Patterns in prime numbers 
*

*$13$ is prime

*$\frac {14}2$ is prime

*$\frac {15}3$ is prime


The same thing also holds for $2017$, and I've found cases where this holds even up to $6$, others have shown it works even up to $10$.
My question:

Is there a way to show that there will always exist a prime number such that 
  $p$, $\frac{p+1}{2}, \frac{p+2}{3}, \frac{p+3}{4},\dots, \frac{p+n}{n+1}$ are all prime numbers?

What is the largest value of $n$ for which this is possible?

Below is just a proof of a link between $p$ and $n$, thus, not an answer.

Primes larger than $5$ can be partitioned into 2 subsets of type $6n+1$ and $6n+5$. It's easy to see that $p+1=6n+5+1=6(n+1) \Rightarrow \frac{p+1}{2}=3(n+1)$ which is not a prime. So, we can discard $6n+5$ class of primes. Particularly, $13=6\cdot2+1$.

We will be looking at the primes $p=6n+1$. We can write $6n=2^{k}n_1$, where $n_1$ is odd, i.e.
$$p=2^{k}n_1+1 \tag{1}$$
and $n_1>1$, since there is $3$ in $6n$.

Proposition $$n<2^k-1 \tag{2}$$
Easy to see from:
$$q_1=\frac{p+1}{2}$$
$$q_2=\frac{p+2}{3}$$
$$...$$
$$q_{2^k-1}=\frac{p+2^k-1}{2^k}\overset{(1)}{=}\frac{2^{k}n_1+1+2^k-1}{2^k}=n_1+1$$
But $n_1$ is odd, thus $q_{2^k-1}=n_1+1$ is even and, thus, definitely not prime.
