I found in MO the following equivalent form of the Riemann Hypothesis (RH), that I copy for your convenience:
Let $D(N)$ denote the absolute value of the difference between the number of square free integers divisible by an even number of primes and the number of those divisible by an odd number of primes less or equal than some positive integer $N$.
RH says that $D(N)$ comes close to the square root of $N$. More precisely: for any $\epsilon>0$ there is $N_0$ such that any $N>N_0$ verifies $D(N)≤N^{1/2+\epsilon}$.
I have tried to formulate $D(N)$ in terms of the Prime counting function $\pi(x)$. Therefore, I would like to know if the following identity is correct:
$$D(N)=\pi(N)-\left(\sum_{p_i<\sqrt{N}}{\pi\left(\frac{N}{p_i}\right)}-\left(\frac{\pi\left(\sqrt{N}\right)^{2}+\pi\left(\sqrt{N}\right)}{2}\right)\right)+\left(\sum_{p_i<p_j<\sqrt[3]{N}}{\pi\left(\frac{N}{p_ip_j}\right)}-\left(\frac{\pi\left(\sqrt[3]{N}\right)^{2}+\pi\left(\sqrt[3]{N}\right)}{2}\right)\right)-\dots$$
Basically, it pretends to be an application of the inclusion-exclusion principle. Firstly, they are counted the square free integers less than $N$ divisible by only one prime (which can be counted directly by the prime counting function). Secondly, to count the square free integers less than $N$ which are divisible by two different prime factors, note that at least one of them must be less than $\sqrt{N}$; subsequently, for each $p_i<\sqrt{N}$, we count the prime numbers less than $\frac{N}{p_i}$, and then, as in this counting we are including the non-square free integers divisible by two prime factors, we substract those. A similar reasoning is used to count the square free integers less than $N$ which are divisible by three different prime factors, and so on...
If there is some error in this formulation of $D(N)$, more than happy to know about it; and in that case, I would be grateful if a fixed expression for $D(N)$ in terms of the Prime counting function could be shared. Then, if correct, Riemann Hypothesis would be equivalent to show that for any $\epsilon>0$ there is $N_0$ such that any $N>N_0$ verifies $$\pi(N)-\left(\sum_{p_i<\sqrt{N}}{\pi\left(\frac{N}{p_i}\right)}-\left(\frac{\pi\left(\sqrt{N}\right)^{2}+\pi\left(\sqrt{N}\right)}{2}\right)\right)+\left(\sum_{p_i<p_j<\sqrt[3]{N}}{\pi\left(\frac{N}{p_ip_j}\right)}-\left(\frac{\pi\left(\sqrt[3]{N}\right)^{2}+\pi\left(\sqrt[3]{N}\right)}{2}\right)\right)-\dots≤N^{1/2+\epsilon}$$
Thanks in advance!
After reviewing more thoroughly the formula I proposed, I have noted that it is flawed. I post my analysis as a partial answer, as I have not been able to derive a formulation as was my intention.
We are interested in studying the behaviour of Merten's function $M\left(n\right)$, which is defined for all positive integers as
$$M\left(n\right)=\sum_{k=1}^{n}\mu\left(k\right)$$
Less formally, $M\left(x\right)$ is the count of square-free integers up to $x$ that have an even number of prime factors, minus the count of those that have an odd number. Therefore, it makes sense trying to reformulate $M\left(x\right)$ in terms of the Prime Counting Function $\pi\left(x\right)$ using the inclusion-exclusion principle.
Moving forward, we denote as $\pi_{k}\left(n\right)$ the number of positive integers less than $n$ which are divisible by $k$ prime factors; as it is commonly agreed in the mathematical community, $\pi_{1}\left(n\right)$ is denoted simply as $\pi\left(n\right)$.
Firstly, they can be counted those integers less than $n$ which are divisible only by one prime factor; that is, prime numbers less than $n$, which can be counted directly with the prime counting function as being $\pi\left(n\right)$.
Secondly, to count the square free integers less than $n$ which are divisible by two different prime factors, note that at least one of them must be less than $\sqrt{n}$; subsequently, for each $p_{i}<\sqrt{n}$, we can count the prime numbers less than $\frac{n}{p_{i}}$. We can denote this counting as $\sum_{p_{i}<\sqrt{n}}\pi\left(\frac{n}{p_{i}}\right)$ , and each $\pi\left(\frac{n}{p_{i}}\right)$ as as the “ith counting”. Note that in each ith counting we are including the non-square free integers divisible by the ith prime factor, as well as some square free integers already counted.
