Appendix: Evidence of Equivalent Conditions for the Riemann Hypothesis

Theorem:

Let \( n \) be the product of the first \( k \) prime numbers:

where \( p_k \) is the \( k \)-th prime number. Then:

Proof:

1. Express \( \log(n) \) as a Sum:

   Using the property of logarithms for products:

   \( \log(n) = \log(p_1 \times p_2 \times \dots \times p_k) = \log(p_1) + \log(p_2) + \dots + \log(p_k) \).
2. Define Chebyshev's Function \( \theta(x) \):

   Chebyshev's function is defined as:

   \( \theta(x) = \sum_{\substack{p \leq x \\ p \text{ is prime}}} \log(p) \).

   For \( x = p_k \):

   \( \theta(p_k) = \sum_{i=1}^{k} \log(p_i) \).
3. Apply Chebyshev's Inequality:

   Chebyshev proved that for all \( x \geq 2 \):

   \( \theta(x) < x \).

   Therefore, for \( x = p_k \):

   \( \theta(p_k) = \sum_{i=1}^{k} \log(p_i) < p_k \).
4. Conclude the Inequality:

   Thus:

   \( \log(n) = \theta(p_k) < p_k \).

Conclusion:

For \( n = p_1 \times p_2 \times \dots \times p_k \), it holds that:

Statement:

Let \( n \) be a highly composite number with prime factorization:

where \( p_i \) are prime numbers in increasing order (\( p_1 = 2, \, p_2 = 3, \, \dots \)), and the exponents satisfy \( j_1 \geq j_2 \geq \dots \geq j_k \geq 1 \), with at least one exponent \( j_i > 1 \).

Then, for highly composite numbers \( n \), it holds that:

Proof:

1. Express \( \log(n) \):

   The natural logarithm of \( n \) is:

   \( \log(n) = \sum\limits_{i=1}^{k} j_i \ln(p_i) \).
2. Split the sum:

   We split the sum into two parts:

   \( \log(n) = \underbrace{\sum\limits_{i=1}^{k} \ln(p_i)}_{\theta(p_k)} + \underbrace{\sum\limits_{i=1}^{k} (j_i - 1) \ln(p_i)}_{\Delta} \),

   where:

   • \( \theta(p_k) = \sum\limits_{i=1}^{k} \ln(p_i) \) is Chebyshev's function.
   • \( \Delta = \sum\limits_{i=1}^{k} (j_i - 1) \ln(p_i) \) represents the contribution from exponents greater than 1.
3. Estimate \( \delta_k = p_k - \theta(p_k) \):

   From the Prime Number Theorem and known estimates, we have:

   \( |\delta_k| \leq C \dfrac{p_k}{\ln p_k} \),

   where \( C > 0 \) is a constant. For our purposes, we choose \( C = 0.1 \).

4. Estimate \( \Delta \):

   Let \( r \) be the number of exponents \( j_i > 1 \), and let \( j_{\text{min}} = \min\{j_i : j_i > 1\} \geq 2 \). Then:

   \( \Delta = \sum\limits_{i=1}^{k} (j_i - 1) \ln(p_i) \geq (j_{\text{min}} - 1) \sum\limits_{i=1}^{r} \ln(p_i) \).

   Since \( p_i \geq 2 \), we have:

   \( \sum\limits_{i=1}^{r} \ln(p_i) \geq r \ln(2) \),

   thus:

   \( \Delta \geq (j_{\text{min}} - 1) r \ln(2) \).
5. Compare \( \Delta \) and \( |\delta_k| \):

   We need to show that:

   \( \Delta > |\delta_k| \implies (j_{\text{min}} - 1) r \ln(2) > C \dfrac{p_k}{\ln p_k} \).
6. Relate \( p_k \) and \( k \):

   The \( k \)-th prime satisfies approximately:

   \( p_k \approx k \ln k \),

   and thus:

   \( \ln p_k \approx \ln(k \ln k) \approx \ln k + \ln \ln k \).
7. Estimate \( \dfrac{p_k}{\ln p_k} \):

   We have:

   \( \dfrac{p_k}{\ln p_k} \approx \dfrac{k \ln k}{\ln k + \ln \ln k} \approx k \).

   Therefore:

   \( |\delta_k| \leq C k \).
8. Compare \( \Delta \) and \( |\delta_k| \):

   From our estimates:

   \( \Delta \geq (j_{\text{min}} - 1) r \ln(2) \).

   To ensure \( \Delta > |\delta_k| \), we need:

   \( (j_{\text{min}} - 1) r \ln(2) > C k \).
9. Relate \( r \) and \( k \):

   Let \( r = \alpha k \), where \( 0 < \alpha \leq 1 \). Then:

   \( (j_{\text{min}} - 1) \alpha k \ln(2) > C k \implies (j_{\text{min}} - 1) \alpha \ln(2) > C \).
10. Set conditions on \( \alpha \) and \( j_{\text{min}} \):

    To satisfy the inequality, we need:

    \( \alpha > \dfrac{C}{(j_{\text{min}} - 1) \ln(2)} \).

    Since \( j_{\text{min}} - 1 \geq 1 \) and \( \ln(2) \approx 0.6931 \), we obtain:

    \( \alpha > \dfrac{0.1}{0.6931} \approx 0.1443 \).
11. Conclusion:

    For highly composite numbers, \( \alpha \) is often greater than the required value, so \( \Delta > |\delta_k| \), and thus:

    \( \log(n) = \theta(p_k) + \Delta = p_k - \delta_k + \Delta > p_k \).

    Therefore, \( \log(n) > p_k \).

Example to Illustrate the Proof:

Let \( n = 2^5 \times 3^4 \times 5^3 \times 7^2 \times 11^2 \times 13 \times 17 \times 19 \).

• Primes (\( p_i \)): \( 2, 3, 5, 7, 11, 13, 17, 19 \).
• Exponents (\( j_i \)): \( 5, 4, 3, 2, 2, 1, 1, 1 \).
• Number of primes (\( k \)): \( 8 \).
• Number of exponents greater than 1 (\( r \)): \( 5 \).
• Largest prime (\( p_k \)): \( 19 \).

Calculate \( \theta(p_k) \):

Calculate \( \Delta \):

Calculate \( \delta_k \):

Compare \( \Delta \) and \( \delta_k \):

Compute \( \log(n) \):

Compare \( \log(n) \) and \( p_k \):

Verify the condition for \( \alpha \):

We have \( \alpha = \dfrac{r}{k} = \dfrac{5}{8} = 0.625 \), which is greater than \( 0.1443 \). Therefore, the condition is satisfied.

Conclusion:

This example illustrates the validity of the proof that for highly composite numbers \( n \), it holds that \( \log(n) > p_k \).