# Cramer’s conjecture implies twin prime conjecture

Let’s define the quantity $G(x):=\sup\{p_{n+1}-p_{n}\mid \frac{p_{n}+p_{n+1}}{2}\le x\}$, and assume Cramer’s conjecture. Then $K_{m}:=\sup\{\frac{G(x)}{\log^2 x}\mid x>2\}<\infty$. Let’s denote by $U(n)$ the event $n$ is $1$-central, and by $H(n)$ the event $(n-1,n+1)$ is a couple of twin primes. Then from de Bayes’ theorem, one gets $P(H(n))=\frac{P(H(n)\cap U(n))}{P(U(n)/H(n))}=P(H(n)\cap U(n))$, since if $n$ is the half sum of a couple of twin primes, then necessarily $n$ is $1$-central. But $P(H(n)\cap U(n))$ is equal to the ratio of the number of quantities $2r_{0}(n)$ such that $n$ is the half sum of twin primes to the number $G(n)$ as defined above, hence $P(H(n))\ge\frac{1}{K_{m}\log^{2} n}$. As $P(H(n))=\frac{1}{2}\frac{\pi_{2}(n)}{n}$, one finally gets $\pi_{2}(x)\ge\frac{2x}{K_{m}\log^{2} x}$. Hence the assumption of Cramer’s conjecture implies that there are infinitely many twin primes.

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# Langlands functoriality conjecture implies GRH

See the attached document in http://www.les-mathematiques.net/phorum/read.php?43,1210447

The assumption of Langlands functoriality conjecture ensures that the considered class of L-functions is closed under both usual and tensor products. This last property holds unconditionally for the class of L-functions generated by the non negative powers of the Riemann Zeta function, which entails by the same arguments the truth of RH.

We defined $r$ to be a primality radius of $n$ iff both $n-r$ and $n+r$ are primes, which requires $r. If we slightly soften this assumption, writing $(\vert n-r\vert,n+r)\in\mathbb{P}^{2}$, we can say that a classical primality radius is a natural primality radius and that $n$ is a relative primality radius of $r$ iff  $r$ is a natural primality radius of $n$.

We thus obtain some kind of duality, where every statement of the form $P(n,r,\ \ natural \ \ primality \ \ radius)$ where $P$ is a ternary predicate, is equivalent to $P(r,n, \ \ relative \ \ primality \ \ radius)$.

As a consequence, we get that the reformulation of Goldbach’s conjecture as « every positive integer $n>1$ admits a natural primality radius $r$ » is equivalent to this weak form of de Polignac’s conjecture: « every positive integer $r>1$ admits a relative primality radius $n$« , which is obviously equivalent to saying that every even integer is the difference of two (non necessarily consecutive) primes.

The strong de Polignac conjecture is equivalent to saying that « every positive even integer is the difference of two consecutive primes in infinitely many ways » or that « every positive integer $r$ is the fundamental (natural) primality radius of infinitely many $1$-central numbers $n$ « .

# A PNT based variance inequality for Cramer’s conjecture

The Prime Number Theorem (PNT for short) says that the average gap between two consecutive primes of size $n$ is $\sim\log n$. Defining the quantity $\rho_{i}(n)$ as $\frac{2r_{i-1}(n)}{\pi(n+r_{i-1}(n))-\pi(n-r_{i-1}(n))}$, where $r_{k-1}(n)$ is the $k$-th typical primality radius of $n$, one can expect $\rho_{i}(n)$ to get closer to $\log n$ as $i$ increases for a given $n$.

Let’s define the « gap-variance » of $n$, denoted by $\sigma^{2}_{\rho}(n)$, as follows:

$\displaystyle{\sigma^{2}_{\rho}(n):=\frac{1}{N_{2}(n)}\sum_{i=1}^{N_{2}(n)}\vert \rho_{i}(n)-\log n\vert^{2}}$

where $N_{2}(n)$ is the total number of typical primality radii of $n$.

From the observation above, one can write $\displaystyle{\sigma^{2}_{\rho}(n)\leq \frac{1}{N_{2}(n)}\sum_{i=1}^{N_{2}(n)}\vert\rho_{1}(n)-\log n\vert^{2}}$

Hence $\sigma^{2}_{\rho}(n)\leq\vert\rho_{1}(n)-\log n\vert^{2}$.

As, whenever $m$ is $1$-central, one has $\rho_{1}(m)=2r_{0}(m)$, one gets for such $m$:

$\sigma^{2}_{\rho}(m)\leq(2r_{0}(m)-\log m)^{2}\leq 4r_{0}(m)^{2}-4r_{0}(m)\log m+\log^{2} m$

So that $\sigma_{\rho}^{2}\leq \log^{2}(m)+4r_{0}(m)(r_{0}(m)-\log m)$

Hence, $r_{0}(m)\leq \log m\Rightarrow\sigma^{2}_{\rho}(m)\leq\log^{2}m$, so that it’s very likely that the strong form of Cramer’s conjecture, namely $\displaystyle{\limsup_{k\to\infty}\frac{p_{k+1}-p_{k}}{\log^{2}p_{k}}=1}$, is true.

