# Would the following conjectures imply Cramer's conjecture?

Assume Goldbach's conjecture. Then for every $n\ge 2$ there exists at least one non-negative integer $r\le n-2$ such that both $n+r$ and $n-r$ are primes. Let's write $r_{0}(n):=\inf\{r\le n-2, (n-r,n+r)\in\mathbb{P}^{2}\}$ and $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$.

I call $r$ a primality radius of $n$, $r_{0}(n)$ the fundamental primality radius of $n$ and $k_{0}(n)$ the order of centrality of $n$. I say that $n$ is a $k$-central number if and only if $k_{0}(n)=k$.

Now let's consider $\rho(n):=\dfrac{2r_{0}(n)}{k_{0}(n)}$ and $\sigma(n):=\dfrac{\log(\rho(n))}{\log\log n}$. Let $\sigma_{+}:=\lim\sup \sigma(n)$ and $\sigma_{-}:=\lim \inf\sigma(n)$ and $\sigma_{m}:=\dfrac{\sigma_{+}+\sigma_{-}}{2}$. Twin prime conjecture implies $\sigma_{-}=0$, while Cramer's conjecture implies $\sigma_{+}=2$. The Prime Number Theorem asserts that, on average, $\sigma(n)=1$, so that $\sigma_{m}$ is very likely to be equal to $1$.

Let's now define the following density: $\delta_{\varepsilon,x}(\sigma):=\dfrac{\vert\{n\leq x, \sigma(n)\in(\sigma-\varepsilon,\sigma+\varepsilon)\}\vert}{x}$.

I formulate the following conjectures:

Symmetric Density Conjecture:

$\displaystyle{\forall \sigma\in(\sigma_{-},\sigma_{+}), \lim_{\varepsilon\to 0}\lim_{x\to \infty}\dfrac{\delta_{\varepsilon,x}(\sigma)}{\delta_{\varepsilon,x}(2\sigma_{m}-\sigma)}=1}$

Increasing Density conjecture:

$\displaystyle{\sigma_{-}\leq \sigma_{1}\leq \sigma_{2}\leq \sigma_{m}\Longrightarrow\lim_{\varepsilon\to 0}\lim_{x\to \infty}\dfrac{\delta_{\varepsilon,x}(\sigma_{1})}{\delta_{\varepsilon,x}(\sigma_{2})}\leq 1}$

Would these last two conjectures plus the twin prime conjecture, if proven, imply Cramer's conjecture? Thanks in advance.

EDIT: I think that the fact that whenever $r$ is a "potential typical primality radius" of $n$, then so is $P_{ord_{C}(n)}-r$ (see About Goldbach's conjecture), implies symmetric density conjecture. I offer a bounty to anyone who'll be able to prove (or disprove) this.

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You don't need the twin prime conjecture to get $\sigma_-=0$; Zhang's proof of the Bounded Gap Conjecture suffices. –  Charles Jun 19 '13 at 19:05
@Charles : Indeed. –  Sylvain JULIEN Jun 19 '13 at 21:18
Let $\displaystyle{\delta(\sigma):=\lim_{\varepsilon\to 0}\lim_{x\to\infty}\delta_{\varepsilon,x}}$. Then Symmetric Density conjecture implies $\sigma_{m}:=\int_{\sigma_{-}}^{\sigma_{+}}\sigma\delta(\sigma)d\sigma$. –  Sylvain JULIEN Jun 21 '13 at 20:49
Read $\displaystyle{\lim_{\varepsilon\to 0}\lim_{x\to\infty}\delta_{\varepsilon,x}(\sigma)}$ in the previous comment. –  Sylvain JULIEN Jun 21 '13 at 20:52