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.