Let $\ \mathbb J_n\,:=\,\{1\ \ldots\ n\}\ $ be the initial interval of natural numbers, and $$2=p_0<p_1<\ldots$$ be the increasing sequence of all primes. Let

$$ \forall_{n=1\ 2\ \ldots}\ \ d_n\, :=\, \frac{p_n-p_{n-1}}{p_{n+1}-p_{n-1}} $$

so that $\ p_n = (1-d_n)\cdot p_{n-1} + d_n\cdot p_{n+1}.\ $Consider arbitrary reals: $$0<a<b<1$$


Can you prove or disprove or provide a correct limit or that there is none or a reference for equality:

$$ \lim_{n=\infty}\ \frac 1n\cdot \left |\{k\in\mathbb J_n: a\le d_k\le b\}\right|\ =\ b-a $$

One may also consider the related questions about the frequency of the inequalities $\ p_n^2 > p_{n-1}\cdot p_{n+1}\ $ or $\ p_{n+1}\cdot p_{n+2} > p_n\cdot p_{n+3},\ $ or even some more precise questions too. (There is a slight bias toward the $\ >\ $ inequality because it holds when the intermediate primes or their arithmetic mean are close to the arithmetic mean of the external pair of primes; but this wouldn't matter in the limit, or would it?)

  • $\begingroup$ In your last limit line, shouldn't it be $a\le d_k \le b$, or have I misunderstood your claim? $\endgroup$ – Alan Jan 25 '16 at 19:42
  • $\begingroup$ I dont understand. Do you mean $d_n$ or $d_k$? $\endgroup$ – Sylvain JULIEN Jan 25 '16 at 19:45
  • $\begingroup$ I am so sorry--I fixed the typo, yes, there should be $d_k$, not $d_n$ at its last apperance of $d_?$. $\endgroup$ – Włodzimierz Holsztyński Jan 25 '16 at 19:49
  • 1
    $\begingroup$ Perhaps you can get insights from the symetric density conjecture I consider in mathoverflow.net/questions/132973/… $\endgroup$ – Sylvain JULIEN Jan 25 '16 at 20:38
  • 1
    $\begingroup$ I am struck by the thought that this is equivalent to the prime k-tuples conjecture. It probably isn't, but I could imagine some strong links between the two. Gerhard "Walk The Imaginary Bridge First" Paseman, 2016.01.25 $\endgroup$ – Gerhard Paseman Jan 26 '16 at 0:53

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.