Can the minimum of two consecutive prime gaps become arbitary large? Here : 
https://oeis.org/A023186
the so-called "Lonely primes" are shown. 
Let $$[a,b,c]$$ be a triple of consecutive primes and define $$d:=\min(c-b,b-a)$$
My question :

Can we prove that $d$ can become arbitary large ? In other words, can we prove that the minimum of two consecutive prime gaps can become arbitary large ?

With the chinese remainder theorem we can construct a positive integer $N$, such that the each number in the range $$[N-d,N+d]$$ except $N$ (d is some positive integer ) must have a prime factor not exceeding $p_{2d}$ But can we show that $N$ can be prime (perhaps by using the Dirichlet theorem) and that also $N-d>p_{2d}$ can be satisfied ?

If you read that OEIS page you linked to, it says

Erdős and Suranyi call these reclusive primes and prove that there are an infinite number of them.

(Emphasis mine.) Each lonely prime is the $b$ of a consecutive prime triple $[a, b, c]$ for which $d = \min(b-a, c-b)$ is larger than for any previous prime triple. Since this sequence of lonely primes continues infinitely, there must be arbitrarily large $d$.
In particular, two consecutive gaps of size at least $2k$ must occur at the latest on either side of lonely prime number $k$ (counting $2$ as the $0$th lonely prime). If you instead take sequence number 120937, then it duplicates the lonely primes such that it happens for the first time exactly at term number $k$.
