EDIT : I copy-paste the beginning of a previous question since Gerry Myerson suggested this question should be self-contained.

"let's consider a composite natural number $n$ greater or equal to $4$. Goldbach's conjecture is equivalent to the following statement: "there is at least one natural number $r$ such as $(n-r)$ and $(n+r)$ are both primes". For obvious reasons $r\leq n-3$. Such a number $r$ will be called a "primality radius" of $n$.

Now let's define the number $ord_{C}(n)$, which depends on $n$, in the following way: $ord_C(n):=\pi(\sqrt{2n-3})$, where $\pi(x)$ is the number of primes less or equal to $x$. $(n+r)$ is a prime only if for all prime $p$ less or equal to $\sqrt{2n-3}$, $p$ doesn't divide $(n+r)$. There are exactly $ord_{C}(n)$ such primes. The number $ord_{C}(n)$ will be called the "natural configuration order" of $n$. Now let's define the "$k$-order configuration" of an integer $m$, denoted $C_{k}(n)$, as the sequence $(m \ \ mod \ \ 2, \ \ m \ \ mod \ \ 3,...,m \ \ mod \ \ p_{k})$. For example $C_{4}(10)=(10\ \ mod \ \ 2,\ \ 10 \ \ mod \ \ 3, \ \ 10 \ \ mod \ \ 5, \ \ 10 \ \ mod \ \ 7)=(0,1,0,3)$. I call $C_{ord_{C}(n)}(n)$ the "natural configuration" of $n$.

A sufficient condition to make $r$ be a primality radius of $n$ is that for all integer $i$ such that $1\leq i\leq ord_{C}(n)$, $(n-r) \ \ mod \ \ p_{i}$ differs from $0$ and $(n+r) \ \ mod \ \ p_{i}$ differs from $0$. If this statement is true, $r$ will be called a "potential typical primality radius" of $n$."

Is it true that Dirichlet's theorem about primes in arithmetic progression implies that there exists $k_n$ depending only on $n$ such that the number $\mathcal{N}_{n}(x)$ of potential typical primality radii of $n$ less than $x$ is such that $\mathcal{N}_{n}(x)\sim k_{n}x$? If so, the expression of $k_{n}$ is quite easy to find out.
Thanks in advance.


You seem to be using certain variable inputs in two different ways. First, $r$ is assumed to be less than $n-3$. Then, you define a notion of potential typical primality radius of $r$, which depends only on remainders of division by primes less than $\sqrt{2n-3}$. It is not clear in this definition that you are assuming $r$ is bounded by $n-3$. Finally, you discuss asymptotics of potential typical primality radii $x$ (or at least I assume that is the meaning of the symbol $\sim$ in the expression $\mathcal{N}_n(x) \sim k_n x$), and this requires $x$ to increase without bound.

If we assume you meant to ignore the bound, then the asymptotic estimate is correct, but it doesn't depend on Dirichlet's theorem. Instead, it only depends on the Chinese remainder theorem, because you are making a statement about counting residue classes modulo the product of all primes less than $\sqrt{2n-3}$. Also, it's not clear how much benefit you will derive from this, since you haven't eliminated the possibility that $k_n = 0$.

| cite | improve this answer | |
  • $\begingroup$ Thanks a lot. Actually $k_n$ can't be equal to $0$, since otherwise the number of potential typical primality radii less than $P_{ord_C(n)}$ (the product of the first $ord_c(n)$ primes) would be equal to $0$ too. But it's not difficult to show that $k_n=\dfrac{\phi(P_{ord_{C}(n)})}{P_{ord_{C}(n)}}\prod_{p}\dfrac{p-2}{p-1}$, where $p$ runs through the set of odd primes not dividing $n$, hence $k_n$ is never equal to $0$ for sufficiently large composite $n$. $\endgroup$ – Sylvain JULIEN May 11 '13 at 16:33
  • $\begingroup$ By the way, $\phi$ denotes Euler's totient function in the previous comment. $\endgroup$ – Sylvain JULIEN May 11 '13 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.