Starting backward from $0$, the only thing that must be proved is: For any $n\in \mathbb{N}$ there exists a prime $p$, such that $p+1,p+2,\dots, p+n-1$ are not primes. It follows from the fact that the density of primes is $0$.

I have a construction as follows. Let's denote a prime $k_m$ be the largest prime less than $n!$. Here is a sequence $a_n$=$k_{n+1}$+$a_{n-1}$. Here, $a_1$=2 and $n$>0. Notice that the 1000th term of this sequence satisfies the condition.

dgrozev wrote: Starting backward from $0$, the only thing that must be proved is: For any $n\in \mathbb{N}$ there exists a prime $p$, such that $p+1,p+2,\dots, p+n-1$ are not primes. It follows from the fact that the density of primes is $0$. can you elaborate it

Ok, what number $N_0$ should we start from, so that to obtain $N_0,N_1,\cdots N_{1000}=0$? What is $N_{999}$? It can be any prime number, say $p_1$. What about $N_{998}$. We search for a prime number $p_2$ and $N_{998}>p_2$, such that $N_{998}-p_2=N_{999}$ and there are no other primes between $p_2+1,p_2+2,\dots, p_2+N_{999}=N_{998}$. We can we find such $p_2$, ok? Keep going, and you'll get to $N_0$.