# A possible way to tackle the twin prime conjecture

Assume that there are only a finite number $m=2N$ of twin primes greater than $4$ sorted in increasing order and let’s denote them $j_{1}=5, j_{2}=7,\cdots, j_{m-1},j_{m}$.

Let’s now formulate the following Prim conjecture:

There exists a primorial $P$ greater than $2j_{m}$ such that both $j_{m-1}$ and $j_{m}$ are primality radii of $P$.

If true, this would entail that there are at least $m+2$ twin primes greater than $4$ and not $m$. Hence the infiniteness of twin primes.

I postulate that $P$ can be expressed as a function of $m$ by solving the following equation:

$\pi(\sqrt{2P-3})=\pi(j_{m})$.

This is just the very beginning of a sketch of proof which obviously needs further investigations.

# Proof that eta(n)>2: case n multiple of 3

Suppose $n>26$ is a multiple of $3$. One can assume without loss of generality that $r_{0}(n)>2$. Hence $2r_{0}(n)$ is a Goldbach gap greater than $5$ and thus there are at least three different Goldbach gaps, namely $2$, $4$ and $2r_{0}(n)$, hence $\eta(n)>2$.

This and the previous articles of this blog entail that every even integer greater than $3$ is the sum of two primes.

# Proof that eta(n)>2: case n coprime with 3

Suppose $n>26$ is coprime with $6$. Then every potential typical primality radius of $n$ is a multiple of $6$. But as $84=6\times 14$ is a multiple of $7$ less than $210$, and $90=6\times 15$ is a multiple of $5$ less than $210$, it follows that $84$ and $90$ can’t be potential typical primality radii of $n$ since each of them shares with $n$ a residue class mod $p$ for $p\in\{5,7\}$.
Hence $\eta(n)>2$ (since one has at least $3$ different Goldbach gaps, namely of size $6$, $12$ or $18$).

The same reasoning can be applied for $n>26$ even and coprime with $3$ by replacing $84$ with $189$ and $90$ with $195$.

# Explicit upper bound for r0(n)

Numerical computations seem to show that $r_{0}(n)$ is always less than $0.07\log^{4} n$ whenever $n>28$. Solving the equation $x=0.07\log^{4} x$, one gets as a threshold the value $x0\approx 0.25216...$, hence every even integer greater than $3$ is the sum of two primes, as the small cases (less than $57$) can be easily checked by hand.

# sharp upper bound for alpha_{n}

As mentioned in the previous article, it appears that the upper bound $\alpha_{n}=O_{\varepsilon}(n^{1/2+\varepsilon})$ would follow from the following reasonable assumption:

$N_{2}(n)$ is « rather close » to $N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}$

This follows from the very definition of what a typical primality radius is. Indeed, writing

$N_{2}(n)=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

one gets:

$\dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

Hence

$1+\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\left(\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}\right)+O(f(n)\dfrac{P_{ord_{C}(n)}}{n.N_{1}(n)})$

So $\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}-\dfrac{n(P_{ord_{C}(n)}-\sqrt{2n-3})}{n(P_{ord_{C}(n)}-\sqrt{2n-3})}+O(f(n)\dfrac{\log^{2} n}{n})$

Thus $\alpha_{n}=\dfrac{(n-P_{ord_{C}(n)})\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n)\log^{2} n)$

Finally $\vert\alpha_{n}\vert=(\sqrt{2n})^{1+\varepsilon}+O(f(n)\log^{2}n)$

Assuming $f(n)=\sqrt{n}$ (which remains to be proven), one gets $\vert\alpha_{n}\vert=O_{\varepsilon}(n^{1/2+\varepsilon})$. Maybe some kind of central limit theorem could shed a new light on the true nature of $f(n)$.

# A conjectural expression for N2(n)

The definition of a « typical » primality radius $r$ of $n$ leads to the following assumption:

the following ratio:

$\dfrac{N_{2}(n)}{N_{1}(n)}$

is asymptotically equal to the ratio:

$\dfrac{n-\sqrt{2n-3}}{P_{ord_{c}(n)}-\sqrt{2n-3}}$

Hence, for $n$ large enough, the quantity $N_{2}(n)$ should be « close » to the expression above times $N_{1}(n)$.

This implies $\alpha_{n}=o(n)$, as one can easily figure out replacing $N_{2}(n)$ by its exact expression given by:

$\frac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})$

in the following limit:

$\frac{N_{1}(n)}{N_{2}(n)}\left(\frac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}-\frac{N_{2}(n)}{N_{1}(n)}\right)\to 0$

We will show in the next article that one might even get:

$\alpha_{n}=O_{\varepsilon}(n^{1/2+\varepsilon})$.