and for composites, ( f(n) ) is defined using the standard deviation of their divisors. The average primeness is given by ( \frac{1}{N} \sum_{r=2}^N f(r) ).

• The number of primes up to ( N ) (denoted ( \pi(N) )) grows asymptotically as ( \frac{N}{\log N} ...

Write the divisor set of n as D(n) = {d : d|n}, its cardinality τ(n), and the divisor-sums σ(n) = Σ_{d|n} d, σ2(n) = Σ_{d|n} d^2. Then the mean and variance of the divisors are μ(n) = σ(n)/τ(n), s_n^2 = σ2(n)/τ(n) − (σ(n)/τ(n))^2, so your function is f(n) = (2/(n−1)) sqrt( σ2(n)/τ(n) − (σ(n)/τ(n))^2 ).

Answer to Question 1 Yes, the average tends to 0, and one can even get a quantitative decay.

T...

This is a fascinating question about measuring "primeness" as a continuous quantity rather than a binary property. Let me analyze both questions:

Question 1: Does the average primeness tend to zero?

As noted, this has been answered affirmatively on MathOverflow. The key insight is that for most large numbers, the standard deviation of divisors grows much more slowly than the number itself.

H...

Throughout we write

 τ(n)=∑{d∣n}1 (number of divisors)
 σ₂(n)=∑{d∣n}d² (sum of the squares of the divisors).

For n≥2 let

 sₙ² = 1/τ(n)·σ₂(n) – ( 1/τ(n)·∑_{d∣n}d )²

(the population variance of the divisors of n) and

 f(n)= 2sₙ /(n−1).

1. The average value of f(n)

Write S(N)=∑_{n≤N}f(n).

Because the second term in sₙ² is non–negative we have

              ...

<think> I need to tackle this problem about the "average primeness" of natural numbers. The user has defined a function \( f(n) = \frac{2s_n}{n-1} \) for \( n \ge 2 \), where \( s_n \) is the standard deviation of the divisors of \( n \). The question is whether the average of this function from 2 to N tends to zero as N goes to infinity, and also if f(n) is injective over composite numbers.

Firs...

This is a fascinating question that combines elementary concepts (divisors, standard deviation) with deep results in analytic number theory. Here is a breakdown of the answers to your questions, explaining the reasoning in detail.

Summary of Answers

• Question 1: Does the average primeness tend to zero? Yes. The limit is indeed zero. The intuitive reason is twofold:
  1. Pri...

We keep the notation

 d(n) = τ(n) = number of positive divisors of n
 σ₁(n)= Σ_{d|n} d , σ₂(n)= Σ_{d|n} d²

and  μ(n)=σ₁(n)/τ(n), s²(n)=σ₂(n)/τ(n) – μ(n)² , so that

 f(n)= 2 s(n)/(n–1)= 2 ( σ₂(n)/τ(n) – (σ₁(n)/τ(n))² )½ /(n–1) . (1)

Question 1 lim_{N→∞}(1/N) Σ_{n≤N} f(n)=0.

1. A set of integers whose natural density is 1 and on which f(n) is small.

For every n let a(n) be the...

Below is a self‐contained sketch of why the “continuous‐primeness’’
average tends to 0, together with a few remarks on injectivity.

1. Notation and restatement
– Let d(n) = the number of (positive) divisors of n, and write those divisors as
D(n)={d₁<d₂<⋯<d_{d(n)}}.
– Let μ_n = (1/d(n))∑_{d|n} d be the arithmetic‐mean of the divisors.
– Define the (population) standard de...

Does the average primeness of natural numbers tend to zero?

This question asks about a measure of "primeness" for natural numbers, defined as:

where is the standard deviation of the divisors of . This function is designed to equal 1 for prime numbers and assign values between 0 and 1 for composite numbers, with the value decreasing as the number of divisor...