Is there a relationship between $\pi(x^2+x) - \pi(x^2)$ and $\pi(x) - \pi\left(\dfrac{x}{2}\right)$
This does not seem intuitively correct but I am not able to find a mistake.
Here's the argument:
Let:
- $x > 2$ be an integer
- $p_n$ be the $n$th prime
- $p\#$ the primorial of $p$
- gcd$(a,b)$ be the greatest common divisor of $a$ and $b$.
- $H_t(x)$ be the count of the integers $i$ such that $0 < i \le x$ and gcd$(i,t\#)=1$ where $t > 0$ is an integer.
- $\pi(x)$ be the prime counting function
Observations:
- If $n=\pi(t)$, then: $$H_{p_n}(x) = H_t(x)$$
- $H_{p_n}(x)$ is equivalent to the following recurrence relation:
- $H_1(x) = \lfloor x\rfloor$
- $H_2(x) = H_1(x) - H_1\left(\dfrac{x}{2}\right)$
- $H_{p_n}(x) = H_{p_{n-1}}(x) - H_{p_{n-1}}\left(\dfrac{x}{p_n}\right)$
Question
As I was thinking about $\pi(i^2+x) - \pi(i^2)$, where $i > 3$ is an integer, it occurred to me that there might be a relationship between this and $\pi(i) - \pi\left(\dfrac{i}{2}\right)$
Here is the argument. Please let me know if I have made a mistake:
(1) Let $i > 3$ be an integer.
(2) Let $n = \pi(i)$
(3) $H_{p_n}(i^2 + i) - H_{p_n}(i^2) = \pi(i^2+i) - \pi(i^2)$
(4) Let $u = \pi\left(\dfrac{i}{2}\right)$
(5) For all prime $q$ where $p_u < q \le p_n$:
- $H_{p_u}\left(\dfrac{i^2+i}{q}\right) - H_{p_u}\left(\dfrac{i^2}{q}\right) < 2$ since:
- $\dfrac{i^2+i}{q} < 2i+2$
- $i < \dfrac{i^2}{q} < 2i$
- $\dfrac{i^2+i}{q} - \dfrac{i^2}{q} = \dfrac{i}{q} < 2$
(6) Using induction, this leads to: $$H_{p_n}(i^2 + i) - H_{p_n}(i^2) \ge H_{p_u}(i^2+i) - H_{p_u}(i^2) - \pi(i) + \pi\left(\frac{i}{2}\right)$$
- Base Case: $i=4$: $p_n = 3, p_u=2$ $$H_3(20) - H_3(16) = 2 \ge H_2(20) - H_2(16) - 3 + 2 = 2 - 1 = 1$$
- Inductive Case: with $u=\pi\left(\dfrac{i+1}{2}\right)$
- We can assume that $i+1 = p_{n+1}$ since if not $p_n, p_u$ are the same for $i+1$ and $i$.
- $H_{p_{n+1}}((i+1)^2+i+1) - H_{p_{n+1}}((i+1)^2) = H_{p_n}((i+1)^2+i) - H_{p_n}((i+1)^2) - \left[H_{p_n}\left(\dfrac{(i+1)^2+i+1}{p_{n+1}}\right) - H_{p_n}\left(\dfrac{(i+1)^2}{p_{n+1}}\right)\right] = H_{p_u}((i+1)^2 + i + 1) - H_{p_h}((i+1)^2) - \left[\sum\limits_{u < q \le n} \left(H_{p_q}\left(\dfrac{(i+1)^2+i+1}{p_{q+1}}\right) - H_{p_q}\left(\dfrac{(i+1)^2}{p_{q+1}}\right)\right) \right] \ge H_{p_u}((i+1)^2 + i + 1) - H_{p_h}((i+1)^2) - n + u$