I have constructed an exact function for counting primes in intervals and am curious to know if it 1) has any importance? 2) Has been derived already? I have no formal education in number theory, and no university affiliation, so my only option is to post on here to get any advice or feedback. $\mathbb{P}$ represents the set of prime numbers. Defining $a\#$ as the product of all primes less than or equal to $a$, it is possible to move forward.
Chebyshev's Theorem states there will exist a prime number between $a$ and $2a$ for all $a \in \mathbb{N}$. Since two is the only even prime number, it follows that for every $a > 1$, there will exist some natural number $b < a$ where $a + b$ is an odd, prime number.
The next question to consider is when the sum of two natural numbers is a prime. Given $a, b \in \mathbb{N}$ where $a \geq b$, the sum $a + b \in \mathbb{P}$ only when $GCD(a + b, a\#) = 1$}.
For the forward conditional where $a, b \in \mathbb{N}$ and $a \geq b$ is given by
$$a + b \in \mathbb{P} \Rightarrow GCD(a + b, a\#) = 1$$
Let $a + b$ be a prime number less than or equal to $2a$. Since $a + b$ is a prime, it will have no divisors other than one and itself, showing $GCD(a + b, a\#) = 1$ must hold. For the reverse condition given by
$$GCD(a + b, a\#) = 1 \Rightarrow a + b\in \mathbb{P}$$
Since all composite numbers between $a$ and $2a$ must be the product of prime numbers less than or equal to $a$, it follows that when $a + b$ is coprime to $a\#$, that number is not divisible by any lesser prime numbers. This forces $a + b \in \mathbb{P}$, thus proving the proposition.
As a remark, it is important to note, that in general, $b$, can take any value up to the value $a^2 + a$. This holds because the first number that is coprime to $a\#$, and composite, is the square of the first prime greater than $a$. Since the smallest value this prime can have, in principle, is $a + 1$, it is clear that $a^2 + 2a < (a + 1)^2$. This means that the maximum value for $b$ is $a^2 + a$.
In order to predict the number of prime numbers between some value $a$ and $a + b$, the number of coprime elements to $a\#$ up to $a + b$ will be needed. The Möbius Function, will be needed and its properties are as follows.
\begin{equation*}
\mu(x)=\begin{cases}
1, & \text{if $x = 1$} \\
0, & \text{if $x$ is not square-free}.\\
(-1)^r, & \text{where $r$ is the number of distinct primes of $x$}.
\end{cases}
\end{equation*}
Recall that $a$ and $b$ sum to a prime only when $GCD(a + b, a\#) = 1$. This is equivalent to stating that the number of prime numbers between $a$ and $a + b$ are those elements up to $a + b$ and greater than $a$ which are coprime to $a\#$.
The number of primes between $a$ and $a + b$ where $b \leq a(a + 1)$, which will be denoted by the prime interval counting function $\pi(a, a + b)$ where $a \in \mathbb{N}$, is given by the relationship $\pi(a, a + b) = \sum_{m | a\#} \mu(m) \big{[}\frac{a + b}{m}\big{]} – 1$. This can be proven below.
The number of primes between $a$ and $a + b$ where $a \in \mathbb{N}$ and $b \leq a(a+1)$, is given by the number of coprime elements to $a\#$ up to $a + b$. Knowing that the totient function is related to the Möbius Function, where $\phi(a\#) = a\# \sum_{m | a\#}\frac{\mu(m)}{m} \; \text{and} \; m \in \mathbb{N}$, gives the total number of coprime elements of $a\#$. Truncation of the excess elements is what is needed. To accomplish this, the number of coprime elements of $a\#$ up to $a + b$ is given by taking the partial sum $\sum_{n \leq a + b} \sum_{m | a\#} \frac{\mu(m)}{m} = \sum_{m | a\#} \mu(m) \big{[} \frac{a + b}{m}\big{]}$. Since this function gives all of the coprime values to $a\#$ up to $a + b$, it is necessary to subtract off the coprime values less than or equal to $a$. Since $a\#$ is simply the product of all prime numbers up to the value $a$, the only $c < a$ where $GCD(c, a\#) = 1$ is when $c = 1$, since any composite or prime less than $a$ will share a factor with $a\#$. With this, the exact number of primes in range of $a$ and $a + b$, written as $\pi(a, a + b)$, is given by $\pi(a, a + b) = \sum_{m | a\#} \mu(m) \big{[} {\frac{a + b}{m}}\big{]} – 1$.
My reason for creating this function was to find the conditions where $\pi(a, a + b) = 1$ in a general sense. Knowing that would essentially allow one to parametrize the primes as $p = a + f(a)$, where $p \in \mathbb{P}$. As I said, I have no formal training, so if it is something that isn't important, so be it.
Best Answer
Using inclusion-exclusion $$\sum_{m | N} \mu(m) \lfloor \frac{x}{m}\rfloor $$ is the number of integers $\le x$ that are coprime with $N$.
If $a+b \le a^2$ with $a \# = \prod_{p \le a}p$ then $$\pi(a+b)-\pi(a) = \sum_{2 \le n \le a+b, \gcd(n,a\#)=1} 1= -1+\sum_{m | \ a\#} \mu(m) \lfloor \frac{a+b}{m}\rfloor $$