It is straight forward to see that:
$$\tag0p_{\pi(n)+1}-p_{\pi(n+1)}=0 \operatorname{iff} \quad n \in {\{p_k-1:k \in \mathbb N}\}$$
however what is not as straightforward firstly, is to show that $n$ must be the lesser of a pair of twin primes if the same expression is equal to $2$:
$$\tag1p_{\pi(n)+1}-p_{\pi(n+1)}=2 \quad\operatorname{iff} \quad \exists k,j \in \mathbb N \quad\operatorname{s.t}\quad n=p_k=p_j-2$$
And furthermore, establish a proof for:
$$\tag2p_{\pi(n)+1}\equiv p_{\pi(n+1)}(\operatorname{mod}2)$$
Also an equality that I have been unable to determine the truth value of:
$$\Bigl\lfloor \frac{\ln(p_n)}{\ln(p_{\pi(n)})} \Bigr\rfloor=\Bigl\lfloor \frac{\lfloor\ln(p_n)\rfloor}{\ln(p_{\pi(n)})} \Bigr\rfloor$$
To simplify $(3)$ I defined a function $f$: $$f(n,k)=\Bigl\lfloor \frac{\ln(n+k)}{\ln(n)} \Bigr\rfloor-\Bigl\lfloor \frac{\lfloor\ln(n+k)\rfloor}{\ln(n)} \Bigr\rfloor$$
$$f(n_0,k-1)-f(n_0,k)=1$$ $$f(n_1,k-1)-f(n_1,k)=1$$ $$f(n_2,k-1)-f(n_2,k)=1$$ $$f(n,k-1)-f(n,k)=0 \quad\quad n_0 \lt n \lt n_1\land \quad\quad n_1 \lt n \lt n_2$$ Because $$n_2-n_1 \gt n_1-n_0$$ Are we able to assert that: $$n_2-n_1 \rightarrow \infty$$
and
$g(n,k)=\Bigl\lfloor \frac{\lfloor\ln(n+k)\rfloor}{\lfloor\ln(n)\rfloor} \Bigr\rfloor-\Bigl\lfloor \frac{\ln(n+k)}{\ln(n)} \Bigr\rfloor$
$g(n,k)=0 \operatorname{for} n \gt 7, k \gt 0$
Many thanks in advance.
$(2)$ can only fail if one of the primes $p_{\pi(n)+1}$, $p_{\pi(n+1)}$ is even and the other is odd. As $\pi(n)\le \pi(n+1)\le \pi(n)+1$, this requires $\pi(n)=\pi(n+1)=1$. So both $n$ and $n+1$ must be $\ge 2$ and at the same time $\le 3$.