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Let   $p_1\ p_2\ \ldots$ be the sequence of all natural prime numbers. There is a slight (just slight) but clear tendency for imitating the number of primes in an interval $(p_k;\ p_n)$   by the number of primes in the double interval   $(p_k\!+p_{k+1};\ p_{n-1}\!+p_n)$; possibly by   $(2\cdot p_k; 2\cdot p_n)$   too. Let me ask two open questions along this line. The first one will be most likely hopeless while the second one may lead to a discussion and at least to numerical computations.

P1.   Does there exist a natural number   $d$   such that for every natural number   $n$   the real interval

$$ (2\cdot p_n;\ 2\cdot p_{n+d})$$

contains at least one prime?

P2.   (when P1 fails): Given a natural number   $d$,   let   $w(d)$   be the least natural number such that the interval of P1 (see above) does not contain any prime number. What is the growth of the sequence

$$w(1)\ \ w(2)\ \ w(3)\ \ldots$$


The above notions got shifted from my original definition by a half of a prime. The question Q1 below is still equivalent to question P1 above:

Q1.   Does there exist a natural number   $d>1$   such that for every natural number   $n$   the real interval

$$ (p_n\!+p_{n+1};\ p_{n+d-1}\!+p_{n+d})$$

contains at least one prime?

Q2.   (when Q1 fails): Given a natural number   $d>1$,   let   $v(d)$   be the least natural number such that the interval of Q1 (see above) does not contain any prime number. What is the growth of the sequence

$$v(1)\ \ v(2)\ \ v(3)\ \ldots$$

EXAMPLE   Consider the consecutive primes

$$p_{360} = 1901 \qquad p_{361}=1907 \qquad p_{362}=1913$$

Then the real interval

$$(p_{360}\!+p_{361};\ p_{361}\!+p_{362})\ \ =\ \ (3802; 3820)$$

contains no primes, i.e.   $v(2)\le 360$.


In general, I'd be interested in similar relative properties of primes, where primes are studied in relations to other primes, and the relation is not trivial, meaning not reduced to general properties between integers.

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1  
You might consider the growth of maximal prime gaps. I expect v exists and is superexponential. See Tomas Silva and Nicely web pages on gaps to start. –  The Masked Avenger Sep 3 '13 at 5:44
    
@TMA, I'll get hold of the paper. Thank you. –  Włodzimierz Holsztyński Sep 3 '13 at 18:27
1  
I also read of a conjecture which asserts a pair of twin primes between $p_n^2$ and $p_{n+1}^2$. –  The Masked Avenger Sep 3 '13 at 23:20
    
Twins?! -- sounds ambitious :-) –  Włodzimierz Holsztyński Sep 3 '13 at 23:47
1  
Perhaps. Not as ambitious as the (failed) conjecture that there are a pair of twin primes between $n^2$ and $(n+1)^2$. –  The Masked Avenger Sep 3 '13 at 23:55

2 Answers 2

up vote 18 down vote accepted

The answer to P1 is negative, thanks to the recent work of Maynard on bounded gaps between primes.

What Maynard shows is that given any $d$, there exists a k-tuple $h_1,\dots,h_k$ such that for infinitely many $n$, at least $d+1$ of $n+h_1,\dots,n+h_k$ are prime. In fact, the argument shows that for sufficiently large $x$, the number of $n \in [x,2x]$ such that at least $m$ of $n+h_1,\dots,n+h_k$ are prime, and the rest are almost prime in the sense that they have no prime factor less than $x^\varepsilon$ for some small fixed $\varepsilon>0$, is $\gg \frac{x}{\log^k x}$. (Such strengthenings of Zhang/Maynard type theorems are discussed in this paper of Pintz, and also in this Polymath8b preprint.) A standard upper bound sieve (e.g. Selberg sieve or beta sieve) then shows that after removing about $O( x/\log^{k+1} x)$ of these $n$, one can also ensure that none of the numbers between $2(n+h_1)$ and $2(n+h_k)$ are prime. If we let $p_i$ be the first prime greater than or equal to $n+h_1$, then we have $2(n+h_1) \leq 2p_i \leq 2p_{i+d} \leq 2(n+h_k)$, and so we obtain a counterexample to P1 for any $d$.

Unfortunately we don't get an effective rate on P2 this way due to the reliance on the Bombieri-Vinogradov theorem in the work of Maynard etc. However this can likely be removed by following the ideas mentioned near the end of this paper of Pintz, though I did not attempt this. [EDIT: it looks likely that the quantitative version of Maynard's results in this recent paper of Banks, Freiberg, and Maynard will do the trick, after some small modification.]

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The answer to your first question is most surely no. Certainly this follows from the Hardy-Littlewood $k$-tuple conjectures. Choose $n$ so that the primes $p_n$, $\ldots$, $p_{n+d}$ are all very close to each other. Hardy-Littlewood says that there are lots of such tuples. Then you are asking for one more prime in the very short interval $[2p_n,2p_{n+d}]$ (essentially) and the sieve could be used to show that this happens rarely. This will also give a conjectural way of tackling question 2 -- just find a small admissible $d$-tuple (which has been discussed before on MO).

It may even be possible to prove this unconditionally, although I don't fully see this. By Westzynthius/Erdos-Rankin you can construct largish intervals without primes, and if you halve that interval usually you will get a reasonable number of primes there.

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