Let $G_n(X)$ denote the $n^{th}$ record largest gap between consecutive primes below $X$, which some call maximal gap.

Let $G_{n+1}(Y)$ denote the size of next larger maximal record gap in size from $G_n(X)$ with consecutive primes below Y.

For example, we know the forth maximal prime gap uses the primes 23 and 29. So, $G_4(23) = 29-23 = 6$. The next prime gap is 2, but the one following it is also 37-31 = 6. Yes, a gap of 2 is smaller than $29-23$, but $G_4(29) = 6$, and $G_4(31) = 6$ while not being a maximal. This holds until $G_5(89) = 8$ at the $5^{th}$ maximal of $97-89$. In other words, $G_n(X)$ is the worst case function for gaps until the next worst case, which happens at the next maximal.

Denote the primes $p_x$, $p_{x+1}$ and $p_y$, $p_{y+1}$ as the primes starting and ending the largest maximal gap less than $X$ and $Y$ respectfully.

Note: All gaps $g_i = p_{i+1} - p_i$, where $p_i < p_y$ are $ g_i <= G_n(X)$. This includes $g_{y-1} = p_y - p_{y-1}$.

Can these statements be proved:

$$\frac{G_{n+1}(Y)}{G_n(X)} \le 2?$$


$$\lim_{n \to \infty} \frac{G_{n+1}(Y)}{G_n(X)} = 1?$$

Here is my try on the first:

Start with Ramanujan primes, see https://en.wikipedia.org/wiki/Ramanujan_prime.

"The $n^{th}$ Ramanujan prime is the least integer $R_n$ for which $\pi(x) - \pi(x/2) \ge n$, for all $x \ge R_n$."

And by the RPC, https://en.wikipedia.org/wiki/Ramanujan_prime#Ramanujan_prime_corollary,

$2p_{i-n} > p_i$ for $i > k$ where $k = \pi(p_k) = \pi(R_n)$, i.e. $p_k$ is the $k^{th}$ prime and the $n^{th}$ Ramanujan prime.

Let $R_b$ be the largest Ramanujan prime $R_b < p_y \le Y$.

Let $p_i = p_y$, then the prime $p_{i+1} = p_{y+1}$ is bounded by $2p_{i-n+1} = 2p_{y-b+1}$ by the RPC. Under the assumption that the smaller gap is $g_{y-b} = G_n(p_{y-b})$, we see that the larger gap is $p_{y+1} - p_y = g_y = G_{n+1}(p_y) \le 2G_n(X)$.

Does this make sense? Can someone verify this proof?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.