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. Prime numbers, where f(p) = 1, become increasingly rare. Their contribution to the average tends to zero.
  2. Most composite numbers have a large number of divisors. For a number n, the value f(n) gets smaller as the number of divisors d(n) gets larger. Since "almost all" integers have a large number of divisors, their f(n) values are close to zero, pulling the average down.

• Question 2: Is f(n) injective over composites? This is an open problem. It is extremely difficult to prove or disprove. It boils down to a complex Diophantine-like equation involving the divisor functions, for which no straightforward solution method is known. Finding a collision f(m) = f(n) for composites m ≠ n would require finding two numbers whose divisor distributions have a very specific and unlikely relationship.

Detailed Explanation for Question 1

Let's prove that .

The average can be split into a sum over primes and a sum over composites:

We will analyze the two parts separately.

Part 1: The Contribution from Prime Numbers

For any prime number p, its divisors are just 1 and p.

• Number of divisors: d(p) = 2.
• Mean of divisors: μ_p = (1+p)/2.
• Standard deviation of divisors (s_p): So, .
• The "primeness" function f(p) is: This confirms your definition: f(n) = 1 if and only if n is prime.

The sum over primes is simply the count of primes up to N, which is denoted by π(N). The contribution of primes to the average A_N is .

By the Prime Number Theorem, we know that . Therefore: So, the average contribution from prime numbers tends to zero.

Part 2: The Contribution from Composite Numbers

This is the more complex part. We need to show that .

The key is to find a good upper bound for f(n) in terms of its number of divisors, d(n). As shown in the MathOverflow answer, a useful (though not immediately obvious) inequality is: This gives the following bound on f(n): As , . For simplicity, we can say for large n: This shows that if a number n has many divisors, its f(n) value is small. For example, if d(n) = 100, then f(n) is approximately 2/√100 = 0.2.

Now we can use a powerful idea from number theory: almost all integers have a large number of divisors. More formally, the natural density of the set of integers {n | d(n) ≤ K} approaches 0 as K approaches infinity.

Let's make this rigorous to show the limit is 0.

1. Pick an arbitrary small number ε > 0. We want to show that for a large enough N, the average A_N is less than ε.

2. Split the numbers into two groups:

   • Group 1: Numbers with "many" divisors.
   • Group 2: Numbers with "few" divisors.

   Let's choose a threshold K. A number n has "many" divisors if d(n) > K and "few" if d(n) ≤ K. We choose K to be large enough so that .

3. Analyze the average A_N:

4. Bound the sum for the "many divisors" group: For any r with d(r) > K, we have . The sum for this group is: So, their contribution to the average is less than .

5. Bound the sum for the "few divisors" group: For any r, f(r) ≤ 1. So, the sum for this group is at most the number of elements in it. Let be the natural density of integers n with d(n) ≤ K. We know that . This means for our chosen ε, we can pick our K even larger (if necessary) so that the density is less than . For large N, the proportion of numbers with d(r) ≤ K will approach this density:

6. Combine the bounds: For a sufficiently large N, we have: Since we can make the average A_N smaller than any ε > 0 by choosing N large enough, the limit must be 0.

Detailed Explanation for Question 2

The question asks if there exist two distinct composite numbers m and n (say 3 < m < n) such that f(m) = f(n). This is an open research-level question. Here’s why it's so hard.

The condition f(m) = f(n) is equivalent to: Using the formula for variance, , the equation becomes: where:

• d(k) is the number of divisors of k.
• σ(k) is the sum of the divisors of k.
• σ₂(k) is the sum of the squares of the divisors of k.

These divisor functions depend on the prime factorization of a number in a complex, multiplicative way. For :

To find a collision f(m) = f(n), you would need to find two different composites m and n whose prime factorizations result in the two sides of the equation being exactly equal.

Why this is hard:

1. Erratic Nature: The values of d(k), σ(k), and σ₂(k) behave erratically as k changes. There is no simple pattern.
2. Transcendental-like Equation: The equation mixes the numbers m and n themselves (in the (m-1)^2 term) with functions of their prime factors. This makes it extremely difficult to solve analytically.
3. Numerical Search is Challenging: While one could search for a solution computationally, the values of f(n) are likely irrational numbers involving square roots of integers. Proving that two such numbers are equal is harder than showing two floating-point approximations are close. You would need to find m and n such that s_m^2 / (m-1)^2 = s_n^2 / (n-1)^2 holds true for the exact rational values.

There is no theoretical reason to believe a solution does not exist. It would be a form of "numerical accident." However, finding such an accident or proving one must exist is currently beyond known mathematical techniques.