## Sunday, July 19, 2015

### Neue Empirische Daten Über Die Zahlentheoretische Funktion sigma(n), von Sterneck, 1913. Part I

In the paper in the title, von Sterneck performs numerical tests on the inequality
\label{Neueeq:1}
| M (n) | \le \sqrt{n}

where $M(n) = \displaystyle\sum_{k=1}^n \mu(n)$ is the Mertens function. The inequality (1) is of historical interest, because it implies the correctness of the Riemann hypothesis. In 1985 however, Odlyzko and te Riele, showed that (1) is incorrect, although an explicit number $n$ for which (1) fails is still not known. Nevertheless, I find von Sterneck's paper still interesting because I want to see how $M(n)$ was calculated for large values of $n$ before the advent of (electronic) computers.

In more detail, von Sterneck says that before 1901 he had constructed a table of values of  $M(n)$ for all $n$ up to 500,000. In that table he observed that the stronger inequality
\label{eq:2}
| M (n) | \le \frac{1}{2} \sqrt{n}

is valid for all $n \le 500,000$ (excluding some exceptions for small $n$ ) . In the paper mentioned in the title, he says that he has calculated 16 more values of  $M(n)$ with  $n$ up to 5,000,000.

 Values calculated by von Sterneck; he uses the notation $\sigma (n)$ instead of $M (n)$.

Von Sterneck had to hire "Rechnern" for this calculation of these 16 values. Now, in 2015, it takes Mathematica 32 seconds on my laptop to calculate all values $M (n)$ up to 5,000,000.

 Calculation of all values of $M(n)$ up to $n = 5,000,000$ in Mathematica

9 out of 16 values calculated by von Sterneck are incorrect however, see the next table with the values calculated in Mathematica.

Sterneck notices that the inequality (1) is valid for his 16 calculated values of $M (n )$, and he conjectures that (1) is valid for all $n \le 5,000,000$ (excluding some exceptions for small $n$ ) . With Mathematica, this is now easy to verify; I check that (2) is indeed valid for all $201 \le n \le 5,000,000$

Further links

• In this paper, te Riele gives more historical background on the Mertens conjecture.
• In the next post I will prove an equation which von Serneck used for his calculation.