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Consider the function $a(n)$ defined in formula (1) below and it's summatory function $f(x)$ defined in formula (2) below where $f(x)$ is related to the Riemann zeta function $\zeta(s)$ as illustrated in formula (3) below.

$$a(n)=\sum\limits_{d|n}\mu(d)\ \mu\left(\frac{n}{d}\right)\tag{1}$$

$$f(x)=\sum\limits_{n=1}^x a(n)\tag{2}$$

$$F(s)=s\int_0^\infty f(x)\ x^{-s-1}\ ds=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a(n)\ n^{-s}\right)=\frac{1}{\zeta(s)^2}\ ,\quad\Re(s)>1\tag{3}$$

The following list illustrates the values taken on by $f(x)$ at the first $100$ positive integer values of $x$. Note that $f(x)$ jumps around more than Mertens function $M(x)=\sum\limits_{n=1}^x \mu(n)$ which only ever takes a step of $\pm 1$.

$\{1,-1,-3,-2,-4,0,-2,-2,-1,3,1,-1,-3,1,5,5,3,1,-1,-3,1,5,3,3,4,8,8,6,4,-4,-6,-6,-2,2,6,7,5,9,13,13,11,3,1,-1,-3,1,-1,-1,0,-2,2,0,-2,-2,2,2,6,10,8,12,10,14,12,12,16,8,6,4,8,0,-2,-2,-4,0,-2,-4,0,-8,-10,-10,-10,-6,-8,-4,0,4,8,8,6,10,14,12,16,20,24,24,22,20,18,19\}$

Figure (1) below illustrates a discrete plot of the number of times $f(x)=k$ when $x$ is an integer in the range $0<x\le 10000$. In Figure (1) below, $k$ is the horizontal axis and the number of times $f(x)=k$ is the vertical axis.

Illustration of counts of f(x)=k

Figure (1): Illustration of counts of $f(x)=k$ when $0<x\le 10000$

Assuming $x\in\mathbb{Z}_{>0}$, I'm wondering if $f(x)=k$ an infinite number of times for every integer $k$ as $x\to\infty$. I'm particularly interested in whether $f(x)=0$ and $f(x)=4$ an infinite number of times as $x\to\infty$ (for reasons that will become apparent in a question I'm planning on posting in the near future).

Question: Is it true that $f(x)=k$ at an infinite number of positive integers $x$ for every integer $k$? If this is too hard a question, is it true that $f(x)=0$ and $f(x)=4$ both at an infinite number of positive integers $x$?

Steven Clark
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    Two observations that I hope are useful: (1) it follows from Landau's theorem that $f(n)$ is both unbounded above and unbounded below; (2) if $n$ is divisible by the cube of a prime then $a(n)=0$, otherwise $a(n)=(-2)^k$ where $k$ is the number of primes that divide $n$ to the 1st power (primes that divide $n$ to the second power do not affect the value of $a(n)$). – Greg Martin May 24 '21 at 22:17
  • @GregMartin Thanks for confirming my intuition that $f(n)$ is unbounded both above and below. Your second observation is also new information to me. I searched OEIS in the hopes of finding more information on $a(n)$ and $f(n)$ and was a bit surprised I couldn't seem to find an OEIS entry for $f(n)$. – Steven Clark May 24 '21 at 22:40
  • @GregMartin I don't undersand well the proven lower bound for the number of sign changes (of $M(x)$ but it should be the same for this function), apparently it doesn't grow fast enough to prove that the function $f$ takes some value infinitely often. Is it improved under some standard conjecture? – reuns May 24 '21 at 22:47
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    Since $M(x)$ changes by $\pm1$, the infinitude of sign changes does imply that $M(x)$ takes the value $0$ infinitely often; indeed, the unboundedness of $M(x)$ implies that $M(x)$ takes every integer value infinitely often. (To answer your question, the natural conjecture seems to be that the number of sign changes of $M(x)$ up to $X$ is on the order of $\sqrt X$, by analogy with random walks.) – Greg Martin May 24 '21 at 23:01

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