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The Mertens function $M(x)$ is defined as follows.

(1) $\quad M(x)=\sum\limits_{n=1}^x\mu(n)$

I've noticed the Merten's function can also be evaluated as follows which is related to OEIS entry A087003 and the Collatz conjecture.

(2) $\quad M(x)=\sum\limits_{n=\left\lfloor\frac{x-2}{4}\right\rfloor+1}^{\left\lfloor\frac{x-1}{2}\right\rfloor}\mu(2\,n+1)$

I've verified formula (2) above for the first $10,000$ positive integer values of $x$.

Question (1): Has formula (2) above been proven (or disproven) and if not, can it be?

I've read the following conjecture on the growth of $M(x)$ for any $\epsilon<1/2$ is equivalent to the Riemann hypothesis (see Weisstein, Eric W. "Mertens Conjecture." From MathWorld--A Wolfram Web Resource).

(3) $\quad M(x)=O\left(x^{1/2+\epsilon}\right)$

Question (2): Assuming formula (2) above is correct, does it show any promise with respect to improving upon the fastest known algorithm for computing $M(x)$ and/or the tightest known error bound on the growth of $M(x)$?

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  • $\begingroup$ Yes you are right, my bad $\endgroup$
    – Kinheadpump
    Commented Jul 19, 2019 at 20:01

1 Answer 1

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There is nothing special happening here. Let $$F(s) = \sum_{n=0}^\infty \mu(2n+1)(2n+1)^{-s}, \qquad \frac{1}{\zeta(s)} = (1-2^{-s}) F(s)$$ $$f(x) = \sum_{2n+1 \le x} \mu(2n+1), \qquad M(x) = f(x)-f(x/2)$$

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    $\begingroup$ can do it directly by separating the $\mu(2(2k+1))=-\mu(2k+1)$ since $\mu$ is multiplicative, $\mu(4k)=0$ and cancelling out the odd $\mu$ for the first half, but using RZ is cooler! $\endgroup$
    – Conrad
    Commented Jul 20, 2019 at 1:49

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