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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.

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  • $\begingroup$ Can you write matlab function that do this? $\endgroup$ – Mendi Barel Aug 17 at 0:45
  • $\begingroup$ I have written python code to verify it to certain values. Using it iteratively, it gives the exact number of primes beyond 10,000,000. I also wrote a code to choose random values of $a$ and $b$ based on the constraints to calculate the number of primes in those intervals, and it works perfectly. The proof I think is sound, but it is always nice to verify. $\endgroup$ – jmath Aug 17 at 0:49
  • $\begingroup$ Can you post the pyton code in your question? $\endgroup$ – Mendi Barel Aug 17 at 1:50
  • $\begingroup$ I may decide to post the code on a different exchange dealing with computation. For the purposes of this question the code is not needed, as the proof should suffice. $\endgroup$ – jmath Aug 17 at 2:09
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    $\begingroup$ @Mendi In the formula $\pi(a,a+b)=\sum_{m|a\#}\mu(m)\big{[}{\frac{a+b}{m}}\big{]}-1$, $\pi(a,a+b)$ indicates the number of primes greater than $a$ and less than or equal to $b$, $m|a\#$ indicates $m$ divides $a\#$ where $a\#$ is the product of all primes less than or equal to $a$, $\mu(m)$ is the Möbius function, and $\big{[}{\frac{a+b}{m}}\big{]}$ is the floor function of $\frac{a+b}{m}$. $\endgroup$ – Steven Clark Aug 18 at 2:18
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Assume the following definitions where $n$ is a positive integer, $p$ is a prime, $\mu(n)$ is the Möbius function, and $M(x)$ is Mertens function:

(1) $\quad \pi(x)=\sum\limits_{n\le x}a(n)\,,\qquad a(n)=\cases{1,& $n\in\mathbb{P}$ \\ 0,& otherwise}\quad$ (see https://oeis.org/A010051 for $a(n)$)

(2) $\quad M(x)=\sum\limits_{n\le x}\mu(n)$

(3) $\quad Q(x)=\sum\limits_{n\le x}|\mu(n)|$

The prime-counting function $\pi(x)$ can be expressed in a number of ways such as (4) to (6) below where:

  • $\nu(n)$ is the number of distinct primes in the factorization of $n$,
  • $\omega(n)$ is the number of prime factors counting multiplicities in the factorization of $n$, and
  • $rad(n)$ is the radical of $n$ (also referred to as the square-free kernel of $n$),

(4) $\quad \pi(x)=\sum\limits_{n\le x}b(n)\,\left\lfloor\frac{x}{n}\right\rfloor\,,\qquad b(n)=\sum\limits_{d|n} a(d)\,\mu\left(\frac{n}{d}\right)\quad$ (see https://oeis.org/A143519 for $b(n)$)

(5) $\quad \pi(x)=\sum\limits_{n\le x}\nu(n)\,M\left(\frac{x}{n}\right)$

(6) $\quad \pi(x)=\sum\limits_{n\le x} (-1)^{\Omega(n)+1}\, \nu(n)\,Q\left(\frac{x}{n}\right)$

Any of the formulas (4) to (6) above can be used to evaluate $\pi(a+b)-\pi(a)$. Note all three of these formulas involve the Moebius function $\mu(n)$ (either directly or indirectly).

The $b(n)$ coefficient function referenced in formula (4) above can also be evaluated as follows.

(7) $\qquad b(n)=\sum\limits_{p|n}\mu\left(\frac{n}{p}\right)\qquad\quad(p\in\mathbb{P})$

(8) $\qquad b(n)=\cases{-\mu(n)\,\nu(n) & $\Omega(n)=\nu(n)$ \\ \mu(rad(n)) & $\Omega(n)=\nu(n)+1$ \\ 0,& otherwise}$

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  • $\begingroup$ So, it sounds like this function is not of any real importance. $\endgroup$ – jmath Aug 17 at 3:15
  • $\begingroup$ Proofs ? And there is no Fourier representation of $\pi(x)$ $\endgroup$ – reuns Aug 17 at 4:26
  • $\begingroup$ @jmath Your formula is interesting but only valid in an interval. I thought I'd give you a few more formulas to consider that are more generally useful. $\endgroup$ – Steven Clark Aug 17 at 4:36
  • $\begingroup$ I see, thank you for those. I will read more about them. $\endgroup$ – jmath Aug 17 at 5:29
  • $\begingroup$ @reuns My website documents derivation of some of the formulas above, but you deleted the reference to my website. In general, I believe all functions of the form $f(x)=\sum_{n\le x} a(n)$ where $a(n)\in\mathbb{C}$ have a Fourier series representation (see primefourierseries.com/?page_id=7595). There formulas are generally only valid for $x>0$, and convergence of the Fourier series is dependent upon the growth rate of $f(x)$. $\endgroup$ – Steven Clark Aug 18 at 19:28
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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 $$\pi(a+b)-\pi(a) = \sum_{2 \le n \le a+b, \gcd(n,a\#)=1} 1= -1+\sum_{m | \ a\#} \mu(m) \lfloor \frac{a+b}{m}\rfloor $$

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