Bounding Maximal gaps with Ramanujan primes

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Gaps and Maximal Gaps

We define terms used in this article. A prime gap as $g_n := p_{n+1} - p_n$, and we define $g_n$ as a maximal gap, if $g_i < g_n$ for all $i < n$. Define $M_{m,g}$ to be the $m$-th record maximal gap between primes $M_{m,p}$ and the next prime, so that $M_{m,g} = g_n$. The prior maximal gap is $M_{m-1,g} = h_n$. Clearly, in the interval [$M_{m-1,p},M_{m,p}$) all gaps are

$$g_n \le h_n.\qquad\qquad\text{ (1)}$$

Ramanujan Primes

The definition of Ramanujan primes from Wikipedia:$^{[2]}$

The $n$-th Ramanujan prime is the least integer $R_n$ for which $$\pi(x) - \pi(x/2) \ge n,$$ for all $x \ge R_n$. In other words: Ramanujan primes are the least integers $R_n$ for which there are at least $n$ primes between $x$ and $x/2$ for all $x \ge R_n$.

Obvious Observations with Ramanujan Primes

Every prime $p$ is bounded by a pair Ramanujan primes $R_n, R_{n+1}$. If $p = 2$, then $R_1 = p = 2 < R_{1+1} = 11$; for other primes, $ R_n < p \le R_{n+1}$.

Every prime gap prime pair, $p_k, p_{k+1}$ for $g_k = p_{k+1} - p_k$, is bounded by a pair Ramanujan primes $R_n, R_{n+1}$$.^{[1]}$

$$p_s = R_n \le p_k,\space p_{k+1} \le R_{n+1} = p_t.$$

Clearly, with the above statements, the bounding of maximal gaps with Ramanujan primes happens.

Primes Associated with Ramanujan Primes

There are four primes associated to each Ramanujan prime; two are for finding the Ramanujan prime itself $p_{\pi(p_{(s/2)})}$ and $p_{\lfloor2n\alpha\rfloor}$. The other two primes associated to each Ramanujan prime by the Ramanujan Prime Corollary, $p_{i-n}$ (the smaller) and $p_i$ (the larger).

Ramanujan Prime Corollary

They hold to the following inequality where $s = \pi(R_n)\text{ for }s < i$,

$$p_i < 2p_{i-n}.\qquad\qquad\text{ (2)}^{[1,2]}$$

It is important to note where these primes are on a number line (assuming $i=s + 1$):

$$p_{s-n} = p_{\pi(p_s/2)} < p_{s}/2 < p_{i-n} < p_{2n} < p_s < p_i < p_{\lfloor2n\alpha\rfloor} < p_{3n}.$$

For larger values of $i$, $p_{i-n} = p_{2n}$ or $p_i = p_{\lfloor2n\alpha\rfloor}$ can happen, but when $p_i = p_t$ happens, it is clear that the next Ramanujan prime can make the results sharper.

Theorem 1

Because of (1) and (2), all gaps up to the next maximal gap which is between the $[R_n, R_{n+1})$, $$p_s = R_n < p_{i},\space p_{i+1} \le R_{n+1} = p_t.$$ $$g_{i-n} \le h_s.$$ So, $g_{i-n}/h_s < 2$.

Under assumption that $\frac{M_{n+1}}{M_n} \le 2$, what is true?

[1] https://www.emis.de/journals/INTEGERS/papers/p52/p52.pdf

[2] https://en.wikipedia.org/wiki/Ramanujan_prime

What work do I need to do on this proof? Does anyone get what I am saying?