Lower bound for $p_{n^3}-p_{(n-1)^3}$?

Question

The difference between two primes is at least $2$ so $p_{n^3}-p_{(n-1)^3} \geq 6n^2$. Is there any known sharper bound?

Answer 1

Looking at simple sieving, there are at most 2 primes in $[n, n+6)$, 8 primes in $[n, n+30)$, and so on for the product of for first $k$ primes, the number being $\prod_{i=1}^k (p_i-1)$.

Therefore if $r_k =\prod_{i=1}^k \frac{p_i}{p_i-1}$, $p_b-p_a \ge (b-a)r_k $.

Your case has $b-a = 3n^2+3n+1$ and $k = 1$ with $r_1 = \frac{2}{1} =2$.

With $k=2$, $r_2 =\frac{2\cdot 3}{1\cdot 2} =3$, so we get $9n^2$.

With $k=3$, $r_3 =\frac{2\cdot 3\cdot 5}{1\cdot 2\cdot 4} =\frac{15}{4}$, so we get $\frac{45}{4}n^2$.

In general, the difference is at least $3r_kn^2$.

By Merten's third theoren (https://en.wikipedia.org/wiki/Mertens%27_theorems), $\lim_{n \to \infty} \ln(n)\prod_{p \le n} (1-\frac1{p}) =e^{-\gamma}$, where $\gamma$ is the Euler–Mascheroni constant. Therefore $\prod_{p \le n} \frac{p}{p-1} \approx e^{\gamma}\ln(n) $.

Therefore, since $p_k \approx k \ln(k)$, $r_k \approx e^{\gamma}\ln(k \ln k) $.

To use $k$ primes, you must have $n^3 \gt \prod_{i=1}^k p_i$. Since $ \prod_{p\le n} \ln p \approx n$ (see https://en.wikipedia.org/wiki/Chebyshev_function), this means that $n^3 \gt e^{k\ln k+O(k \ln\ln k)} $ so, arguing non-rigorously, $k\ln k+O(k \ln\ln k) \le 3\ln(n) $.

Choosing this expression for the max $k$, we can get $r_k \approx e^{\gamma}\ln(3\ln(n)) $.