For instance, let $n=50$. As $7<\sqrt{50}$, then $\sum_{p_{i}<\sqrt{n}}\pi\left(\frac{n}{p_{i}}\right)=\pi\left(\frac{50}{2}\right)+\pi\left(\frac{50}{3}\right)+\pi\left(\frac{50}{5}\right)+\pi\left(\frac{50}{7}\right)$. With the term $\pi\left(\frac{50}{2}\right)=9$, we are counting the two-prime-square-free integers $\left\{ 2*2,2*3,2*5,2*7,2*11,2*13,2*17,2*19,2*23\right\}$ . As we are counting the non-square-free integer $2*2$, we would need to “discount” it. With the term $\pi\left(\frac{50}{3}\right)=6$, we are counting the two-prime-square-free integers $\left\{ 3*2,3*3,3*5,3*7,3*11,3*13\right\}$ . As we are counting the non-square-free integer $3*3$, and the square free integer $3*2$, which was already counted, we would need to “discount” both.
Note that for each “ith counting” we need to discount all the already discounted square free integers, plus the non-square-free integer divisible by the ith factor, plus an additional square free integer which was counted in the “(i-1)th” counting. Therefore, we need to discount in total $\sum_{i=1}^{\pi\left(\sqrt{n}\right)}i$ integers that were counted more than once. Or which is the same, we need to discount $\left(\frac{\pi\left(\sqrt{n}\right)^{2}+\pi\left(\sqrt{n}\right)}{2}\right)$ integers.
Thus, at the end we have that we can count the square free integers less than $n$ which are divisible by two different prime factors as
$$\left(\sum_{p_{i}<\sqrt{n}}\pi\left(\frac{n}{p_{i}}\right)\right)-\left(\frac{\pi\left(\sqrt{n}\right)^{2}+\pi\left(\sqrt{n}\right)}{2}\right)$$
Until this point, and if I am not mistaken in my reasoning the formula proposed was correct:
$$M\left(n\right)=\pi\left(n\right)-\left(\sum_{p_{i}<\sqrt{n}}\pi\left(\frac{n}{p_{i}}\right)-\left(\frac{\pi\left(\sqrt{n}\right)^{2}+\pi\left(\sqrt{n}\right)}{2}\right)\right)+\dots$$
However, things get messy when we proceed with the next counting. To count the square free integers less than $n$ which are divisible by three different prime factors, note that at least one of them must be less than $\sqrt[3]{n}$; subsequently, for each $p_{i}<\sqrt[3]{n}$, we can count the square free integers which are divisible by two different prime factors and are less than $\frac{n}{p_{i}}$. We can denote this counting as $\sum_{p_{i}<\sqrt[3]{n}}\pi_{2}\left(\frac{n}{p_{i}}\right)$ , and each $\pi_{2}\left(\frac{n}{p_{i}}\right)$ as as the “ith counting”. Note that in each counting we are including the non-square free integers divisible by the ith prime factor, as well as some square free integers already counted. And these are difficult to exclude "smoothly", in a "simple formulation" way to do it.
For instance, let $n=350$. As $7<\sqrt[3]{350}$, then $\sum_{p_{i}<\sqrt[3]{n}}\pi_{2}\left(\frac{n}{p_{i}}\right)=\pi_{2}\left(\frac{350}{2}\right)+\pi_{2}\left(\frac{350}{3}\right)+\pi_{2}\left(\frac{350}{5}\right)+\pi_{2}\left(\frac{350}{7}\right)$. With the term $\pi_{2}\left(\frac{350}{2}\right)$, note that we are counting $\pi\left(\frac{350}{4}\right)$ non-square-free positive integers which are multiples of $2^{2}$, and $\pi\left(\sqrt{\frac{350}{2}}\right)$ non-square-free positive integers which are multiples of each of the prime numbers less than $\sqrt{\frac{350}{2}}$ squared. With the term $\pi_{2}\left(\frac{350}{3}\right)$, we are counting $\pi\left(\frac{350}{6}\right)$ positive integers which are multiples of $2*3$, and which were already counted in the previous term; also, $\pi\left(\sqrt{\frac{350}{3}}\right)$ non-square-free positive integers which are multiples of each of the prime numbers less than $\sqrt{\frac{350}{3}}$ squared; and also, $\pi\left(\frac{350}{9}\right)$ non-square-free positive integers which are multiples of $3^{2}$. And so on...
I have not been able to derive a formulation in terms of $\pi\left(n\right)$ to count from the three-prime-square-free integers on, and seeing how things get complicated, I am skeptical.
The purpose of this formulation was trying to bound $M\left(n\right)$ noting that $$\pi\left(n\right)-\sum_{p_{i}<\sqrt{n}}\pi\left(\frac{n}{p_{i}}\right)\approx \frac{\left(\lfloor n \rfloor-\sum_{p_{i}<\sqrt{n}}\lfloor\frac{n}{p_{i}}\rfloor\right)}{\log\left(n\right)}$$ and that the RHS numerator is the beginning of Legendre's identity. If the rest of terms of the formulation of $M\left(n\right)$ were "simple" enough to bound them...