The Möbius Function and Mertens Function

I had come across the Möbius function before but didn't see it as all that important. However, I was reminded of it recently when I encountered a Numberphile video on Mertens conjecture (uploaded 23rd January 2020).

In number theory, we define the Mertens function as:$$M(n) = \sum_{1\le k \le n} \mu(k)$$where \( \mu (k)\) is the Möbius function. The Mertens conjecture is that for all \(n > 1\): $$\left| M(n) \right| < \sqrt { n }$$To quote from Wikipedia:

In mathematics, the Mertens conjecture is the disproven statement that the Mertens function \(M(n)\) is bounded by \( \sqrt {n}\), which implies the Riemann hypothesis. It was conjectured by Thomas Joannes Stieltjes in an 1885 letter to Charles Hermite (reprinted in Stieltjes (1905)) and Franz Mertens (1897), and disproved by Andrew Odlyzko and Herman te Riele (1985). It is a striking example of a mathematical proof contradicting a large amount of computational evidence in favour of a conjecture.

I set about writing a program in SageMath to plot the first million values of the Mertens function. I managed to execute it in SageMathCell but the code was clunky as I had to get the program to determine the values of -1, 0 and 1 based on the factorisation. Later I realised that the Möbius function would replace the need for this. In SageMath, the spelling is moebius and moebius(1) --> 1 etc. So let's define the Möbius function:

For any positive integer n, \(μ(n)\) has values in {−1, 0, 1} depending on the factorisation of \(n\) into prime factors:$$\mu(n) =
  \begin{cases}
    1       & \quad \text{if } n \text{ is a square-free positive integer with an even number of prime factors}\\
   -1  & \quad \text{if } n \text{ is a square-free positive integer with an odd number of prime factors}\\
0  & \quad \text{if } n \text{ has a squared prime factor}
  \end{cases}$$Figure 1 shows the results of plotting the first one million values of the Möbius function. Here is the permalink to the SageMathCell calculation.

Figure 1

The square root of \( \pm \) 1,000,000 is \( \pm \) 1,000 and as can be seen the function is well within those bounds. However, as mentioned earlier, the bounds are eventually exceeded. Sometimes, instead of \(M(n) \), the function \(m(n) \) is used where:$$m(n)=\frac{M(n)}{\sqrt{n}}$$For the Mertens conjecture to hold, the following condition would be necessary: \(-1 <m(n)<1\). However, for enormously large values of \(n\), it has been shown that \(m(n)<-1.837625 \) and \(m(n)>1.826054\) are possible.