"Primes are the most arbitrary and ornery objects studied by mathematicians: they grow like weeds among the natural numbers, seeming to obey no other law than that of chance, and nobody can predict where the next one will sprout..." - D. Zagier
"Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate." - Leonhard Euler
Claim: The Pattern of Prime Numbers
For $k \geq 3$, the $k$-th prime $p_k$ can be predicted from the previous primes $p_1, p_2, \dots, p_{k-1}$ using:
$$ p_k = \left\lceil \left( 1 - \frac{1}{\zeta(s) \cdot \prod_{j=1}^{k-1} \left( 1 - p_j^{-s} \right)} \right)^{-\frac{1}{s}} \right\rceil $$
where $s = k \cdot \ln\bigl(k \ln(k)\bigr)$ and $\zeta(s)$ is the Riemann zeta function.
Examples
- Given the primes $2,3,5,7,11$, the $\text{6th prime}$ is
$$ \boxed{\small p_{6} = \left\lceil \left(1 - \frac{1}{\zeta\left(6 \cdot \ln\left(6 \cdot \ln(6)\right)\right) \cdot \prod_{j=1}^{5} \left(1 - p_j^{-6 \cdot \ln\left(6 \cdot \ln(6)\right)}\right)}\right)^{-\frac{1}{6 \cdot \ln\left(6 \cdot \ln(6)\right)}} \right\rceil = 13} $$
- Given the primes $2,3,5,7,11,13,17,19,23$, the $\text{10th prime}$ is
$$ \boxed{\small p_{10} = \left\lceil \left(1 - \frac{1}{\zeta\left(10 \cdot \ln\left(10 \cdot \ln(10)\right)\right) \cdot \prod_{j=1}^{9} \left(1 - p_j^{-10 \cdot \ln\left(10 \cdot \ln(10)\right)}\right)}\right)^{-\frac{1}{10 \cdot \ln\left(10 \cdot \ln(10)\right)}} \right\rceil = 29} $$
- Given the primes $2,3,5,7,...,17359,17377,17383,17387$, the $\text{2000th prime}$ is
$$ \boxed{\small p_{2000} = \left\lceil \left(1 - \frac{1}{\zeta\left(2000 \cdot \ln\left(2000 \cdot \ln(2000)\right)\right) \cdot \prod_{j=1}^{1999} \left(1 - p_j^{-2000 \cdot \ln\left(2000 \cdot \ln(2000)\right)}\right)}\right)^{-\frac{1}{2000 \cdot \ln\left(2000 \cdot \ln(2000)\right)}} \right\rceil = 17389} $$
- Given the primes $2,3,5,7,...,37781,37783,37799,37811$, the $\text{4000th prime}$ is
$$ \boxed{\small p_{4000} = \left\lceil \left(1 - \frac{1}{\zeta\left(4000 \cdot \ln\left(4000 \cdot \ln(4000)\right)\right) \cdot \prod_{j=1}^{3999} \left(1 - p_j^{-4000 \cdot \ln\left(4000 \cdot \ln(4000)\right)}\right)}\right)^{-\frac{1}{4000 \cdot \ln\left(4000 \cdot \ln(4000)\right)}} \right\rceil = 37813} $$
- The Answer to the Ultimate Question of Life, the Universe, and Everything is the $42^{\text{nd}}$ prime.
Proof
The Euler product formula for the Riemann zeta function states:
$$ \zeta(s) = \prod_{p \, \text{prime}} \left( 1 - p^{-s} \right)^{-1} \quad ❤ $$
Its reciprocal is:
$$ \prod_{p \text{ prime}} \left( 1 - p^{-s} \right) = \frac{1}{\zeta(s)} $$
We split the product at the $k$-th prime $p_k$:
$$ \prod_{p \,\text{prime}} (1 - p^{-s}) = \prod_{j=1}^{k-1} \bigl(1 - p_j^{-s}\bigr) \cdot \prod_{p \ge p_k} \bigl(1 - p^{-s}\bigr) $$
Hence,
$$ \prod_{j=1}^{k-1} \bigl(1 - p_j^{-s}\bigr) \cdot \prod_{p \ge p_k} \bigl(1 - p^{-s}\bigr) = \frac{1}{\zeta(s)} $$
For large $s$, approximate the tail product:
$$ \prod_{p \geq p_k} \left( 1 - p^{-s} \right) \approx 1 - p_k^{-s} \tag{1} $$
Substituting,
$$ \left( \prod_{j=1}^{k-1} \left(1 - p_j^{-s}\right) \right) \cdot \left(1 - p_k^{-s}\right) \approx \frac{1}{\zeta(s)} $$
Thus,
$$ 1 - p_k^{-s} \approx \frac{1}{\zeta(s) \cdot\prod_{j=1}^{k-1} \bigl(1 - p_j^{-s}\bigr)} $$
Solving for $p_k$ gives:
$$ p_k \approx \left( 1 - \frac{1}{\zeta(s) \cdot\prod_{j=1}^{k-1} \bigl(1 - p_j^{-s}\bigr)} \right)^{-1/s} $$
We now apply the ceiling function for exactness:
$$ p_k = \left\lceil \left( 1 - \frac{1}{\zeta(s) \cdot \prod_{j=1}^{k-1} \left( 1 - p_j^{-s} \right)} \right)^{-\frac{1}{s}} \right\rceil \tag{2} $$
Derivation of $s$ using the Prime Number Theorem:
$$ p_k \sim k \ln k $$
Substituting into $p_k^{-s}$:
$$ p_k^{-s} \sim (k \ln(k))^{-s} $$
Taking the natural logarithm:
$$ \ln\left( p_k^{-s} \right) \sim \ln\left( (k \ln(k))^{-s} \right) $$
yields:
$$ \ln\left( p_k^{-s} \right) \sim -s \cdot \ln(k \ln(k)) $$
For large $s$, this implies:
$$ -s \cdot \ln(k \ln(k)) \ll 0 $$
or equivalently:
$$ s \gg \frac{1}{\ln(k \ln(k))} $$
Choosing $s$ proportional to $k$, we arrive at:
$$ s \sim k \cdot \ln(k \ln(k)) \tag{3} $$
$\blacksquare$
Questions
I've marked the above proof with three heuristic steps $(1),(2),(3)$ that need rigorous justification.
- Step $(1)$ Approximating the infinite tail by a single factor:
$$ \prod_{p \geq p_k} \left( 1 - p^{-s} \right) \approx 1 - p_k^{-s} $$
It seems very blunt to replace the entire product by just one factor?
- Step $(2)$ Taking the ceiling is exact only if the real-valued approximation lies within
$$ (p_k - 1,\,p_k] $$
Is there a way to ensure the approximation falls strictly in that interval?
- Step $(3)$ Suboptimal value for $s$
$$ s \sim k \cdot \ln(k \ln(k)) \tag{3} $$
Experimenting with other values such as $s \sim 2 \cdot k \cdot \ln(k \ln(k))$ also appears to work?
I welcome any feedback, suggestions, or corrections. Thanks!
