This question was hinted upon with the still open question at [1].

Let $M_n = $ A005250($n$) of the OEIS. That is to say, $M_n = p_{i+1}-p_i$, where $p_i$ is the smallest prime such that $p_{i+1} - p_i > p_{j+1} - p_j$ for all $j < i$.

It was stated at [1] "While the extra information of all of the answers is nice, it appears that proving this conjecture would lead to disprove something with the "heuristic analysis using Cramér's model" and how its is used." Well, this is that question.

What can be proved if the conjecture that $$\frac{M_{n+1}}{M_n} \le 2 \text{ and that}\lim_{n\to \infty}\frac{M_{n+1}}{M_n}= 1$$ is assumed to be true? In particular,does this conjecture prove Shanks in [2], the "Cramér's model"[3], or something else?

[1] Is there a conjecture with maximal prime gaps.

[2] Shanks, Daniel (1964), "On Maximal Gaps between Successive Primes", Mathematics of Computation (American Mathematical Society) 18 (88): 646–651;


For one the conjecture implies that between any two integers

$p$ and $2p$ where p is a prime belonging to M there is guaranteed to be another prime in M.

Furthermore the right side implies that for any multiplicative constant $h > 1$

There exists infinitely many primes P in M whose density goes up as the magnitudes considered increases such that between

$P$ and $hP$

There exists another prime

  • These of course are trivial derivations – frogeyedpeas Jun 12 '14 at 14:49
  • Is that it? Seems like there would be more to it because of the reaction at [1]. – user160140 Jun 26 '14 at 13:33
  • From , "However, from the prime number theorem one can also deduce an upper bound on the length of prime gaps: for every ε > 0, there is a number N such that g_n < εp_n for all n > N." So, because g_n = p_{n+1} - p_n, p_{n+1} < p_n(1+ε). Because ε > 0 ε +1 can equal h of frogeyedpeas that part is proved. – John Nicholson Nov 30 '17 at 1:39

This conjecture has a sequence. See OEIS A053695.

This conjecture has also has a solution: Bounding Maximal gaps with Ramanujan primes

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