# Subset of the Set of Prime Gaps whose elements are always even

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

$$g(n,k)=\Bigl\lfloor \frac{\lfloor\ln(n+k)\rfloor}{\lfloor\ln(n)\rfloor} \Bigr\rfloor-\Bigl\lfloor \frac{\ln(n+k)}{\ln(n)} \Bigr\rfloor$$

$$g(n,k)=0 \operatorname{for} n \gt 7, k \gt 0$$

• Not sure I follow. Aside from $p=2$, all primes are odd. – lulu Nov 7 '18 at 21:13
• Correct yes I meant to express $\tag1p_{\pi(n)+1}-p_{\pi(n+1)}$ appears to be always even – Adam Nov 7 '18 at 21:15
• I'm sorry I go very crazy when ever I think of Goldbach's so yes $(2)$ is very obvious I've just drunk too much tonight – Adam Nov 7 '18 at 21:35
$$(2)$$ 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 $$\le 3$$.