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.

1more comment