Generalization of Bertrand's Postulate

Bertrand's postulate states that there is a prime $p$ between $n$ and $2n-2$ for $n>3$. According to Dirichlet's theorem we have that a sequaence $$a\cdot n+b$$ has infinite primes iff $a$ and $b$ are relatively prime. So in some sense, Bertrand's postulate gives a maximum of time for encountering a prime in the sequence $$2\cdot n+1$$ So, the question is: there is a generalization of Bertrand's Postulate for sequences $a\cdot n+b$ that accomplish the Dirichlet's theorem?

EDIT: (For a more concise explanation of the particular generalization.) We know that given $$a_n=2\cdot n+1$$ we have that for all $m$ there is a prime in the sequence greater than $a_m$ and less than $a_{2m}$. So, the thing is that if there is some generalization of Bertrand's Postulate using the sequence form, for an arbitrary sequence $$c_n=a\cdot n+b$$ with $a$ and $b$ coprime. Something as, for every relatively prime $a$ and $b$, there is a $k\leq a\cdot b$, such that for all $m$ there is a prime in the sequence between $c_m$ and $c_{k\cdot m}$.

Such kind of thing is what I am looking for.

Solution 1:

One generalization of Bertrand's postulate I know is a theorem of Sylvester and Schur. See for example http://www.math.uiuc.edu/~pppollac/sschur.pdf

The theorem says that for any positive integer $k$ the product of $k$ consecutive integers greater than $k$ contains a prime factor greater than $k$.

I hope this helps you somehow.

Solution 2:

The answer is that there is such a generalization of Bertrand Postulate. For fizing constants, in the rest of the answer $a$ and $b$ will be coprime numbers, and $c_n$ will be given the sequence by

$$c_n=a+bn$$

We consider the prime number theorem for arithmetic progressions, this say that

$$\pi_{a,b}(x)\sim \frac{1}{\varphi(b)}\frac{x}{\log x}\text{,}$$

where

$$\pi_{a,b}(x)=\text{Card}\{p\leq x\,|\,p\text{ is prime and }p\equiv a\text{ (mod }b\text{)}\}\text{,}$$

$\phi$ is the Euler's totient function and $\sim$ denote that the limit of the quotient of the two functions tends to $1$ as $x$ tends to infinity. Let now $\rho_{a,b}$ be given by

$$\rho_{a,b}(n)=\text{Card}\{k\leq n\,|c_n\text{ is prime}\}\text{,}$$

we can show by straighforward calculation that

$$\rho_{a,b}(n)=\pi_{a,b}(a+bn)-\pi_{a,b}(a)\text{.}$$

From this, we have by the previous theorem that

$$\rho_{a,b}(n)\sim \frac{1}{\varphi(b)}\frac{a+bn}{\log (a+bn)}\text{,}$$

and so, by standard methods of analysis,

$$\rho_{a,b}(n)\sim \frac{b}{\varphi(b)}\frac{n}{\log n}\text{.}$$

Here, we can show, that for all $\varepsilon>0$,

$$\lim_{n\to \infty}\rho_{a,b}((1+\varepsilon)n)-\rho_{a,b}(n)=\infty$$

and thus define $N_\varepsilon$ as the first $n$ for which the previous difference is positive for all natural numbers greater or equal than it, i.e. $\rho_{a,b}((1+\varepsilon)n)-\rho_{a,b}(n)$ is positive for all $n\geq N_\varepsilon$ and negative for $n=N_\varepsilon-1$.

Therefore, we conclude that given $\varepsilon>0$, for all $n\geq N_\varepsilon$ there is a $k$ between $n$ and $(1+\varepsilon)n$ such that $c_k$ is prime. Furthermore, we can make $\varepsilon$ big enough such that $N_\epsilon$ is zero, since

$$\varepsilon\mapsto N_\varepsilon$$

is monotonous decreasing function. In this way, we obtain Bertrand's postulate for arithmetic progressions.

Solution 3:

Here is another generalization

Let $n$ and $k$ be positive integers such that $n > 1.1\log(2.5k)$ then there are at least $k-1$ primes between $n$ and $kn$.

Ref: https://arxiv.org/pdf/1710.09891.pdf