Problems & Puzzles: Conjectures

Conjecture 74. What is true assuming valid this...

John W. Nicholson pointed out the following question posed in the following web page, that I simply copy here for our readers:

What can be proved if the conjecture that

Mn+1/Mn <=2 and lim[(Mn+1/Mn)=1] as n->infinite

is assumed to be true?

Q. In particular, does this conjecture prove Shanks in [2], the "Cramér's model"[3], or something else?

See  the definition of Mn in the mentioned page.

On Sept 8, 2022 John W. Nicholson wrote:

I have not publish this solution and would like some review of it. In Srinivasan and Nicholson,  INTEGERS, 2015 paper in particular Lemma 1.ii. LINK: http://emis.muni.cz/journals/INTEGERS/papers/p52/p52.pdf

I show that  p_(s+k) < 2*p_(s+k-n) for all positive integers k. Where p_s is the n-th Ramanujan prime, so R_n = p_s. If you let i = s+k, the inequality becomes p_i < 2 p_(i-n).

Note that with k=0, we have Lemma 1.i, and note that p_pi(R_n/2) = p_pi(p_s/2) = p_(s-n) where pi(x) is the prime count function.

Anyways, let us pick the largest Ramanujan prime < next unknown maximal prime gap prime, R_n < P_m where P_m is the m-th record prime of p_(g+1) - p_g < p_(h+1) - p_h for all g < h and G_m is the value of p_(h+1) - p_h at P_m.

Because all of the primes < next unknown maximal prime gap prime, p < P_m, have prime gaps <= the largest known maximal prime gap, all of the primes p are p_(s-n) < p_(i-n) < p_s < P_m and have prime gaps g_(i-n) so that for all s-n < i <= s are g_(i-n) <= G_(m-1).

Now each of the primes p_(i-n) bounds the primes p_i by p_i < 2 p_(i-n), but p_(s-n) < p_(i-n) < p_s, so p_(i-n+1) - p_(i-n) is < G_(m-1). This means that the gaps are found to be 2 <= g_i < G_(m-1) unless it is a maximal gap. And if it is a maximal gap then G_(m-1) <  g_i = G_m < 2 * G_(m-1). Which is what we wanted to prove.

This might seem to be over simplified, but it is what I have got for now.

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