Bertrand's Postulate states that "for all $n \geq 2$, there is a prime $p$ between $n$ and $2n-2$". So a consequence is that, for all $n \geq 2$, there exists a prime $p \in (n, 2n)$.
Let be $\delta: \mathbb{N} \to \mathbb{N}$ the function defined by $\delta(1) = 0$, and $\delta(n) = p_1 + \ldots + p_r$ if $n = p_1 \cdot \ldots \cdot p_r$ is the canonical prime descomposition of $n$. For example $\delta(20) = \delta(2^2 \cdot 5) = 2 + 2 + 5 = 9$, $\delta(p) = p$ for all primes $p$.
Define $n$ is refined is there is a prime in the interval $(n, n + \delta(n))$. Otherwise we say that $n$ is unrefined. I choose this name because $n$ improves, refines, the Bertrand Postulate. The first unrefined number is 5120.
Chebyshev theorem states that for all $\epsilon > 0 \in \mathbb(R)$, there is $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$, there is a prime number $p$ in the interval $(n, n+\epsilon n)$. Chebyshev theorem is consequence of Prime Number Theorem.
Consequence to Chebyshev theorem, it's easy to see that if $(i_n)_{n \in \mathbb{N}}$ is the sequence of unrefined numbers, $\delta(i_n)/i_n$ tends to $0$. So, for every fixed $m \geq 1$, there is just a finite unrefined numbers in the set $I_m = \{m\cdot p \, | \, p \text{ primer number} \}$. Note that $I_1$ is the set of prime numbers.
I have two questions:
- are unrefined numbers finite?
- where is the counting function of unrefined numbers? That is, if we define $\Pi(x) = |\{ n \text{ unrefined } | n \leq x \}|$, how is $\Pi(x)$?
I conjecture that unrefined numbers are infinite but its counting function is very low.
This question is connected with Sierpinski conjecture: For any $n\geq 2$ and any $1 < k \leq n$ there exists a prime number between $(k-1)n$ and $kn$.
This question is also related to Legendre's conjecture: there is a prime between $n$ and $n^2$. When $n=p$, $n$ is refined if, and only if, $n$ satisfies Legendre's conjecture.