We can define a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that:
- $f$ is bijective.
- If $f(n)=m$, then $f(m)=n$.
- If $f(n)=m$ then $\mu(n)+\mu(m)=0$
We want to define $f$ such that the distance between $n$ and $f(n)$ is the least possible. In particular, if $\mu(n)=0 \Rightarrow f(n)=n$. Therefore, we define $f$ by induction, if $n$ is the first number with $f(n)$ undefined, we look for the first unassigned $m$ such that $\mu(n)+\mu(m)=0$, and we then define $f(n)=m$. Now we obtain a succesion $f(1),f(2),...$ which is: $$2,1,6,4,10,3,14,8,...$$ I've been able to create a program in both Haskell and Mathematica which can calculate a list of the pairs $(n,f(n))$ for about $n=10^5$ which has given me some insight on the growth of $f(n)$, $f(n)-n$ and $F(x)=\sum_{n \leq x} (f(n)-n)$. Here are a few pictures I sent before my computer broke.
What I'm looking for here is some conjecture on the growth of these functions and how to prove it, if possible. With respect to the Mertens function $\left( M(x) := \sum_{n \leq x} \mu(n)\right)$ I've seen that it's equivalent to the Riemann Hypothesis iff $M(x)=\mathcal{O}(x^{\frac{1}{2} + \epsilon}) $ for any $\epsilon >0$. Is it possible to find a similar equivalence with $f(n)$?
My main problem is being able to manipulate $f(n)$ in a way similar to how one can manipulate a usual function or at least an arithmetic function like $\mu(n)$. Finding an expression for it or some property of $f$ would probably be really helpful.
This one is $f(n)$
https://i.stack.imgur.com/d6Rp0.jpg
This one is $f(n)-n$
https://i.stack.imgur.com/joA6R.jpg
And this one is $F(x)$
