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.






Langlands functoriality conjecture implies GRH

See the attached document in,1210447

Just click on Ouvrir (open) or télécharger (download) to read the pdf.

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.

Relative primality radius

We defined r to be a primality radius of n iff both n-r and n+r are primes, which requires r<n. 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:


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

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.

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


one gets:




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:


is asymptotically equal to the ratio:


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:


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: