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 http://www.les-mathematiques.net/phorum/read.php?43,1210447

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:

\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 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

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}).