∑d≤√x′ M(xd)+∑d≤√xμ(d)ωP(xd)−ωP(√x)M(√x)=0
where
I calculate an example for x = 50 and P = \{ 2,3 \} . The first term is
\begin{equation} \sum_{d \le 7}\!\phantom{}^{\prime}\ M\left( \frac{50}{d} \right) = M(50) + M(10) + M(7) \end{equation}
The second term is
\begin{equation} \sum_{d \le 7} \mu(d) \omega_P\left( \frac{50}{d} \right) \end{equation}
If I use
\begin{equation} \omega_P(x) = \lfloor x \rfloor - \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{3}\right\rfloor + \left\lfloor \frac{x}{6} \right\rfloor \end{equation}
then it can be calculated that the second term is zero. The third term is \omega(7) M(7) = 3 M(7) . All together, von Sterneck's formula gives the recursive relation
\begin{equation} M(50) + M(10) +M(7) + 0 - 3 M(7) =0 \end{equation}
Because M(10) = -1 and M(7) = -2 this gives M(50) = -3 .
Sketch of the proof of the formula
The proof proceeds as follows. Define the function e_P
\begin{equation} e_P(m) = \begin{cases} 1 \quad\text{if}\quad p_1 \nmid m \text{ and } \cdots \text{ and } p_j \nmid m\\ 0 \quad\text{otherwise} \end{cases} \end{equation}
then calculate the Dirichlet convolution
\begin{equation} e_P \ast\mu = 1_{\langle P \rangle} \cdot \mu
\end{equation}
where 1_{\langle P \rangle} is the characteristic function on the set \langle P \rangle =
\left\{ n \in \mathbb{N} \middle|\quad \forall p: p | n \implies p \in P \right\} . Because \sum_{ k \le x}e_P(k) = \omega_P(x) , Dirichlet's hyperbola method gives- μ the Möbius function
- M(x)=∑n≤xμ(n) the Mertens function
- ωP(x)={k∈N|1≤k≤x and p1∤ with P = \left\{ p_1, \ldots p_j \right\} a non-empty set of primes
- \sum_{k \le x}^{\prime} the sum over integers k with p_1 \nmid k \text{ and } \cdots \text{ and } p_j \nmid k
- x \ge p_1 p_2 \cdots\ p_j
I calculate an example for x = 50 and P = \{ 2,3 \} . The first term is
\begin{equation} \sum_{d \le 7}\!\phantom{}^{\prime}\ M\left( \frac{50}{d} \right) = M(50) + M(10) + M(7) \end{equation}
The second term is
\begin{equation} \sum_{d \le 7} \mu(d) \omega_P\left( \frac{50}{d} \right) \end{equation}
If I use
\begin{equation} \omega_P(x) = \lfloor x \rfloor - \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{3}\right\rfloor + \left\lfloor \frac{x}{6} \right\rfloor \end{equation}
then it can be calculated that the second term is zero. The third term is \omega(7) M(7) = 3 M(7) . All together, von Sterneck's formula gives the recursive relation
\begin{equation} M(50) + M(10) +M(7) + 0 - 3 M(7) =0 \end{equation}
Because M(10) = -1 and M(7) = -2 this gives M(50) = -3 .
Sketch of the proof of the formula
The proof proceeds as follows. Define the function e_P
\begin{equation} e_P(m) = \begin{cases} 1 \quad\text{if}\quad p_1 \nmid m \text{ and } \cdots \text{ and } p_j \nmid m\\ 0 \quad\text{otherwise} \end{cases} \end{equation}
then calculate the Dirichlet convolution
\begin{equation} \sum_{n \le x} 1_{\langle P \rangle} (n) \mu(n) = \sum_{d \le \sqrt{x}}e_P(d) \ M\left( \frac{x}{d} \right) + \sum_{d \le \sqrt{x}} \mu(d) \omega_P\left( \frac{x}{d} \right) - \omega_P(\sqrt{x})M(\sqrt{x}) \end{equation}
The result then follows because the left hand side is zero for x \ge p_1 p_2 \cdots\ p_j
Von Sterneck used this formula with P = \{ 2,3,5,7 \} . In that way he could calculate values of M(n) for n up to 5,000,000 based on a pre-computed table of M(n) for n up to 500,000.